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For each integer $1\le j\le 2017$, let $S_j$ denote the set of integers $0\le i\le 2^{2017} - 1$ such that $\left\lfloor \frac{i}{2^{j-1}} \right\rfloor$ is an odd integer. Let $P$ be a polynomial such that
\[P\left(x_0, x_1, \ldots, x_{2^{2017} - 1}\right) = \prod_{1\le j\le 2017} \left(1 - \prod_{i\in S_j} x_i\right).\]
Compute the remainder when
\[ \sum_{\left(x_0, \ldots, x_{2^{2017} - 1}\right)\in\{0, 1\}^{2^{2017}}} P\left(x_0, \ldots, x_{2^{2017} - 1}\right)\]
is divided by $2017$.
[i]Proposed by Ashwin Sah[/i]
|
1. **Define the sets \( S_j \):**
For each integer \( 1 \le j \le 2017 \), the set \( S_j \) consists of integers \( 0 \le i \le 2^{2017} - 1 \) such that \( \left\lfloor \frac{i}{2^{j-1}} \right\rfloor \) is an odd integer. This can be written as:
\[
S_j = \{ i \mid 0 \le i \le 2^{2017} - 1, \left\lfloor \frac{i}{2^{j-1}} \right\rfloor \text{ is odd} \}
\]
2. **Express the polynomial \( P \):**
The polynomial \( P \) is given by:
\[
P(x_0, x_1, \ldots, x_{2^{2017} - 1}) = \prod_{1 \le j \le 2017} \left(1 - \prod_{i \in S_j} x_i \right)
\]
Let \( T_j = \prod_{i \in S_j} x_i \). Then:
\[
P(x_0, x_1, \ldots, x_{2^{2017} - 1}) = \prod_{1 \le j \le 2017} (1 - T_j)
\]
3. **Evaluate the polynomial \( P \):**
The polynomial \( P \) evaluates to 1 if all \( T_j \) are 0, and 0 if at least one \( T_j \) is 1. We need to count the number of tuples \( (x_0, x_1, \ldots, x_{2^{2017} - 1}) \in \{0, 1\}^{2^{2017}} \) such that all \( T_j \) are 0.
4. **Use complementary counting and the Principle of Inclusion-Exclusion (PIE):**
Let \( A_j \) denote the set of tuples where \( T_j = 1 \). We want to compute:
\[
X = |P| - \sum |A_j| + \sum |A_i \cap A_j| - \sum |A_i \cap A_j \cap A_k| + \cdots
\]
where \( |P| = 2^{2^{2017}} \).
5. **Calculate the sizes of intersections:**
- \( |A_j| = 2^{2^{2016}} \) because \( |S_j| = 2^{2016} \).
- For \( |A_i \cap A_j| \) with \( i < j \), \( S_j \) fills half of the gaps in \( S_i \), so \( |A_i \cap A_j| = 2^{2^{2015}} \).
- Generally, \( |A_{a_1} \cap A_{a_2} \cap \cdots \cap A_{a_k}| = 2^{2^{2017 - k}} \).
6. **Sum using PIE:**
\[
X = \sum_{k=0}^{2017} (-1)^k \binom{2017}{k} 2^{2^{2017 - k}}
\]
Simplifying modulo 2017:
\[
X \equiv 2^{2^{2017}} - 2 \pmod{2017}
\]
7. **Compute \( 2^{2^{2017}} \mod 2017 \):**
Using Fermat's Little Theorem, \( 2^{2016} \equiv 1 \pmod{2017} \). We need to find \( 2^{2017} \mod 2016 \):
\[
2^{2017} \equiv 2 \pmod{2016}
\]
Thus:
\[
2^{2^{2017}} \equiv 2^2 = 4 \pmod{2017}
\]
8. **Final calculation:**
\[
X \equiv 4 - 2 = 2 \pmod{2017}
\]
The final answer is \(\boxed{2}\)
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For any positive integer $n$, let $S_n$ denote the set of positive integers which cannot be written in the form $an+2017b$ for nonnegative integers $a$ and $b$. Let $A_n$ denote the average of the elements of $S_n$ if the cardinality of $S_n$ is positive and finite, and $0$ otherwise. Compute \[\left\lfloor\displaystyle\sum_{n=1}^{\infty}\frac{A_n}{2^n}\right\rfloor.\]
[i]Proposed by Tristan Shin[/i]
|
1. **Understanding the Problem:**
We need to compute the sum \(\left\lfloor \sum_{n=1}^{\infty} \frac{A_n}{2^n} \right\rfloor\), where \(A_n\) is the average of the elements of \(S_n\). The set \(S_n\) consists of positive integers that cannot be written in the form \(an + 2017b\) for nonnegative integers \(a\) and \(b\).
2. **Identifying \(S_n\):**
- For \(n = 1\), \(S_1 = \emptyset\) because any positive integer can be written as \(a \cdot 1 + 2017b\).
- For \(n = 2\), \(S_2\) consists of all odd numbers less than \(2017\) because even numbers can be written as \(2a + 2017b\).
3. **Generalizing \(S_n\):**
- For \(n \geq 2\), \(S_n\) consists of numbers that cannot be written as \(an + 2017b\). The largest number that cannot be written in this form is given by the Frobenius number \(g(n, 2017) = n \cdot 2016 - n - 2017\).
4. **Cardinality of \(S_n\):**
- The number of elements in \(S_n\) is given by \(\frac{(n-1)(2016)}{2}\) using the Chicken McNugget theorem.
5. **Sum of Elements in \(S_n\):**
- The sum of the elements in \(S_n\) can be modeled by a quadratic polynomial \(f(x) = ax^2 + bx + c\). Given the sums for \(S_2\), \(S_3\), and \(S_4\), we can determine \(a\), \(b\), and \(c\).
6. **Finding the Polynomial:**
- Given \(f(1) = 1016064\), \(f(2) = 3387216\), and \(f(3) = 7113456\), we solve for \(a\), \(b\), and \(c\):
\[
\begin{cases}
a + b + c = 1016064 \\
4a + 2b + c = 3387216 \\
9a + 3b + c = 7113456
\end{cases}
\]
- Solving these equations, we find \(a = 677544\), \(b = 338520\), and \(c = 0\).
7. **Sum of Elements in \(S_k\):**
- The sum of the elements in \(S_k\) is \(f(k-1) = 677544(k-1)^2 + 338520(k-1)\).
8. **Average \(A_n\):**
- The average \(A_n\) is given by:
\[
A_n = \frac{677544(n-1)^2 + 338520(n-1)}{1008(n-1)}
\]
- Simplifying, we get:
\[
A_n = \frac{677544(n-1) + 338520}{1008}
\]
9. **Summing the Series:**
- We need to compute:
\[
\left\lfloor \sum_{n=1}^{\infty} \frac{A_n}{2^n} \right\rfloor
\]
- Ignoring terms where \(2017 \mid n\) as they are negligible, we simplify the fraction:
\[
\left\lfloor \sum_{n=1}^{\infty} \frac{4033n - 2018}{6 \cdot 2^n} \right\rfloor
\]
10. **Final Calculation:**
- The series converges to a value, and after performing the summation, we get:
\[
\boxed{840}
\]
|
840
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $N$ be the number of functions $f: \mathbb{Z}/16\mathbb{Z} \to \mathbb{Z}/16\mathbb{Z}$ such that for all $a,b \in \mathbb{Z}/16\mathbb{Z}$: \[f(a)^2+f(b)^2+f(a+b)^2 \equiv 1+2f(a)f(b)f(a+b) \pmod{16}.\] Find the remainder when $N$ is divided by 2017.
[i]Proposed by Zack Chroman[/i]
|
1. **Understanding the problem**: We need to find the number of functions \( f: \mathbb{Z}/16\mathbb{Z} \to \mathbb{Z}/16\mathbb{Z} \) that satisfy the given functional equation for all \( a, b \in \mathbb{Z}/16\mathbb{Z} \):
\[
f(a)^2 + f(b)^2 + f(a+b)^2 \equiv 1 + 2f(a)f(b)f(a+b) \pmod{16}.
\]
2. **Initial observations**: We can check that either \( f(\text{odds}) \) are all even or \( f(\text{odds}) \) are all odd. Moreover, \( f(\text{evens}) \) are all odd. We need to find all \( f: \mathbb{Z}/16\mathbb{Z} \to \mathbb{Z}/8\mathbb{Z} \) since the equation modulo 16 implies the same equation modulo 8.
3. **Case 1: \( f(\text{odds}) \) are all even**:
- Let \( P(a, b) \) denote the assertion \( f(a)^2 + f(b)^2 + f(a \pm b)^2 \equiv 1 + 2f(a)f(b)f(a \pm b) \pmod{16} \).
- Suppose \( f(2n+1) \) are all even. Then \( f(2n+1)^2 \equiv 0 \) or \( 4 \pmod{16} \).
- If \( f(2n+1)^2 \equiv 0 \pmod{16} \), then \( P(2n+1, 2n+1) \) gives \( f(4n+2)^2 \equiv 1 \pmod{16} \).
- If \( f(2n+1)^2 \equiv 4 \pmod{16} \), then \( f(4n+2)^2 + 8 \equiv 1 + 8f(4n+2)^2 \pmod{16} \), so \( f(4n+2)^2 \equiv 1 \pmod{16} \).
- It follows that \( f(4n+2) \) is either \( 1 \) or \( 7 \) for all \( n \).
4. **Consistency check**:
- Suppose \( f(2m+1)^2 \equiv 0 \pmod{16} \) and \( f(2n+1)^2 \equiv 4 \pmod{16} \). Then \( P(2m+1, 2n+1) \) gives \( 4 + f(2m+2n+2)^2 \equiv 1 \pmod{16} \), a contradiction since \( -3 \) is not a quadratic residue.
- Thus, either \( f(\text{odds}) \in \{2, 6\} \) for all odds, or \( f(\text{odds}) \in \{0, 4\} \) for all odds.
5. **Further analysis**:
- Take \( P(4n+2, 4n+2) \). If \( f(4n+2) = 1 \), then this gives \( f(8n+4)^2 - 2f(8n+4) + 1 \equiv 0 \pmod{16} \), so \( f(8n+4) \equiv 1 \pmod{4} \).
- If \( f(4n+2) = 7 \), then once again \( f(8n+4) \equiv 1 \pmod{4} \).
- Now, take two odd numbers \( a, b \) with \( a + b = 8n + 4 \). Either \( f(a)^2 = f(b)^2 \equiv 0 \) or \( f(a)^2 = f(b)^2 \equiv 4 \). In both cases, \( f(8n+4)^2 \equiv 1 \pmod{16} \). Combined with \( f(8n+4) \equiv 1 \pmod{4} \), we get \( f(8n+4) = 1 \).
6. **Conclusion for Case 1**:
- If \( f(4n+2) = 1 \) but \( f(4m+2) = 7 \), and WLOG \( m+n \) is even, then \( P(4m+2, 4n+2) \) gives \( 2 + f(4m+4n+4)^2 \equiv 1 - 2f(4m+4n+4) \pmod{16} \), which is false.
- Thus, either \( f(4n+2) \) is \( 1 \) for all \( n \), or \( 7 \) for all \( n \).
7. **Final check for Case 1**:
- Take any \( 4n = a + b \) with \( a, b \) multiples of two but not 4. Then \( P(a, b) \) gives \( 2 + f(4n)^2 \equiv 1 + 2f(4n) \), or \( f(4n) \equiv 1 \pmod{4} \).
- Take \( 4n = a + b \) with \( a, b \) odd. Then \( P(a, b) \) gives \( 8 + f(4n)^2 \equiv 1 + 8f(4n) \), or \( f(4n)^2 \equiv 1 \).
- Thus, \( f(4n) = 1 \) for all \( n \).
8. **Number of solutions for Case 1**:
- \( f(2n+1) \in \{0, 4\} \) for all \( n \) or \( f(2n+1) \in \{2, 6\} \) for all \( n \).
- \( f(4n+2) = 1 \) for all \( n \) or \( 7 \) for all \( n \).
- \( f(4n) = 1 \) for all \( n \).
- Total solutions: \( 2 \cdot 2^8 \cdot 2 = 2^{10} \).
9. **Case 2: \( f(\text{odds}) \) are all odd**:
- If \( f(x) \) is odd, then \( f(2x) \equiv 1 \pmod{4} \).
- Thus, \( f(\text{evens}) \equiv 1 \pmod{4} \).
10. **Consistency check for Case 2**:
- For \( x, y, z \) odd, \( x^2 + y^2 + z^2 \equiv 1 + 2xyz \pmod{16} \iff (x+4)^2 + y^2 + z^2 \equiv 1 + 2(x+4)yz \pmod{16} \).
- Suppose \( f(a) = 1 \) and \( f(b) = 3 \) where \( a, b \) are odd. Then \( f(a+b) = 1 \), so \( 1^2 + 1^2 + 3^2 \equiv 1 + 2 \cdot 1 \cdot 1 \cdot 3 \), which is false.
- Thus, either \( f(\text{odds}) = 1 \) or \( f(\text{odds}) = 3 \) for all odds.
11. **Number of solutions for Case 2**:
- \( 2 \) for whether \( f(\text{odds}) \) is \( 1 \) or \( 3 \).
- \( 2^{16} \) for the two solutions mod 8 for each solution mod 4.
- Total solutions: \( 2 \cdot 2^{16} = 2^{17} \).
12. **Final calculation**:
- Total solutions: \( 2^{10} + 2^{17} = 2^{10}(1 + 2^7) = 2^{10} \cdot 129 = 2^{10} \cdot 129 \).
- \( 2^{26} + 2^{33} \equiv 2^{26} + 2^{26} \cdot 128 \equiv 2^{26}(1 + 128) \equiv 2^{26} \cdot 129 \pmod{2017} \).
- Calculate \( 2^{26} \mod 2017 \):
\[
2^{26} \equiv 793 \pmod{2017}.
\]
- Thus, \( 2^{26} \cdot 129 \equiv 793 \cdot 129 \equiv 793 \pmod{2017} \).
The final answer is \(\boxed{793}\).
|
793
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A [i]simple hyperplane[/i] in $\mathbb{R}^4$ has the form \[k_1x_1+k_2x_2+k_3x_3+k_4x_4=0\] for some integers $k_1,k_2,k_3,k_4\in \{-1,0,1\}$ that are not all zero. Find the number of regions that the set of all simple hyperplanes divide the unit ball $x_1^2+x_2^2+x_3^2+x_4^2\leq 1$ into.
[i]Proposed by Yannick Yao[/i]
|
1. **Understanding the Problem:**
We need to find the number of regions that the set of all simple hyperplanes in \(\mathbb{R}^4\) divides the unit ball \(x_1^2 + x_2^2 + x_3^2 + x_4^2 \leq 1\) into. A simple hyperplane in \(\mathbb{R}^4\) is given by the equation \(k_1x_1 + k_2x_2 + k_3x_3 + k_4x_4 = 0\) where \(k_1, k_2, k_3, k_4 \in \{-1, 0, 1\}\) and not all are zero.
2. **Counting the Simple Hyperplanes:**
Each coefficient \(k_i\) can be \(-1\), \(0\), or \(1\). Since not all coefficients can be zero, we need to subtract the case where all coefficients are zero from the total number of combinations:
\[
3^4 - 1 = 81 - 1 = 80
\]
So, there are 80 simple hyperplanes.
3. **Regions Created by Hyperplanes:**
According to a known result in geometry, the number of regions \(R(n, k)\) into which \(k\) hyperplanes divide \(\mathbb{R}^n\) is given by:
\[
R(n, k) = \sum_{i=0}^{n} \binom{k}{i}
\]
Here, \(n = 4\) and \(k = 80\).
4. **Calculating the Number of Regions:**
We need to sum the binomial coefficients from \(i = 0\) to \(i = 4\):
\[
R(4, 80) = \binom{80}{0} + \binom{80}{1} + \binom{80}{2} + \binom{80}{3} + \binom{80}{4}
\]
Calculating each term:
\[
\binom{80}{0} = 1
\]
\[
\binom{80}{1} = 80
\]
\[
\binom{80}{2} = \frac{80 \cdot 79}{2} = 3160
\]
\[
\binom{80}{3} = \frac{80 \cdot 79 \cdot 78}{6} = 82160
\]
\[
\binom{80}{4} = \frac{80 \cdot 79 \cdot 78 \cdot 77}{24} = 1581580
\]
Summing these values:
\[
R(4, 80) = 1 + 80 + 3160 + 82160 + 1581580 = 1661981
\]
5. **Conclusion:**
The number of regions that the set of all simple hyperplanes divides the unit ball in \(\mathbb{R}^4\) into is \(\boxed{1661981}\).
|
1661981
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The USAMO is a $6$ question test. For each question, you submit a positive integer number $p$ of pages on which your solution is written. On the $i$th page of this question, you write the fraction $i/p$ to denote that this is the $i$th page out of $p$ for this question. When you turned in your submissions for the $2017$ USAMO, the bored proctor computed the sum of the fractions for all of the pages which you turned in. Surprisingly, this number turned out to be $2017$. How many pages did you turn in?
[i]Proposed by Tristan Shin[/i]
|
1. Let \( a_k \) be the number of pages submitted for the \( k \)-th question, where \( k = 1, 2, \ldots, 6 \).
2. For each question, the sum of the fractions written on the pages is given by:
\[
\sum_{i=1}^{a_k} \frac{i}{a_k}
\]
3. We can simplify this sum as follows:
\[
\sum_{i=1}^{a_k} \frac{i}{a_k} = \frac{1}{a_k} \sum_{i=1}^{a_k} i
\]
4. The sum of the first \( a_k \) positive integers is:
\[
\sum_{i=1}^{a_k} i = \frac{a_k(a_k + 1)}{2}
\]
5. Substituting this back into our expression, we get:
\[
\sum_{i=1}^{a_k} \frac{i}{a_k} = \frac{1}{a_k} \cdot \frac{a_k(a_k + 1)}{2} = \frac{a_k + 1}{2}
\]
6. Summing this result over all 6 questions, we have:
\[
\sum_{k=1}^6 \sum_{i=1}^{a_k} \frac{i}{a_k} = \sum_{k=1}^6 \frac{a_k + 1}{2}
\]
7. This sum is given to be 2017:
\[
\sum_{k=1}^6 \frac{a_k + 1}{2} = 2017
\]
8. Simplifying the left-hand side, we get:
\[
\frac{1}{2} \sum_{k=1}^6 (a_k + 1) = 2017
\]
9. Distributing the sum inside the parentheses:
\[
\frac{1}{2} \left( \sum_{k=1}^6 a_k + \sum_{k=1}^6 1 \right) = 2017
\]
10. Since there are 6 questions, \(\sum_{k=1}^6 1 = 6\):
\[
\frac{1}{2} \left( \sum_{k=1}^6 a_k + 6 \right) = 2017
\]
11. Multiplying both sides by 2 to clear the fraction:
\[
\sum_{k=1}^6 a_k + 6 = 4034
\]
12. Subtracting 6 from both sides:
\[
\sum_{k=1}^6 a_k = 4028
\]
The final answer is \(\boxed{4028}\).
|
4028
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A convex equilateral pentagon with side length $2$ has two right angles. The greatest possible area of the pentagon is $m+\sqrt{n}$, where $m$ and $n$ are positive integers. Find $100m+n$.
[i]Proposed by Yannick Yao[/i]
|
1. **Identify the structure of the pentagon:**
- We are given a convex equilateral pentagon with side length \(2\) and two right angles.
- Let the pentagon be labeled as \(ABCDE\).
2. **Case 1: Adjacent right angles:**
- Suppose \(\angle A = \angle B = 90^\circ\).
- Quadrilateral \(ABCE\) has three equal sides, making it an isosceles trapezoid. Given \(\angle A = \angle B = 90^\circ\), \(ABCE\) is actually a square with side length \(2\).
- The area of the square \(ABCE\) is:
\[
\text{Area of square} = 2 \times 2 = 4
\]
- The remaining part of the pentagon, \(CDE\), forms an equilateral triangle with side length \(2\).
- The area of an equilateral triangle with side length \(a\) is given by:
\[
\text{Area of equilateral triangle} = \frac{\sqrt{3}}{4} a^2
\]
Substituting \(a = 2\):
\[
\text{Area of } CDE = \frac{\sqrt{3}}{4} \times 2^2 = \sqrt{3}
\]
- The total area of the pentagon in this case is:
\[
\text{Total area} = 4 + \sqrt{3}
\]
3. **Case 2: Non-adjacent right angles:**
- Suppose \(\angle A = \angle C = 90^\circ\).
- Triangles \(BAD\) and \(BCE\) are both isosceles right triangles with legs of length \(2\).
- The hypotenuse of each isosceles right triangle is:
\[
\text{Hypotenuse} = 2\sqrt{2}
\]
- The combined area of \(BAD\) and \(BCE\) forms a square with side length \(2\sqrt{2}\), but we can calculate the area directly:
\[
\text{Area of } BAD = \text{Area of } BCE = \frac{1}{2} \times 2 \times 2 = 2
\]
\[
\text{Total area of } BAD \text{ and } BCE = 2 + 2 = 4
\]
- Triangle \(BDE\) is an isosceles triangle with base \(DE = 2\) and height calculated using the Pythagorean theorem:
\[
\text{Height} = \sqrt{(2\sqrt{2})^2 - 1^2} = \sqrt{8 - 1} = \sqrt{7}
\]
- The area of triangle \(BDE\) is:
\[
\text{Area of } BDE = \frac{1}{2} \times 2 \times \sqrt{7} = \sqrt{7}
\]
- The total area of the pentagon in this case is:
\[
\text{Total area} = 4 + \sqrt{7}
\]
4. **Conclusion:**
- Comparing the two cases, the greatest possible area is \(4 + \sqrt{7}\).
- Therefore, \(m = 4\) and \(n = 7\).
- The final answer is \(100m + n = 100 \times 4 + 7 = 407\).
The final answer is \(\boxed{407}\)
|
407
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a$ and $b$ be positive integers such that $(2a+b)(2b+a)=4752$. Find the value of $ab$.
[i]Proposed by James Lin[/i]
|
1. **Expand the given equation:**
\[
(2a + b)(2b + a) = 4752
\]
Expanding the left-hand side, we get:
\[
4ab + 2a^2 + 2b^2 + ab = 4752
\]
Simplifying, we have:
\[
2a^2 + 5ab + 2b^2 = 4752
\]
2. **Divide the equation by 2:**
\[
a^2 + \frac{5}{2}ab + b^2 = 2376
\]
To simplify further, multiply the entire equation by 2:
\[
2a^2 + 5ab + 2b^2 = 4752
\]
3. **Rewrite the equation in a more manageable form:**
\[
(a + b)^2 + \frac{ab}{2} = 2376
\]
Let \( s = a + b \) and \( p = ab \). Then the equation becomes:
\[
s^2 + \frac{p}{2} = 2376
\]
4. **Estimate \( s \):**
\[
\sqrt{2376} \approx 48.7
\]
Since \( s \) must be an integer, we start testing values around 48.7.
5. **Test values for \( s \):**
- For \( s = 44 \):
\[
44^2 + \frac{p}{2} = 2376 \implies 1936 + \frac{p}{2} = 2376 \implies \frac{p}{2} = 440 \implies p = 880
\]
Check if \( a \) and \( b \) are integers:
\[
t^2 - 44t + 880 = 0
\]
The discriminant is:
\[
44^2 - 4 \cdot 880 = 1936 - 3520 = -1584 \quad (\text{not a perfect square})
\]
- For \( s = 45 \):
\[
45^2 + \frac{p}{2} = 2376 \implies 2025 + \frac{p}{2} = 2376 \implies \frac{p}{2} = 351 \implies p = 702
\]
Check if \( a \) and \( b \) are integers:
\[
t^2 - 45t + 702 = 0
\]
The discriminant is:
\[
45^2 - 4 \cdot 702 = 2025 - 2808 = -783 \quad (\text{not a perfect square})
\]
- For \( s = 46 \):
\[
46^2 + \frac{p}{2} = 2376 \implies 2116 + \frac{p}{2} = 2376 \implies \frac{p}{2} = 260 \implies p = 520
\]
Check if \( a \) and \( b \) are integers:
\[
t^2 - 46t + 520 = 0
\]
The discriminant is:
\[
46^2 - 4 \cdot 520 = 2116 - 2080 = 36 \quad (\text{a perfect square})
\]
Solving for \( t \):
\[
t = \frac{46 \pm \sqrt{36}}{2} = \frac{46 \pm 6}{2}
\]
Thus, \( t = 26 \) or \( t = 20 \). Therefore, \( a = 20 \) and \( b = 26 \) (or vice versa).
6. **Calculate \( ab \):**
\[
ab = 20 \times 26 = 520
\]
The final answer is \(\boxed{520}\)
|
520
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Senators Sernie Banders and Cedric "Ced" Truz of OMOrica are running for the office of Price Dent. The election works as follows: There are $66$ states, each composed of many adults and $2017$ children, with only the latter eligible to vote. On election day, the children each cast their vote with equal probability to Banders or Truz. A majority of votes in the state towards a candidate means they "win" the state, and the candidate with the majority of won states becomes the new Price Dent. Should both candidates win an equal number of states, then whoever had the most votes cast for him wins.
Let the probability that Banders and Truz have an unresolvable election, i.e., that they tie on both the state count and the popular vote, be $\frac{p}{q}$ in lowest terms, and let $m, n$ be the remainders when $p, q$, respectively, are divided by $1009$. Find $m + n$.
[i]Proposed by Ashwin Sah[/i]
|
1. **Understanding the Problem:**
- There are 66 states, each with 2017 children eligible to vote.
- Each child votes for either Banders or Truz with equal probability.
- A candidate wins a state if they receive the majority of votes in that state.
- The candidate who wins the majority of states becomes the new Price Dent.
- If both candidates win an equal number of states, the candidate with the most total votes wins.
- We need to find the probability that the election results in a tie in both the state count and the popular vote.
2. **Probability of a Tie in a Single State:**
- Each state has 2017 children, so the number of votes for each candidate follows a binomial distribution.
- The probability of a tie in a single state is zero because 2017 is odd, so one candidate will always have more votes than the other.
3. **Probability of a Tie in the State Count:**
- We need to find the probability that each candidate wins exactly 33 states.
- The number of ways to choose 33 states out of 66 is given by \(\binom{66}{33}\).
- Each state is won by either candidate with equal probability, so the probability of winning exactly 33 states is \(\binom{66}{33} \left(\frac{1}{2}\right)^{66}\).
4. **Probability of a Tie in the Popular Vote:**
- Given that each candidate wins exactly 33 states, we need to find the probability that the total number of votes for each candidate is the same.
- The total number of votes is \(66 \times 2017 = 132122\).
- The probability of a tie in the popular vote is the coefficient of \(x^{66061}\) in the expansion of \(\left( \sum_{k=0}^{1008} \binom{2017}{1008-k} x^k \right)^{33}\).
5. **Calculating the Coefficient:**
- The coefficient of \(x^{66061}\) in the expansion of \(\left( \sum_{k=0}^{1008} \binom{2017}{1008-k} x^k \right)^{33}\) is the same as the coefficient of \(x^{0}\) in the expansion of \(\left( \sum_{k=0}^{1008} \binom{2017}{1008-k} x^k \right)^{33} \left( \sum_{k=0}^{1008} \binom{2017}{1008-k} x^{-k} \right)^{33}\).
6. **Simplifying the Expression:**
- The expression simplifies to \(\left( \sum_{k=0}^{1008} \binom{2017}{1008-k} x^k \right)^{33} \left( \sum_{k=0}^{1008} \binom{2017}{1008-k} x^{-k} \right)^{33}\).
- This is equivalent to \(\left( \sum_{k=0}^{1008} \binom{2017}{1008-k} x^k \right)^{66}\).
7. **Modulo Calculation:**
- We need to calculate the coefficient modulo 1009.
- Using properties of binomial coefficients and Lucas' theorem, we find that the coefficient modulo 1009 is 0.
8. **Final Calculation:**
- The probability of a tie in both the state count and the popular vote is \(\frac{p}{q}\) in lowest terms.
- We need to find the remainders of \(p\) and \(q\) when divided by 1009.
- Since the coefficient modulo 1009 is 0, \(m = 0\) and \(n = 96\).
The final answer is \(m + n = \boxed{96}\).
|
96
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For a graph $G$ on $n$ vertices, let $P_G(x)$ be the unique polynomial of degree at most $n$ such that for each $i=0,1,2,\dots,n$, $P_G (i)$ equals the number of ways to color the vertices of the graph $G$ with $i$ distinct colors such that no two vertices connected by an edge have the same color. For each integer $3\le k \le 2017$, define a $k$-[i]tasty[/i] graph to be a connected graph on $2017$ vertices with $2017$ edges and a cycle of length $k$. Let the [i]tastiness[/i] of a $k$-tasty graph $G$ be the number of coefficients in $P_G(x)$ that are odd integers, and let $t$ be the minimal tastiness over all $k$-tasty graphs with $3\le k \le 2017$. Determine the sum of all integers $b$ between $3$ and $2017$ inclusive for which there exists a $b$-tasty graph with tastiness $t$.
[i]Proposed by Vincent Huang[/i]
|
1. Recognize that \( P_G(x) \) is the chromatic polynomial \( P(G, x) \). The chromatic polynomial \( P(G, x) \) of a graph \( G \) counts the number of ways to color the vertices of \( G \) with \( x \) colors such that no two adjacent vertices share the same color.
2. Use the well-known fact that for an edge \( uv \in E(G) \), the chromatic polynomial satisfies the relation:
\[
P(G, x) = P(G - uv, x) - P(G/uv, x)
\]
where \( G - uv \) is the graph obtained by removing the edge \( uv \) and \( G/uv \) is the graph obtained by contracting the edge \( uv \).
3. Let \( T_{n, k} \) be a \( k \)-tasty graph on \( n \) vertices. Note that \( T_{n, 2} \) is a tree \( T_n \), with the well-known chromatic polynomial:
\[
P(T_n, x) = x(x-1)^{n-1}
\]
4. Pick an edge in the cycle of \( T_{n, k} \) and use the above identity. Removing the edge produces a tree, and contracting the edge produces a \( T_{n-1, k-1} \) graph. Thus:
\[
P(T_{n, k}, x) = P(T_n, x) - P(T_{n-1, k-1}, x)
\]
5. Recurse this relation:
\[
P(T_{n, k}, x) = P(T_n, x) - P(T_{n-1}, x) + \cdots + (-1)^{k-2}P(T_{n-(k-2), k-(k-2)}, x)
\]
which simplifies to:
\[
P(T_{n, k}, x) = x(x-1)^{n-1} - x(x-1)^{n-2} + \cdots + (-1)^{k-2} x(x-1)^{n-k+2}
\]
6. Work in \( \mathbb{F}_2 \) (the field with two elements). In \( \mathbb{F}_2 \), we can ignore negative signs and the factor of \( x \) multiplied over everything since that doesn’t affect the number of odd coefficients:
\[
(x-1)^{n-1} + (x-1)^{n-2} + \cdots + (x-1)^{n-k+2}
\]
7. Factor out \( (x-1)^{n-k+2} \):
\[
(x-1)^{n-k+2} \left((x-1)^{k-3} + (x-1)^{k-4} + \cdots + 1\right)
\]
8. Recognize the geometric series and simplify:
\[
(x-1)^{n-k+2} \left(\frac{(x-1)^{k-2} - 1}{x-2} \right)
\]
9. In \( \mathbb{F}_2 \), \( x-2 \) is just \( x \), and we can ignore dividing by \( x \) since that doesn’t affect the count:
\[
(x-1)^{n-k+2}\left((x-1)^{k-2} - 1\right) = (x-1)^n - (x-1)^{n-k+2}
\]
10. Convert to \( (x+1) \) in \( \mathbb{F}_2 \):
\[
(x+1)^n + (x+1)^{n-k+2}
\]
11. Use Lucas's theorem to count the number of odd coefficients in \( (x+1)^n + (x+1)^{n-k+2} \).
The final answer is \( \boxed{2017} \).
|
2017
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
We define the bulldozer of triangle $ABC$ as the segment between points $P$ and $Q$, distinct points in the plane of $ABC$ such that $PA\cdot BC=PB\cdot CA=PC\cdot AB$ and $QA\cdot BC=QB\cdot CA=QC\cdot AB$. Let $XY$ be a segment of unit length in a plane $\mathcal{P}$, and let $\mathcal{S}$ be the region of $\mathcal P$ that the bulldozer of $XYZ$ sweeps through as $Z$ varies across the points in $\mathcal{P}$ satisfying $XZ=2YZ$. Find the greatest integer that is less than $100$ times the area of $\mathcal S$.
[i]Proposed by Michael Ren[/i]
|
1. **Part 1: A synthetic observation**
**Lemma 1**: \( O, L \) lie on line \( PQ \).
**Proof**: Recall the \( A \)-Apollonius circle \( \omega_A \), which is the locus of points \( X \) such that \( \frac{BX}{XC} = \frac{BA}{AC} \). Define \( \omega_B, \omega_C \) similarly.
It's easy to prove (by angle chasing) that the center \( T_A \) of \( \omega_A \) is the point where the \( A \)-tangent line of \( \odot(ABC) \) meets \( BC \). Furthermore, points \( P, Q \) lie on all \( \omega_B, \omega_C \).
Let \( AL, BL, CL \) cut \( \odot(ABC) \) at \( A_1, B_1, C_1 \). It's easy to see that \( A_1 \in \omega_A, B_1 \in \omega_B, C_1 \in \omega_C \). But \( AL \cdot A_1L = BL \cdot B_1L = CL \cdot C_1L \), therefore \( L \) is the radical center of \( \omega_A, \omega_B, \omega_C \) or they are coaxial.
Since \( \angle T_AAO = 90^\circ \), \( \omega_A \) and \( \odot(ABC) \) are orthogonal. Therefore, the power of \( O \) with respect to \( \omega_A, \omega_B, \omega_C \) is \( OA^2 = OB^2 = OC^2 \). Hence, \( O \) lies on the radical axis of the three circles too.
But \( O \ne L \), therefore \( \omega_A, \omega_B, \omega_C \) are coaxial with radical axis \( OL \). The lemma now follows.
2. **Part 2: Initial Coordinates**
It suffices to analyze the locus which the line \( OL \) sweeps through the \( Z \)-Apollonius circle of \( \Delta XYZ \) (which is fixed).
Set \( X = \left(\frac{-4}{3}, 0\right), Y = \left(\frac{-1}{3}, 0\right) \) (This gets the center \( T \) of the \( Z \)-Apollonius circle at \( (0,0) \)). Let \( Z = (a, b) \) where \( a^2 + b^2 = \frac{4}{9} \).
We now compute \( L \). The trick is to transfer Barycentric Coordinates with respect to \( \Delta ZXY \) to Cartesian coordinates using vector definition. We first compute
\[
YZ^2 = \left(a + \frac{1}{3}\right)^2 + b^2 = \frac{6a + 5}{9}
\]
\[
XZ^2 = \left(a + \frac{4}{3}\right)^2 + b^2 = \frac{24a + 20}{9}
\]
Therefore, the barycentric coordinate of \( L \) is \( (9 : 6a + 5 : 24a + 20) = \left(\frac{9}{30a + 34}, \frac{6a + 5}{30a + 34}, \frac{24a + 20}{30a + 34}\right) \). Hence, the Cartesian coordinate of \( L \) is
\[
\frac{9}{30a + 34} \cdot (a, b) + \frac{6a + 5}{30a + 34} \cdot \left(\frac{-4}{3}, 0\right) + \frac{24a + 20}{30a + 34} \cdot \left(\frac{-1}{3}, 0\right) = \left(\frac{-(21a + 40)}{3(30a + 34)}, \frac{9b}{30a + 34}\right)
\]
Now we compute \( O \). Clearly, the \( x \)-coordinate is \( \frac{-5}{6} \). Let the \( y \)-coordinate be \( y \). Since \( \angle TZO = 90^\circ \), we have
\[
\frac{y - b}{\frac{-5}{6} - a} \cdot \frac{b}{a} = -1 \implies y = b + \frac{a^2 + \frac{5}{6}a}{b} = \frac{15a + 8}{18b}
\]
Therefore, \( O = \left(\frac{-5}{6}, \frac{15a + 8}{18b}\right) \).
Now using Shoelace, the equation of \( OL \) is
\[
y = -\frac{51a + 20}{27b} x - \frac{60a + 14}{81b}
\]
3. **Part 3: Analyzing Locus**
Now that we have the equation of lines, we can see they are a combination of 2 types of lines:
\[
y = \frac{(153x + 60)a + (60x + 14)}{-81\sqrt{\frac{4}{9} - a^2}} \text{ and } y = \frac{(153x + 60)a + (60x + 14)}{81\sqrt{\frac{4}{9} - a^2}}
\]
Note that if \( x \in \left[-\frac{2}{3}, -\frac{13}{21}\right) \cup \left(-\frac{1}{3}, \frac{2}{3}\right] \), then \( \frac{2}{3}(153x + 60) + (60x + 14) \) and \( -\frac{2}{3}(153x + 60) + (60x + 14) \) have different signs. Hence,
\[
\lim_{a \rightarrow \frac{2}{3}^-} \frac{(153x + 60)a + (60x + 14)}{81\sqrt{\frac{4}{9} - a^2}} = -\lim_{a \rightarrow -\frac{2}{3}^+} \frac{(153x + 60)a + (60x + 14)}{81\sqrt{\frac{4}{9} - a^2}} = \pm \infty
\]
which means that \( y \) can take any real value.
When \( x = -\frac{1}{3}, -\frac{13}{21} \), we can easily show by limit that \( \frac{(153x + 60)a + (60x + 14)}{-81\sqrt{\frac{4}{9} - a^2}} \) and \( \frac{(153x + 60)a + (60x + 14)}{81\sqrt{\frac{4}{9} - a^2}} \) can take any nonzero value.
Now the most important part is when \( -\frac{13}{21} < x < -\frac{1}{3} \), we can show by calculus that the only area that the bulldozer can't sweep is the area \( y^2 < \frac{(21x + 13)(-3x - 1)}{27} \), which is the ellipse \( 27y^2 + 63(x + \frac{10}{21})^2 < \frac{9}{7} \) which has area
\[
\pi \times \sqrt{\frac{\frac{9}{7}}{27}} \times \sqrt{\frac{\frac{9}{7}}{63}} = \frac{\pi}{7\sqrt{21}}
\]
Hence,
\[
S = \left[100 \left(\frac{4\pi}{9} - \frac{\pi}{7\sqrt{21}}\right)\right] = [129.8\ldots] = \boxed{129}
\]
|
129
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Steven draws a line segment between every two of the points \[A(2,2), B(-2,2), C(-2,-2), D(2,-2), E(1,0), F(0,1), G(-1,0), H(0,-1).\] How many regions does he divide the square $ABCD$ into?
[i]Proposed by Michael Ren
|
1. **Identify the points and the square:**
The points given are \(A(2,2)\), \(B(-2,2)\), \(C(-2,-2)\), \(D(2,-2)\), \(E(1,0)\), \(F(0,1)\), \(G(-1,0)\), and \(H(0,-1)\). The square \(ABCD\) has vertices at \(A\), \(B\), \(C\), and \(D\).
2. **Draw the square and the points:**
Plot the points on a coordinate plane and draw the square \(ABCD\). The points \(E\), \(F\), \(G\), and \(H\) are inside the square.
3. **Draw line segments between every pair of points:**
There are \(\binom{8}{2} = 28\) line segments to be drawn, as there are 8 points and we are drawing a line segment between every pair of points.
4. **Count the regions formed:**
To find the number of regions formed by these line segments, we can use Euler's formula for planar graphs, which states:
\[
V - E + F = 2
\]
where \(V\) is the number of vertices, \(E\) is the number of edges, and \(F\) is the number of faces (regions).
5. **Calculate the number of vertices and edges:**
- The number of vertices \(V\) is 8 (the given points).
- The number of edges \(E\) is 28 (the line segments between every pair of points).
6. **Determine the number of regions \(F\):**
Using Euler's formula:
\[
8 - 28 + F = 2
\]
Solving for \(F\):
\[
F = 28 - 8 + 2 = 22
\]
7. **Consider the intersections inside the square:**
The intersections of the line segments inside the square will create additional regions. Each pair of intersecting lines inside the square adds a new region. The exact number of these intersections can be complex to calculate directly, but we can use the given solution's symmetry and visual inspection to determine the total number of regions.
8. **Verify the given solution:**
The given solution states that the total number of regions is 60. This can be verified by drawing the diagram and counting the regions formed by the intersections of the line segments.
The final answer is \(\boxed{60}\).
|
60
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $S$ be a set of $13$ distinct, pairwise relatively prime, positive integers. What is the smallest possible value of $\max_{s \in S} s- \min_{s \in S}s$?
[i]Proposed by James Lin
|
1. **Initial Consideration**: We need to find the smallest possible value of $\max_{s \in S} s - \min_{s \in S} s$ for a set $S$ of 13 distinct, pairwise relatively prime, positive integers.
2. **Lower Bound**: Since the integers are pairwise relatively prime, they must be distinct and cannot share any common factors other than 1. The smallest 13 distinct positive integers are $1, 2, 3, \ldots, 13$. However, these are not pairwise relatively prime. We need to find a set of 13 such integers that are pairwise relatively prime and calculate the difference between the maximum and minimum values in this set.
3. **Odd Integers and Multiples**: Consider the set of odd integers. Among any string of 12 consecutive odd integers, at least 2 are multiples of 5. Thus, we need to add 2 to remove one of them. There are also at least 4 odd multiples of 3. To remove 3 of them, we must add 6. However, one can be a multiple of 15, so we can subtract 2 back. For multiples of 7, there is at least 1 of them. We don't need to remove this one. But note that there is at least 1 multiple of 21, so we must subtract 2 from our count.
4. **Tracking Odd Numbers**: Since we moved up 4 odd numbers total, we keep track of 12 + 4 = 16 odd numbers total. Within 16 odd numbers, at least 3 are multiples of 5, so we must add an additional 2. Similarly, there are at least 5 multiples of 3, so we must add another additional 2. Since there is now an additional odd multiple of 7, we must remove one of them to add 2.
5. **Final Calculation**: Summarizing the adjustments:
\[
22 + 2 + 6 - 2 - 2 + 2 + 2 + 2 = 32
\]
6. **Example Set**: An example set that satisfies the conditions is $S = \{17, 19, 23, 25, 27, 29, 31, 32, 37, 41, 43, 47, 49\}$.
Conclusion:
\[
\boxed{32}
\]
|
32
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Henry starts with a list of the first 1000 positive integers, and performs a series of steps on the list. At each step, he erases any nonpositive integers or any integers that have a repeated digit, and then decreases everything in the list by 1. How many steps does it take for Henry's list to be empty?
[i]Proposed by Michael Ren[/i]
|
1. **Initial List and Conditions**:
- Henry starts with the list of the first 1000 positive integers: \( \{1, 2, 3, \ldots, 1000\} \).
- At each step, he erases any nonpositive integers or any integers that have a repeated digit.
- After erasing, he decreases every remaining integer in the list by 1.
2. **Understanding the Erasure Condition**:
- Nonpositive integers are erased immediately.
- Integers with repeated digits are also erased. For example, numbers like 11, 22, 101, etc., will be erased.
3. **Effect of Decreasing by 1**:
- After each step, every integer in the list is decreased by 1. This means that the list will eventually contain nonpositive integers, which will be erased.
4. **Key Observation**:
- The critical observation is to determine the maximum number of steps it takes for any integer to be erased.
- Consider the number 10. It will be decreased to 9, 8, 7, ..., 1, 0, and then -1. This process takes exactly 11 steps.
5. **Erasure of Numbers with Repeated Digits**:
- Numbers like 11, 22, 33, ..., 99, 101, 111, etc., will be erased as soon as they appear in the list.
- For example, 11 will be erased in the first step, 22 in the second step, and so on.
6. **General Case**:
- For any number \( n \) in the list, if it does not have repeated digits, it will be decreased step by step until it becomes nonpositive.
- The maximum number of steps required for any number to become nonpositive is determined by the number 10, which takes 11 steps.
7. **Conclusion**:
- Since the number 10 takes exactly 11 steps to be erased, and all other numbers will be erased in fewer or equal steps, the entire list will be empty after 11 steps.
\[
\boxed{11}
\]
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
We define the sets of lattice points $S_0,S_1,\ldots$ as $S_0=\{(0,0)\}$ and $S_k$ consisting of all lattice points that are exactly one unit away from exactly one point in $S_{k-1}$. Determine the number of points in $S_{2017}$.
[i]Proposed by Michael Ren
|
1. **Understanding the Problem:**
- We start with the set \( S_0 = \{(0,0)\} \).
- Each subsequent set \( S_k \) consists of all lattice points that are exactly one unit away from exactly one point in \( S_{k-1} \).
2. **Visualizing the Growth of Sets:**
- \( S_0 \) contains the single point \((0,0)\).
- \( S_1 \) will contain the points \((1,0)\), \((-1,0)\), \((0,1)\), and \((0,-1)\), which are the four points exactly one unit away from \((0,0)\).
3. **Pattern Recognition:**
- Each point in \( S_k \) is exactly one unit away from exactly one point in \( S_{k-1} \).
- This implies that \( S_k \) forms a boundary around \( S_{k-1} \).
4. **Counting the Points:**
- The number of points in \( S_k \) forms a pattern that can be related to the binomial coefficients.
- Specifically, the number of points in \( S_k \) is given by the binomial coefficient \(\binom{2k}{k}\).
5. **Applying Lucas' Theorem:**
- Lucas' Theorem helps in determining the number of odd binomial coefficients.
- For a binomial coefficient \(\binom{n}{k}\), the number of odd coefficients is given by the product of the number of 1's in the binary representation of \( n \).
6. **Binary Representation:**
- The binary representation of 2017 is \(11111000001_2\).
- This representation has 6 ones.
7. **Calculating the Number of Odd Binomial Coefficients:**
- The number of odd binomial coefficients \(\binom{2017}{n}\) is \(2^6 = 64\).
8. **Final Calculation:**
- Squaring the number of odd binomial coefficients gives us \(64^2 = 4096\).
The final answer is \(\boxed{4096}\).
|
4096
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\{a,b,c,d,e,f,g,h,i\}$ be a permutation of $\{1,2,3,4,5,6,7,8,9\}$ such that $\gcd(c,d)=\gcd(f,g)=1$ and
\[(10a+b)^{c/d}=e^{f/g}.\]
Given that $h>i$, evaluate $10h+i$.
[i]Proposed by James Lin[/i]
|
1. Given the permutation $\{a,b,c,d,e,f,g,h,i\}$ of $\{1,2,3,4,5,6,7,8,9\}$, we need to satisfy the conditions $\gcd(c,d)=1$, $\gcd(f,g)=1$, and the equation:
\[
(10a + b)^{c/d} = e^{f/g}.
\]
2. We need to find values for $a, b, c, d, e, f, g, h, i$ that satisfy the given conditions. Let's start by considering the equation:
\[
(10a + b)^{c/d} = e^{f/g}.
\]
3. To simplify the problem, we can look for simple rational exponents that might work. Consider the example provided in the alternative answer:
\[
27^{1/4} = 9^{3/8}.
\]
4. We can rewrite the numbers 27 and 9 in terms of their prime factors:
\[
27 = 3^3 \quad \text{and} \quad 9 = 3^2.
\]
5. Substituting these into the equation, we get:
\[
(3^3)^{1/4} = (3^2)^{3/8}.
\]
6. Simplifying the exponents, we have:
\[
3^{3/4} = 3^{6/8} = 3^{3/4}.
\]
7. This confirms that the equation holds true. Now, we need to map these values back to the original variables:
\[
10a + b = 27, \quad c = 1, \quad d = 4, \quad e = 9, \quad f = 3, \quad g = 8.
\]
8. We need to ensure that $\gcd(c,d) = \gcd(1,4) = 1$ and $\gcd(f,g) = \gcd(3,8) = 1$, which are both true.
9. Now, we need to find the remaining values $h$ and $i$ such that $h > i$. The remaining numbers are $\{2, 4, 5, 6, 8\}$.
10. Since $h > i$, we can choose the largest and second largest numbers from the remaining set. Thus, $h = 8$ and $i = 6$.
11. Finally, we calculate $10h + i$:
\[
10h + i = 10 \cdot 8 + 6 = 80 + 6 = 86.
\]
The final answer is $\boxed{86}$.
|
86
|
Number Theory
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\mathcal{P}_1$ and $\mathcal{P}_2$ be two parabolas with distinct directrices $\ell_1$ and $\ell_2$ and distinct foci $F_1$ and $F_2$ respectively. It is known that $F_1F_2||\ell_1||\ell_2$, $F_1$ lies on $\mathcal{P}_2$, and $F_2$ lies on $\mathcal{P}_1$. The two parabolas intersect at distinct points $A$ and $B$. Given that $F_1F_2=1$, the value of $AB^2$ can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Find $100m+n$.
[i]Proposed by Yannick Yao
|
1. **Assume the equations of the parabolas:**
Without loss of generality, we can assume one of the parabolas, $\mathcal{P}_1$, is given by the equation \( y = \frac{1}{2}x^2 \). The focus of this parabola is at \( F_1 = (0, \frac{1}{2}) \) and the directrix is \( y = -\frac{1}{2} \).
2. **Determine the properties of the second parabola:**
Since \( F_1F_2 \parallel \ell_1 \parallel \ell_2 \) and \( F_1 \) lies on \( \mathcal{P}_2 \), the focus \( F_2 \) of \( \mathcal{P}_2 \) must have the same y-coordinate as \( F_1 \), which is \( \frac{1}{2} \). Let \( F_2 = (h, \frac{1}{2}) \).
3. **Use the given conditions to find \( h \):**
Since \( F_1F_2 = 1 \), the distance between \( F_1 \) and \( F_2 \) is 1. Therefore, \( h = 1 \) or \( h = -1 \). Without loss of generality, we can choose \( h = 1 \). Thus, \( F_2 = (1, \frac{1}{2}) \).
4. **Find the equation of the second parabola:**
The second parabola \( \mathcal{P}_2 \) has its focus at \( (1, \frac{1}{2}) \) and its directrix parallel to \( y = -\frac{1}{2} \). Since \( F_1 \) lies on \( \mathcal{P}_2 \), we use the definition of a parabola:
\[
\text{Distance from } (0, \frac{1}{2}) \text{ to } (1, \frac{1}{2}) = \text{Distance from } (0, \frac{1}{2}) \text{ to the directrix}
\]
The distance from \( (0, \frac{1}{2}) \) to \( (1, \frac{1}{2}) \) is 1. Let the directrix of \( \mathcal{P}_2 \) be \( y = k \). The distance from \( (0, \frac{1}{2}) \) to the directrix is \( \left| \frac{1}{2} - k \right| \). Therefore,
\[
1 = \left| \frac{1}{2} - k \right|
\]
Solving this, we get \( k = -\frac{1}{2} \) or \( k = \frac{3}{2} \). Since the directrix must be below the focus, we choose \( k = -\frac{1}{2} \).
5. **Equation of the second parabola:**
The equation of \( \mathcal{P}_2 \) is:
\[
-2(y - 1) = (x - 1)^2
\]
Simplifying, we get:
\[
y = -\frac{1}{2}(x - 1)^2 + 1
\]
6. **Find the intersection points of the parabolas:**
Set the equations equal to each other:
\[
\frac{1}{2}x^2 = -\frac{1}{2}(x - 1)^2 + 1
\]
Simplify and solve for \( x \):
\[
\frac{1}{2}x^2 = -\frac{1}{2}(x^2 - 2x + 1) + 1
\]
\[
\frac{1}{2}x^2 = -\frac{1}{2}x^2 + x - \frac{1}{2} + 1
\]
\[
\frac{1}{2}x^2 + \frac{1}{2}x^2 = x + \frac{1}{2}
\]
\[
x^2 = x + \frac{1}{2}
\]
\[
x^2 - x - \frac{1}{2} = 0
\]
Solve the quadratic equation:
\[
x = \frac{1 \pm \sqrt{1 + 2}}{2} = \frac{1 \pm \sqrt{3}}{2}
\]
Thus, the intersection points are:
\[
x_1 = \frac{1 - \sqrt{3}}{2}, \quad x_2 = \frac{1 + \sqrt{3}}{2}
\]
7. **Find the corresponding y-coordinates:**
Substitute \( x_1 \) and \( x_2 \) back into \( y = \frac{1}{2}x^2 \):
\[
y_1 = \frac{1}{2} \left( \frac{1 - \sqrt{3}}{2} \right)^2 = \frac{2 - \sqrt{3}}{4}
\]
\[
y_2 = \frac{1}{2} \left( \frac{1 + \sqrt{3}}{2} \right)^2 = \frac{2 + \sqrt{3}}{4}
\]
8. **Calculate \( AB^2 \):**
The points of intersection are \( A = \left( \frac{1 - \sqrt{3}}{2}, \frac{2 - \sqrt{3}}{4} \right) \) and \( B = \left( \frac{1 + \sqrt{3}}{2}, \frac{2 + \sqrt{3}}{4} \right) \). The distance \( AB \) is:
\[
AB^2 = \left( \frac{1 + \sqrt{3}}{2} - \frac{1 - \sqrt{3}}{2} \right)^2 + \left( \frac{2 + \sqrt{3}}{4} - \frac{2 - \sqrt{3}}{4} \right)^2
\]
\[
= \left( \sqrt{3} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2
\]
\[
= 3 + \frac{3}{4} = \frac{15}{4}
\]
9. **Find \( 100m + n \):**
Since \( AB^2 = \frac{15}{4} \), we have \( m = 15 \) and \( n = 4 \). Therefore, \( 100m + n = 100 \times 15 + 4 = 1504 \).
The final answer is \( \boxed{1504} \)
|
1504
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Tessa the hyper-ant is at the origin of the four-dimensional Euclidean space $\mathbb R^4$. For each step she moves to another lattice point that is $2$ units away from the point she is currently on. How many ways can she return to the origin for the first time after exactly $6$ steps?
[i]Proposed by Yannick Yao
|
To solve this problem, we need to determine the number of ways Tessa the hyper-ant can return to the origin in $\mathbb{R}^4$ after exactly 6 steps, where each step is 2 units away from the current point. We will use the principle of inclusion-exclusion (PIE) to count the valid paths.
1. **Understanding the Problem**:
- Tessa can move in any of the 4 dimensions.
- Each move changes one of the coordinates by $\pm 2$.
- We need to count the number of valid sequences of 6 moves that return Tessa to the origin.
2. **Setting Up the Problem**:
- Each move can be represented as a vector in $\mathbb{R}^4$ with one coordinate being $\pm 2$ and the others being 0.
- To return to the origin after 6 steps, the sum of the vectors must be zero in each coordinate.
3. **Using the Principle of Inclusion-Exclusion (PIE)**:
- We will count the number of ways to distribute the 6 steps among the 4 coordinates such that the sum in each coordinate is zero.
- We denote the number of steps in each coordinate as $x_1, x_2, x_3, x_4$ where $x_i$ is the number of steps in the $i$-th coordinate.
4. **Casework on the Distribution of Steps**:
- We need to consider different cases based on how the 6 steps are distributed among the 4 coordinates.
5. **Calculating the Number of Ways**:
- **Case [6]**: All 6 steps in one coordinate.
\[
4 \cdot \binom{6}{3} = 4 \cdot 20 = 80
\]
- **Case [4, 2]**: 4 steps in one coordinate and 2 steps in another.
\[
4 \cdot 3 \cdot \binom{6}{2} \cdot \binom{4}{2} = 4 \cdot 3 \cdot 15 \cdot 6 = 1080
\]
- **Case [3, 3]**: 3 steps in two coordinates.
\[
\binom{4}{2} \cdot \frac{6!}{3!3!} = 6 \cdot 20 = 120
\]
- **Case [2, 2, 2]**: 2 steps in each of the three coordinates.
\[
\frac{6!}{2!2!2!} = 90
\]
6. **Applying PIE**:
- Using PIE, we combine the counts from the different cases:
\[
[6] - 2[4,2] - [3,3] + [2,2,2]
\]
Substituting the values:
\[
80 - 2 \cdot 1080 - 120 + 90 = 80 - 2160 - 120 + 90 = -2110
\]
7. **Final Calculation**:
- The final number of ways Tessa can return to the origin after 6 steps is:
\[
911040 - 2 \cdot 81216 - 36864 + 13824 = 911040 - 162432 - 36864 + 13824 = 725568
\]
The final answer is $\boxed{725568}$.
|
725568
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For a positive integer $n$, define $f(n)=\sum_{i=0}^{\infty}\frac{\gcd(i,n)}{2^i}$ and let $g:\mathbb N\rightarrow \mathbb Q$ be a function such that $\sum_{d\mid n}g(d)=f(n)$ for all positive integers $n$. Given that $g(12321)=\frac{p}{q}$ for relatively prime integers $p$ and $q$, find $v_2(p)$.
[i]Proposed by Michael Ren[/i]
|
1. **Applying Möbius Inversion**:
Given the function \( f(n) = \sum_{i=0}^{\infty} \frac{\gcd(i,n)}{2^i} \) and the function \( g \) such that \( \sum_{d \mid n} g(d) = f(n) \), we can use Möbius inversion to find \( g(n) \):
\[
g(n) = \sum_{d \mid n} f(d) \mu\left(\frac{n}{d}\right)
\]
where \( \mu \) is the Möbius function.
2. **Prime Factorization**:
Given \( 12321 = 3^2 \cdot 37^2 \), we need to find \( g(12321) \):
\[
g(12321) = f(12321) - f(4107) - f(333) + f(111)
\]
3. **General Form**:
We solve the more general problem of finding \( g(p^2 q^2) \):
\[
g(p^2 q^2) = f(p^2 q^2) - f(p^2 q) - f(p q^2) + f(p q)
\]
The coefficient in front of \( \frac{1}{2^j} \) in the expansion will be:
\[
\gcd(j, p^2 q^2) - \gcd(j, p^2 q) - \gcd(j, p q^2) + \gcd(j, p q)
\]
This is only nonzero if \( p^2 q^2 \mid j \), in which case it is \( p^2 q^2 - p^2 q - p q^2 + p q \).
4. **Summation**:
Thus, we get:
\[
g(p^2 q^2) = \sum_{j=0}^{\infty} \frac{p^2 q^2 - p^2 q - p q^2 + p q}{2^{j p^2 q^2}}
\]
This simplifies to:
\[
g(p^2 q^2) = (p^2 q^2 - p^2 q - p q^2 + p q) \left( \frac{1}{1 - \frac{1}{2^{p^2 q^2}}} \right)
\]
\[
g(p^2 q^2) = (p^2 q^2 - p^2 q - p q^2 + p q) \left( \frac{2^{p^2 q^2}}{2^{p^2 q^2} - 1} \right)
\]
5. **Substitution**:
For \( p = 3 \) and \( q = 37 \):
\[
g(12321) = (3^2 \cdot 37^2 - 3^2 \cdot 37 - 3 \cdot 37^2 + 3 \cdot 37) \left( \frac{2^{12321}}{2^{12321} - 1} \right)
\]
\[
g(12321) = (9 \cdot 1369 - 9 \cdot 37 - 3 \cdot 1369 + 3 \cdot 37) \left( \frac{2^{12321}}{2^{12321} - 1} \right)
\]
\[
g(12321) = (12321 - 333 - 4107 + 111) \left( \frac{2^{12321}}{2^{12321} - 1} \right)
\]
\[
g(12321) = 7992 \left( \frac{2^{12321}}{2^{12321} - 1} \right)
\]
6. **Finding \( v_2(p) \)**:
The number of factors of 2 in the numerator is:
\[
v_2(7992) + 12321 = 3 + 12321 = 12324
\]
The final answer is \( \boxed{ 12324 } \).
|
12324
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Given a sequence of positive integers $a_1, a_2, a_3, \dots, a_{n}$, define the \emph{power tower function} \[f(a_1, a_2, a_3, \dots, a_{n})=a_1^{a_2^{a_3^{\mathstrut^{ .^{.^{.^{a_{n}}}}}}}}.\] Let $b_1, b_2, b_3, \dots, b_{2017}$ be positive integers such that for any $i$ between 1 and 2017 inclusive, \[f(a_1, a_2, a_3, \dots, a_i, \dots, a_{2017})\equiv f(a_1, a_2, a_3, \dots, a_i+b_i, \dots, a_{2017}) \pmod{2017}\] for all sequences $a_1, a_2, a_3, \dots, a_{2017}$ of positive integers greater than 2017. Find the smallest possible value of $b_1+b_2+b_3+\dots+b_{2017}$.
[i]Proposed by Yannick Yao
|
1. **Understanding the Problem:**
We need to find the smallest possible value of \( b_1 + b_2 + b_3 + \dots + b_{2017} \) such that for any sequence of positive integers \( a_1, a_2, \dots, a_{2017} \) greater than 2017, the power tower function \( f(a_1, a_2, \dots, a_i, \dots, a_{2017}) \equiv f(a_1, a_2, \dots, a_i + b_i, \dots, a_{2017}) \pmod{2017} \) holds for all \( i \) from 1 to 2017.
2. **Analyzing the Power Tower Function:**
The power tower function is defined as:
\[
f(a_1, a_2, a_3, \dots, a_{n}) = a_1^{a_2^{a_3^{\cdots^{a_{n}}}}}
\]
We need to ensure that adding \( b_i \) to \( a_i \) does not change the value of the power tower function modulo 2017.
3. **Properties of Modulo 2017:**
Since 2017 is a prime number, by Fermat's Little Theorem, for any integer \( a \) such that \( \gcd(a, 2017) = 1 \):
\[
a^{2016} \equiv 1 \pmod{2017}
\]
This implies that the order of any integer modulo 2017 divides 2016.
4. **Determining \( b_1 \):**
For \( a_1 \) to be unaffected by adding \( b_1 \) modulo 2017, \( b_1 \) must be a multiple of 2016 (the order of any integer modulo 2017). Thus, the smallest \( b_1 \) is 2016.
5. **Determining \( b_2 \):**
For \( a_2 \), we need to consider the exponentiation modulo 2016. The maximal order of any integer modulo 2016 is the least common multiple of the orders of its prime factors:
\[
\text{lcm}(2^5, 3^2, 7) = \text{lcm}(32, 9, 7) = 2016
\]
Thus, \( b_2 \) must be a multiple of 672 (the order of the largest cyclic group modulo 2016). The smallest \( b_2 \) is 672.
6. **Determining \( b_3 \):**
For \( a_3 \), we need to consider the exponentiation modulo 672. The maximal order of any integer modulo 672 is:
\[
\text{lcm}(2^5, 3^2, 7) = \text{lcm}(32, 9, 7) = 224
\]
Thus, \( b_3 \) must be a multiple of 224. The smallest \( b_3 \) is 224.
7. **Determining \( b_4 \):**
For \( a_4 \), we need to consider the exponentiation modulo 224. The maximal order of any integer modulo 224 is:
\[
\text{lcm}(2^5, 7) = \text{lcm}(32, 7) = 32
\]
Thus, \( b_4 \) must be a multiple of 32. The smallest \( b_4 \) is 32.
8. **Determining \( b_5 \):**
For \( a_5 \), we need to consider the exponentiation modulo 32. The maximal order of any integer modulo 32 is:
\[
2^5 = 32
\]
Thus, \( b_5 \) must be a multiple of 16. The smallest \( b_5 \) is 16.
9. **Determining \( b_6 \):**
For \( a_6 \), we need to consider the exponentiation modulo 16. The maximal order of any integer modulo 16 is:
\[
2^4 = 16
\]
Thus, \( b_6 \) must be a multiple of 8. The smallest \( b_6 \) is 8.
10. **Determining \( b_7 \):**
For \( a_7 \), we need to consider the exponentiation modulo 8. The maximal order of any integer modulo 8 is:
\[
2^3 = 8
\]
Thus, \( b_7 \) must be a multiple of 4. The smallest \( b_7 \) is 4.
11. **Determining \( b_8 \):**
For \( a_8 \), we need to consider the exponentiation modulo 4. The maximal order of any integer modulo 4 is:
\[
2^2 = 4
\]
Thus, \( b_8 \) must be a multiple of 2. The smallest \( b_8 \) is 2.
12. **Determining \( b_9 \) to \( b_{2017} \):**
For \( a_9 \) to \( a_{2017} \), we need to consider the exponentiation modulo 2. The maximal order of any integer modulo 2 is:
\[
2^1 = 2
\]
Thus, \( b_9 \) to \( b_{2017} \) must be 1.
13. **Summing Up \( b_i \):**
\[
b_1 + b_2 + b_3 + \dots + b_{2017} = 2016 + 672 + 224 + 32 + 16 + 8 + 4 + 2 + (2017 - 8) \cdot 1
\]
\[
= 2016 + 672 + 224 + 32 + 16 + 8 + 4 + 2 + 2009
\]
\[
= 4983
\]
The final answer is \(\boxed{4983}\)
|
4983
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $p=2017$ be a prime. Suppose that the number of ways to place $p$ indistinguishable red marbles, $p$ indistinguishable green marbles, and $p$ indistinguishable blue marbles around a circle such that no red marble is next to a green marble and no blue marble is next to a blue marble is $N$. (Rotations and reflections of the same configuration are considered distinct.) Given that $N=p^m\cdot n$, where $m$ is a nonnegative integer and $n$ is not divisible by $p$, and $r$ is the remainder of $n$ when divided by $p$, compute $pm+r$.
[i]Proposed by Yannick Yao[/i]
|
1. **Fixing a Blue Marble**:
To simplify the problem, we fix one blue marble at the top of the circle. This reduces the problem to arranging the remaining \(2016\) blue marbles, \(2017\) red marbles, and \(2017\) green marbles around the circle. Note that we must multiply by \(3\) at the end to account for the three possible fixed positions of the blue marble.
2. **Placing Blue Marbles**:
Placing the \(2016\) blue marbles around the circle creates \(2017\) gaps (including the gap before the first blue marble and after the last blue marble) where we can place the red and green marbles.
3. **Distributing Red and Green Marbles**:
Let \(i\) of these gaps contain red marbles and the remaining \(2017-i\) gaps contain green marbles. We need to distribute \(2017\) red marbles into \(i\) non-empty gaps and \(2017\) green marbles into \(2017-i\) non-empty gaps.
4. **Counting Arrangements**:
The number of ways to choose \(i\) gaps out of \(2017\) for red marbles is \(\binom{2017}{i}\). The number of ways to distribute \(2017\) red marbles into \(i\) non-empty gaps is \(\binom{2016}{i-1}\) (using the stars and bars method). Similarly, the number of ways to distribute \(2017\) green marbles into \(2017-i\) non-empty gaps is \(\binom{2016}{2016-i}\).
5. **Summing Over All Possible \(i\)**:
Therefore, the total number of ways \(N\) is given by:
\[
N = 3 \sum_{i=1}^{2016} \binom{2017}{i} \binom{2016}{i-1} \binom{2016}{2016-i}
\]
6. **Simplifying the Binomial Coefficients**:
Using the identity \(\binom{2017}{i} = \frac{2017}{i} \binom{2016}{i-1}\), we can rewrite \(N\) as:
\[
N = 3 \cdot 2017 \sum_{i=1}^{2016} \frac{1}{i} \binom{2016}{i-1} \binom{2016}{2016-i}
\]
7. **Modulo Simplification**:
We need to find the value of the summation modulo \(2017\). Using the fact that \(\binom{p-1}{j} \equiv (-1)^j \pmod{p}\) for a prime \(p\), we get:
\[
\sum_{i=1}^{2016} \frac{(-1)^i}{i} \pmod{2017}
\]
8. **Using Harmonic Series Modulo**:
The harmonic series modulo a prime \(p\) can be simplified using the fact that \(\frac{1}{j} \equiv (-1)^{j-1} \cdot \frac{1}{p} \binom{p}{j} \pmod{p}\). This gives us:
\[
\sum_{i=1}^{2016} \frac{(-1)^{2i+1}}{2017} \binom{2017}{i} = \frac{-(2^{2017} - 2)}{2017} \pmod{2017}
\]
9. **Computing the Final Value**:
Using repeated squaring modulo \(2017^2\), we find that:
\[
\frac{-(2^{2017} - 2)}{2017} \equiv 632 \pmod{2017}
\]
10. **Final Calculation**:
Therefore, \(N = 3 \cdot 2017 \cdot 632\). This gives us \(m = 1\) and \(r = 3 \cdot 632 \pmod{2017} = 1896\). Thus, the final answer is:
\[
pm + r = 2017 + 1896 = \boxed{3913}
\]
|
3913
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For an integer $k$ let $T_k$ denote the number of $k$-tuples of integers $(x_1,x_2,...x_k)$ with $0\le x_i < 73$ for each $i$, such that $73|x_1^2+x_2^2+...+x_k^2-1$. Compute the remainder when $T_1+T_2+...+T_{2017}$ is divided by $2017$.
[i]Proposed by Vincent Huang
|
To solve the problem, we need to compute the number of $k$-tuples $(x_1, x_2, \ldots, x_k)$ such that $0 \le x_i < 73$ for each $i$ and $73 \mid x_1^2 + x_2^2 + \cdots + x_k^2 - 1$. We denote this number by $T_k$. We then need to find the remainder when $T_1 + T_2 + \cdots + T_{2017}$ is divided by 2017.
1. **Counting the number of solutions $T_k$:**
We start by considering the function $f(x_1, \ldots, x_k) = x_1^2 + x_2^2 + \cdots + x_k^2 - 1$. We need to count the number of solutions $(x_1, \ldots, x_k)$ in $\mathbb{Z}/73\mathbb{Z}^k$ that satisfy $f(x_1, \ldots, x_k) \equiv 0 \pmod{73}$.
2. **Using the roots of unity filter:**
Let $\zeta = e^{2\pi i / 73}$ be a primitive 73rd root of unity. The number of solutions $N$ can be expressed using the roots of unity filter:
\[
73 \cdot N = \sum_{x_1, \ldots, x_k} \sum_{j=0}^{72} \zeta^{j f(x_1, \ldots, x_k)}.
\]
This simplifies to:
\[
73^k + \sum_{j=1}^{72} \sum_{x_1, \ldots, x_k} \zeta^{j (x_1^2 + x_2^2 + \cdots + x_k^2 - 1)}.
\]
3. **Simplifying the inner sum:**
\[
\sum_{x_1, \ldots, x_k} \zeta^{j (x_1^2 + x_2^2 + \cdots + x_k^2 - 1)} = \zeta^{-j} \left( \sum_{x_1} \zeta^{j x_1^2} \right) \left( \sum_{x_2} \zeta^{j x_2^2} \right) \cdots \left( \sum_{x_k} \zeta^{j x_k^2} \right).
\]
Each sum $\sum_{x_i} \zeta^{j x_i^2}$ is a Gauss sum $G(j)$.
4. **Properties of Gauss sums:**
The Gauss sum $G(j)$ is defined as:
\[
G(j) = \sum_{x=0}^{72} \zeta^{j x^2}.
\]
If $j \neq 0$, $G(j) = \left( \frac{j}{73} \right) G$, where $G = G(1)$ and $\left( \frac{j}{73} \right)$ is the Legendre symbol. For $j = 0$, $G(0) = 73$.
5. **Combining the results:**
\[
N = 73^{k-1} + \frac{1}{73} \sum_{j=1}^{72} \zeta^{-j} G(j)^k.
\]
Using the properties of Gauss sums, we get:
\[
G(j)^k = \left( \frac{j}{73} \right)^k G^k.
\]
For even $k$, the sum $\sum_{j=1}^{72} \zeta^{-j} \left( \frac{j}{73} \right)^k$ evaluates to $-1$ if $73 \nmid 1$, and for odd $k$, it evaluates to $G$.
6. **Final expressions for $T_k$:**
For even $k$:
\[
T_k = 73^{k-1} - 73^{\frac{k-2}{2}} (-1)^{k/2}.
\]
For odd $k$:
\[
T_k = 73^{k-1} + 73^{\frac{k-1}{2}}.
\]
7. **Summing $T_1 + T_2 + \cdots + T_{2017}$:**
We need to compute the sum of these expressions for $k$ from 1 to 2017 and find the remainder when divided by 2017.
\[
\sum_{k=1}^{2017} T_k = \sum_{\text{odd } k} \left( 73^{k-1} + 73^{\frac{k-1}{2}} \right) + \sum_{\text{even } k} \left( 73^{k-1} - 73^{\frac{k-2}{2}} (-1)^{k/2} \right).
\]
8. **Modulo 2017:**
Since 2017 is a prime number, we can use properties of modular arithmetic to simplify the sum. We need to compute the sum modulo 2017.
The final answer is $\boxed{0}$
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Call a nonempty set $V$ of nonzero integers \emph{victorious} if there exists a polynomial $P(x)$ with integer coefficients such that $P(0)=330$ and that $P(v)=2|v|$ holds for all elements $v\in V$. Find the number of victorious sets.
[i]Proposed by Yannick Yao[/i]
|
To solve this problem, we need to determine the number of nonempty sets \( V \) of nonzero integers such that there exists a polynomial \( P(x) \) with integer coefficients satisfying \( P(0) = 330 \) and \( P(v) = 2|v| \) for all \( v \in V \).
### Case 1: \( V \) consists entirely of positive integers
By the Rational Root Theorem, any integer root must divide \( 330 \). Since everything in \( V \) is positive, we have \( P(v) - 2v = 0 \) for all \( v \in V \), and so \( \prod_{v \in V}(x - v) \mid P(v) - 2v \). It follows that \( \prod_{v \in V} v \mid 330 \) must be true.
We show that this is also sufficient; taking \( P(v) = Q(v) \prod_{v \in V}(x - v) + 2v \) suffices, where we choose \( Q(v) \) to be a polynomial whose constant term times \( \prod_{v \in V} v \) is \( 330 \). It suffices to count the number of sets \( V \) consisting of positive integers that multiply to a factor of \( 330 \).
Note that for each nonempty set that doesn't contain \( 1 \), we can add \( 1 \) to get another valid set. Thus, we will count the number of sets that don't contain \( 1 \).
#### Subcase 1: Product is \( 330 \)
- Possible sets: \((2, 3, 5, 11), (6, 5, 11), (30, 11), (330)\)
- Number of sets: \( 14 \)
#### Subcase 2: Product is \( 165, 110, 66, 30 \)
- Possible sets: \((2, 3, 5), (6, 5), (30)\)
- Number of sets: \( 20 \)
#### Subcase 3: Product is \( 6 \)
- Possible sets: \((2, 3), (6)\)
- Number of sets: \( 12 \)
#### Subcase 4: Product is \( 2 \)
- Possible sets: \((2)\)
- Number of sets: \( 4 \)
#### Subcase 5: Empty
- This isn't allowed, but we can add \( 1 \).
Adding subcases 1-4, multiplying by \( 2 \) (we can add 1) and adding subcase \( 5 \) yields \( 101 \).
### Case 2: \( V \) consists entirely of negative integers
This case is analogous to the positive integers case, giving us \( 101 \) sets.
### Case 3: \( V \) contains both positive and negative integers
Assume \( a > 0 > b \) and \( a, b \in V \). Then \( a - b \mid P(a) - P(b) = 2a + 2b \implies 2a + 2b = k(a - b) \implies \frac{a}{b} = \frac{k + 2}{k - 2} < 0 \). Thus \( k \in \{-1, 0, 1\} \) so we must have \( a = -3b, b = -3a, \) or \( a = -b \).
#### Subcase 1: \( V \) contains exactly 1 positive and 1 negative integer
##### Subsubcase 1: \( V = \{a, -a\} \) for some \( a \)
- Then \( P(x) = Q(x)(x - a)(x + a) + 2a \). If the constant term of \( Q \) is \( c \), we need \( -a^2c + 2a = 330 \implies a(-ac + 2) = 330 \). Checking factors of \( 330 \) gives \( a = 1, 3 \implies \{1, -1\}, \{3, -3\} \).
##### Subsubcase 2: \( V = \{a, -3a\} \) for some \( a \)
- Then \( P(x) = Q(x)(x - a)(x + 3a) - x + 3a \), and similarly we get \( -3a^2c + 3a = 330 \implies a(-ac + 1) = 110 \implies a = 1, 2, 10, -1, -2, -11 \). This gives \( \{1, -3\}, \{2, -6\}, \{10, -30\}, \{-1, 3\}, \{-2, 6\}, \{-11, 33\} \).
#### Subcase 2: \( V \) contains two positive integers and one negative integer
- Then \( V \) must be of the form \( \{a, 3a, -a\} \) for some \( a > 0 \). But letting \( P(x) = Q(x)(x - a)(x + 3a)(x + a) + R(x) \), clearly \( \text{deg}(R) \geq 2 \). But \( R(3a) - R(a) = 4a = (3^{\text{deg}(R)} - 1)a^2 \) which means that \( a \) is not an integer, contradiction.
Since every set \( V \) with positive and negative integers are covered in case 1 or is a superset of case 2 (and if \( V \) doesn't work, clearly any superset of \( V \) doesn't work). Thus this case yields \( 8 \) sets.
Adding the two cases gives us \( 210 \) victorious sets.
The final answer is \(\boxed{210}\)
|
210
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ABC$ be a triangle with $AB=7, AC=9, BC=10$, circumcenter $O$, circumradius $R$, and circumcircle $\omega$. Let the tangents to $\omega$ at $B,C$ meet at $X$. A variable line $\ell$ passes through $O$. Let $A_1$ be the projection of $X$ onto $\ell$ and $A_2$ be the reflection of $A_1$ over $O$. Suppose that there exist two points $Y,Z$ on $\ell$ such that $\angle YAB+\angle YBC+\angle YCA=\angle ZAB+\angle ZBC+\angle ZCA=90^{\circ}$, where all angles are directed, and furthermore that $O$ lies inside segment $YZ$ with $OY*OZ=R^2$. Then there are several possible values for the sine of the angle at which the angle bisector of $\angle AA_2O$ meets $BC$. If the product of these values can be expressed in the form $\frac{a\sqrt{b}}{c}$ for positive integers $a,b,c$ with $b$ squarefree and $a,c$ coprime, determine $a+b+c$.
[i]Proposed by Vincent Huang
|
1. **Scaling and Complex Numbers:**
We start by scaling the triangle such that the circumcircle \((ABC)\) is the unit circle. This means that the circumradius \(R = 1\). Let the complex numbers corresponding to points \(A\), \(B\), and \(C\) be \(a\), \(b\), and \(c\) respectively. The line \(\ell\) is the real axis.
2. **Angle Condition:**
We need to find the complex numbers \(x\) on the real axis that satisfy the angle condition \(\angle YAB + \angle YBC + \angle YCA = 90^\circ\). This translates to the condition that \(\frac{(x-a)(x-b)(x-c)}{(b-a)(c-b)(a-c)}\) is purely imaginary. Therefore, its conjugate equals its negation:
\[
\frac{(x-a)(x-b)(x-c)}{(b-a)(c-b)(a-c)} = \frac{(\overline{x}-\frac{1}{a})(\overline{x}-\frac{1}{b})(\overline{x}-\frac{1}{c})}{\frac{b-a}{ab}\frac{c-b}{bc}\frac{a-c}{ac}} = abc \frac{(a\overline{x}-1)(b\overline{x}-1)(c\overline{x}-1)}{(b-a)(c-b)(a-c)}
\]
Since \(x\) is real, \(x = \overline{x}\), so this becomes:
\[
(x-a)(x-b)(x-c) = abc(ax-1)(bx-1)(cx-1)
\]
Expanding both sides, we get:
\[
x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc = a^2b^2c^2x^3 - (a^2b^2c + a^2bc^2 + ab^2c^2)x^2 + (abc^2 + ab^2c + a^2bc)x - abc
\]
Noting the \(x=0\) solution, we find the product of the other two roots to be:
\[
\frac{abc^2 + ab^2c + a^2bc - ab - ac - bc}{a^2b^2c^2 - 1}
\]
Since these two roots should be on opposite sides and have magnitude \(R^2 = 1\), we have:
\[
\frac{abc^2 + ab^2c + a^2bc - ab - ac - bc}{a^2b^2c^2 - 1} = -1
\]
Simplifying, we get:
\[
a^2b^2c^2 + abc^2 + ab^2c + a^2bc - ab - ac - bc - 1 = 0
\]
Moving terms, we have:
\[
1 + ab + ac + bc = (abc)^2 \left(1 + \frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc}\right)
\]
This implies:
\[
\frac{(1 + ab + ac + bc)}{\overline{(1 + ab + ac + bc)}} = \frac{abc}{\overline{abc}}
\]
Therefore, \(abc\) and \(1 + ab + ac + bc\) point in the same or opposite directions.
3. **Angle Bisector and Intersection:**
The angle we want (call it \(x\)) is:
\[
x = \frac{m \angle {A A_2 O}}{2} - m\angle Q
\]
where \(\angle Q\) is the intersection of \(BC\) and \(\ell\). In complex terms, this becomes:
\[
2x = \frac{\angle{AA_2O}}{\angle Q}^2 = \arg\left(\frac{1 + ab + bc + ac}{(b+c)(b-c)^2}\right)
\]
Since \(1 + ab + ac + bc\) is in the same or opposite direction as \(abc\), we get:
\[
2x = \arg\left(\frac{\pm abc}{(b+c)(b-c)^2}\right)
\]
Note that \(b+c\) and \(b-c\) are perpendicular, so:
\[
\arg((b-c)^2) = - \arg((b+c)^2)
\]
Also, \(\frac{bc}{b+c}\) is real, so:
\[
2x = \arg\left(\frac{\pm abc}{(b+c)(b+c)^2}\right) = \arg\left(\frac{\mp a}{b+c}\right) = \pm [\arg(a) - \arg(b+c)]
\]
4. **Finding the Product of Sines:**
The two angles \(k\) that satisfy this argument condition have a sum of \(180^\circ\), so the angles that satisfy for \(x\) will have a sum of \(90^\circ\). Therefore, their sines will be \(\sin\) and \(\cos\) of \(\frac{k}{2}\). We want to find the product of these sines:
\[
\text{Ans} = \sin\left(\frac{k}{2}\right) \cos\left(\frac{k}{2}\right) = \frac{\sin(k)}{2}
\]
This means we wish to find:
\[
\frac{1}{2} \sin(\arg(a) - \arg(b+c))
\]
Defining \(M\) as the midpoint of \(BC\), we get in complex terms:
\[
m = \frac{b+c}{2} \rightarrow \arg(m) = \arg(b+c)
\]
Therefore, we want to find:
\[
\frac{1}{2} \sin(\arg(a) - \arg(b+c)) = \frac{1}{2} \sin(\arg(a) - \arg(m)) = \frac{\sin(\angle{AOM})}{2}
\]
Since \(\angle{AOM}\) is concave, we get:
\[
\sin(\angle{AOM}) = -\frac{\sin(\angle{AOC} + \angle{MOC})}{2}
\]
5. **Calculating the Sine Values:**
Using Heron's formula, the area of \(\triangle ABC\) is:
\[
K = \sqrt{13(13-7)(13-9)(13-10)} = 6\sqrt{26}
\]
The circumradius is:
\[
R = \frac{abc}{4K} = \frac{630}{24\sqrt{26}} = \frac{105}{4\sqrt{26}}
\]
Using the Law of Cosines on \(\angle AOC\):
\[
(2 - 2\cos(\angle{AOC}))R^2 = 9^2 \rightarrow 2 - 2\cos(\angle{AOC}) = \frac{81}{R^2} = \frac{3744}{1225}
\]
Therefore:
\[
2\cos(\angle{AOC}) = 2 - \frac{3744}{1225} = -\frac{1294}{1225} \rightarrow \cos(\angle{AOC}) = -\frac{647}{1225}
\]
Since \(\sin^2 + \cos^2 = 1\) and \(\angle{AOC}\) is convex, we get:
\[
\sin(\angle{AOC}) = \frac{204\sqrt{26}}{1225}
\]
To find \(\sin(\angle{MOC})\), since \(OM \perp BC\), we have:
\[
\sin(\angle{MOC}) = \frac{MC}{OC} = \frac{5}{R} = \frac{4\sqrt{26}}{21}
\]
Using \(\sin^2 + \cos^2 = 1\) and the fact that \(\angle{MOC}\) is acute, we get:
\[
\cos(\angle{MOC}) = \frac{5}{21}
\]
Therefore:
\[
\sin(\angle{AOC} + \angle{MOC}) = \sin(\angle{AOC})\cos(\angle{MOC}) + \cos(\angle{AOC})\sin(\angle{MOC}) = \frac{1020\sqrt{26} - 2588\sqrt{26}}{21 \cdot 1225} = \frac{-32\sqrt{26}}{525}
\]
Our answer is:
\[
\frac{16\sqrt{26}}{525} \rightarrow 16 + 26 + 525 = \boxed{567}
\]
|
567
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Define a sequence of polynomials $P_0,P_1,...$ by the recurrence $P_0(x)=1, P_1(x)=x, P_{n+1}(x) = 2xP_n(x)-P_{n-1}(x)$. Let $S=\left|P_{2017}'\left(\frac{i}{2}\right)\right|$ and $T=\left|P_{17}'\left(\frac{i}{2}\right)\right|$, where $i$ is the imaginary unit. Then $\frac{S}{T}$ is a rational number with fractional part $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m$.
[i]Proposed by Tristan Shin[/i]
|
1. **Define the sequence of polynomials**:
The sequence of polynomials \( P_n(x) \) is defined by the recurrence relation:
\[
P_0(x) = 1, \quad P_1(x) = x, \quad P_{n+1}(x) = 2xP_n(x) - P_{n-1}(x)
\]
2. **Transform the polynomials**:
Let \( P_n(ix) = i^n Q_n(x) \). Then we have:
\[
Q_0(x) = 1, \quad Q_1(x) = x
\]
and the recurrence relation becomes:
\[
Q_{n+1}(x) = 2xQ_n(x) + Q_{n-1}(x)
\]
3. **Evaluate \( Q_n \) at \( x = \frac{1}{2} \)**:
\[
Q_0\left(\frac{1}{2}\right) = 1, \quad Q_1\left(\frac{1}{2}\right) = \frac{1}{2}
\]
Using the recurrence relation:
\[
Q_{n+1}\left(\frac{1}{2}\right) = Q_n\left(\frac{1}{2}\right) + Q_{n-1}\left(\frac{1}{2}\right)
\]
This is the recurrence relation for the Lucas numbers \( L_n \), so:
\[
Q_n\left(\frac{1}{2}\right) = \frac{1}{2} L_n
\]
4. **Differentiate the recurrence relation**:
\[
Q'_0(x) = 0, \quad Q'_1(x) = 1
\]
and:
\[
Q'_{n+1}(x) = 2xQ'_n(x) + Q'_{n-1}(x) + 2Q_n(x)
\]
5. **Evaluate the derivative at \( x = \frac{1}{2} \)**:
\[
Q'_0\left(\frac{1}{2}\right) = 0, \quad Q'_1\left(\frac{1}{2}\right) = 1
\]
Using the recurrence relation:
\[
Q'_{n+1}\left(\frac{1}{2}\right) = Q'_n\left(\frac{1}{2}\right) + Q'_{n-1}\left(\frac{1}{2}\right) + L_n
\]
6. **Relate to Fibonacci numbers**:
Since \( (n+1)F_{n+1} = nF_n + (n-1)F_{n-1} + L_n \) (where \( F_n \) is the Fibonacci sequence), we deduce:
\[
Q'_{n+1}\left(\frac{1}{2}\right) - (n+1)F_{n+1} = (Q'_n\left(\frac{1}{2}\right) - nF_n) + (Q'_{n-1}\left(\frac{1}{2}\right) - (n-1)F_{n-1})
\]
Since \( Q'_n\left(\frac{1}{2}\right) - nF_n = 0 \) for \( n = 0, 1 \), we have:
\[
Q'_n\left(\frac{1}{2}\right) = nF_n
\]
7. **Calculate \( S \) and \( T \)**:
\[
S = 2017F_{2017}, \quad T = 17F_{17}
\]
8. **Compute the fractional part**:
\[
\frac{S}{T} = \frac{2017F_{2017}}{17F_{17}} = \frac{2017}{17} \cdot \frac{F_{2017}}{F_{17}} = 118.64705882352941 \cdot \frac{F_{2017}}{F_{17}}
\]
The fractional part of \( \frac{S}{T} \) is:
\[
\frac{4142}{27149}
\]
The final answer is \( \boxed{4142} \).
|
4142
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $p = 2017$. If $A$ is an $n\times n$ matrix composed of residues $\pmod{p}$ such that $\det A\not\equiv 0\pmod{p}$ then let $\text{ord}(A)$ be the minimum integer $d > 0$ such that $A^d\equiv I\pmod{p}$, where $I$ is the $n\times n$ identity matrix. Let the maximum such order be $a_n$ for every positive integer $n$. Compute the sum of the digits when $\sum_{k = 1}^{p + 1} a_k$ is expressed in base $p$.
[i]Proposed by Ashwin Sah[/i]
|
1. **Understanding the Problem:**
We are given a prime \( p = 2017 \) and an \( n \times n \) matrix \( A \) with entries in \( \mathbb{F}_p \) (the field of integers modulo \( p \)). The matrix \( A \) is invertible, i.e., \( \det A \not\equiv 0 \pmod{p} \). We need to find the maximum order \( a_n \) of such a matrix \( A \), where the order \( \text{ord}(A) \) is the smallest positive integer \( d \) such that \( A^d \equiv I \pmod{p} \). Finally, we need to compute the sum of the digits of \( \sum_{k=1}^{p+1} a_k \) when expressed in base \( p \).
2. **Finding the Maximum Order \( a_n \):**
- Let \( d \) be the order of \( A \), i.e., \( A^d \equiv I \pmod{p} \).
- The minimal polynomial \( f(x) \) of \( A \) divides \( x^d - 1 \). Since \( A \) is invertible, \( f(x) \) has no repeated roots.
- The characteristic polynomial of \( A \) is a product of irreducible polynomials over \( \mathbb{F}_p \). The roots of these polynomials are eigenvalues of \( A \) in some extension field \( \mathbb{F}_{p^k} \).
- The order of any eigenvalue \( \lambda \) of \( A \) must divide \( p^k - 1 \), where \( k \) is the degree of the minimal polynomial of \( \lambda \).
3. **Maximum Possible Order:**
- The maximum order \( d \) of \( A \) is achieved when the characteristic polynomial of \( A \) has roots that are primitive elements of \( \mathbb{F}_{p^n} \). The order of such an element is \( p^n - 1 \).
- Therefore, the maximum order \( a_n \) is \( p^n - 1 \).
4. **Summing the Orders:**
- We need to compute \( \sum_{k=1}^{p+1} a_k \).
- Since \( a_k = p^k - 1 \), we have:
\[
\sum_{k=1}^{p+1} a_k = \sum_{k=1}^{p+1} (p^k - 1) = \sum_{k=1}^{p+1} p^k - \sum_{k=1}^{p+1} 1
\]
- The first sum is a geometric series:
\[
\sum_{k=1}^{p+1} p^k = p + p^2 + p^3 + \cdots + p^{p+1} = p \frac{p^{p+1} - 1}{p - 1}
\]
- The second sum is simply \( p + 1 \):
\[
\sum_{k=1}^{p+1} 1 = p + 1
\]
- Therefore:
\[
\sum_{k=1}^{p+1} a_k = p \frac{p^{p+1} - 1}{p - 1} - (p + 1)
\]
5. **Expressing in Base \( p \):**
- The expression \( p \frac{p^{p+1} - 1}{p - 1} - (p + 1) \) simplifies to:
\[
\frac{p^{p+2} - p}{p - 1} - (p + 1)
\]
- This is a large number, but when expressed in base \( p \), it simplifies to a series of digits. The sum of the digits of \( p^{p+1} - 1 \) in base \( p \) is \( p \) times the sum of the digits of \( p^{p+1} - 1 \) in base \( p \), which is \( p \).
The final answer is \( \boxed{2015} \)
|
2015
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For any real numbers $x,y$ that satisfies the equation $$x+y-xy=155$$ and $$x^2+y^2=325$$, Find $|x^3-y^3|$
|
1. Given the equations:
\[
x + y - xy = 155
\]
and
\[
x^2 + y^2 = 325
\]
2. From the first equation, we can express \(x + y\) in terms of \(xy\):
\[
x + y = 155 + xy
\]
3. From the second equation, we use the identity for the sum of squares:
\[
x^2 + y^2 = (x + y)^2 - 2xy
\]
Substituting the given value:
\[
325 = (x + y)^2 - 2xy
\]
4. Let \(xy = t\). Then:
\[
x + y = 155 + t
\]
Substituting into the sum of squares equation:
\[
325 = (155 + t)^2 - 2t
\]
5. Expanding and simplifying:
\[
325 = 24025 + 310t + t^2 - 2t
\]
\[
325 = 24025 + 308t + t^2
\]
\[
t^2 + 308t + 23700 = 0
\]
6. Solving the quadratic equation using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[
t = \frac{-308 \pm \sqrt{308^2 - 4 \cdot 1 \cdot 23700}}{2 \cdot 1}
\]
\[
t = \frac{-308 \pm \sqrt{94864 - 94800}}{2}
\]
\[
t = \frac{-308 \pm \sqrt{64}}{2}
\]
\[
t = \frac{-308 \pm 8}{2}
\]
\[
t = \frac{-300}{2} = -150 \quad \text{or} \quad t = \frac{-316}{2} = -158
\]
7. Case 1: \(t = -150\)
\[
x + y = 155 + t = 155 - 150 = 5
\]
Solving for \(x\) and \(y\) using the quadratic equation \(t = xy\):
\[
x^2 - (x+y)x + xy = 0
\]
\[
x^2 - 5x - 150 = 0
\]
Solving this quadratic equation:
\[
x = \frac{5 \pm \sqrt{25 + 600}}{2}
\]
\[
x = \frac{5 \pm 25}{2}
\]
\[
x = 15 \quad \text{or} \quad x = -10
\]
Thus, \((x, y) = (15, -10)\) or \((x, y) = (-10, 15)\).
8. Calculating \(|x^3 - y^3|\):
\[
|x^3 - y^3| = |15^3 - (-10)^3| = |3375 + 1000| = 4375
\]
9. Case 2: \(t = -158\)
\[
x + y = 155 + t = 155 - 158 = -3
\]
Solving for \(x\) and \(y\) using the quadratic equation \(t = xy\):
\[
x^2 - (x+y)x + xy = 0
\]
\[
x^2 + 3x - 158 = 0
\]
Solving this quadratic equation:
\[
x = \frac{-3 \pm \sqrt{9 + 632}}{2}
\]
\[
x = \frac{-3 \pm \sqrt{641}}{2}
\]
Since \(\sqrt{641}\) is not an integer, \(x\) and \(y\) are not real numbers.
Therefore, the only valid solution is from Case 1.
The final answer is \(\boxed{4375}\)
|
4375
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
What is the last two digits of the number $(11^2 + 15^2 + 19^2 + ... + 2007^2)^2$?
|
1. We start by expressing the given sum in a more manageable form. The sequence of numbers is \(11, 15, 19, \ldots, 2007\). This sequence is an arithmetic sequence with the first term \(a = 11\) and common difference \(d = 4\). The general term of the sequence can be written as:
\[
a_n = 11 + (n-1) \cdot 4 = 4n + 7
\]
Therefore, the sum we need to evaluate is:
\[
\left( \sum_{n=1}^{500} (4n + 7)^2 \right)^2
\]
2. Next, we expand the square inside the sum:
\[
(4n + 7)^2 = 16n^2 + 56n + 49
\]
Thus, the sum becomes:
\[
\sum_{n=1}^{500} (4n + 7)^2 = \sum_{n=1}^{500} (16n^2 + 56n + 49)
\]
3. We can split the sum into three separate sums:
\[
\sum_{n=1}^{500} (16n^2 + 56n + 49) = 16 \sum_{n=1}^{500} n^2 + 56 \sum_{n=1}^{500} n + 49 \sum_{n=1}^{500} 1
\]
4. We use the formulas for the sum of the first \(n\) natural numbers and the sum of the squares of the first \(n\) natural numbers:
\[
\sum_{n=1}^{500} n = \frac{500(500+1)}{2} = 125250
\]
\[
\sum_{n=1}^{500} n^2 = \frac{500(500+1)(2 \cdot 500 + 1)}{6} = 41791750
\]
\[
\sum_{n=1}^{500} 1 = 500
\]
5. Substituting these values back into the sum, we get:
\[
16 \sum_{n=1}^{500} n^2 + 56 \sum_{n=1}^{500} n + 49 \sum_{n=1}^{500} 1 = 16 \cdot 41791750 + 56 \cdot 125250 + 49 \cdot 500
\]
6. We now calculate each term modulo 100:
\[
16 \cdot 41791750 \equiv 16 \cdot 50 \equiv 800 \equiv 0 \pmod{100}
\]
\[
56 \cdot 125250 \equiv 56 \cdot 50 \equiv 2800 \equiv 0 \pmod{100}
\]
\[
49 \cdot 500 \equiv 49 \cdot 0 \equiv 0 \pmod{100}
\]
7. Adding these results modulo 100:
\[
0 + 0 + 0 \equiv 0 \pmod{100}
\]
8. Finally, we square the result:
\[
(0)^2 \equiv 0 \pmod{100}
\]
The final answer is \(\boxed{0}\).
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[help me] Let m and n denote the number of digits in $2^{2007}$ and $5^{2007}$ when expressed in base 10. What is the sum m + n?
|
1. To determine the number of digits in \(2^{2007}\) and \(5^{2007}\), we need to use the formula for the number of digits of a number \(n\) in base 10, which is given by:
\[
d(n) = \lfloor \log_{10} n \rfloor + 1
\]
2. First, we calculate the number of digits in \(2^{2007}\):
\[
d(2^{2007}) = \lfloor \log_{10} (2^{2007}) \rfloor + 1 = \lfloor 2007 \log_{10} 2 \rfloor + 1
\]
Using the approximation \(\log_{10} 2 \approx 0.3010\):
\[
2007 \log_{10} 2 \approx 2007 \times 0.3010 = 603.107
\]
Therefore:
\[
d(2^{2007}) = \lfloor 603.107 \rfloor + 1 = 603 + 1 = 604
\]
3. Next, we calculate the number of digits in \(5^{2007}\):
\[
d(5^{2007}) = \lfloor \log_{10} (5^{2007}) \rfloor + 1 = \lfloor 2007 \log_{10} 5 \rfloor + 1
\]
Using the approximation \(\log_{10} 5 \approx 0.6990\):
\[
2007 \log_{10} 5 \approx 2007 \times 0.6990 = 1403.893
\]
Therefore:
\[
d(5^{2007}) = \lfloor 1403.893 \rfloor + 1 = 1403 + 1 = 1404
\]
4. To find the sum \(m + n\), where \(m\) is the number of digits in \(2^{2007}\) and \(n\) is the number of digits in \(5^{2007}\), we add the results from steps 2 and 3:
\[
m + n = 604 + 1404 = 2008
\]
The final answer is \(\boxed{2008}\).
|
2008
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the maximum value of $M =\frac{x}{2x + y} +\frac{y}{2y + z}+\frac{z}{2z + x}$ , $x,y, z > 0$
|
To find the maximum value of \( M = \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \) for \( x, y, z > 0 \), we will use the method of inequalities.
1. **Initial Setup:**
We need to show that:
\[
\frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq 1
\]
2. **Using the AM-GM Inequality:**
Consider the following application of the AM-GM (Arithmetic Mean-Geometric Mean) inequality:
\[
\frac{x}{2x + y} \leq \frac{x}{2x} = \frac{1}{2}
\]
Similarly,
\[
\frac{y}{2y + z} \leq \frac{y}{2y} = \frac{1}{2}
\]
and
\[
\frac{z}{2z + x} \leq \frac{z}{2z} = \frac{1}{2}
\]
3. **Summing the Inequalities:**
Adding these inequalities, we get:
\[
\frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}
\]
However, this approach does not directly help us to prove the desired inequality. We need a different approach.
4. **Using the Titu's Lemma (Cauchy-Schwarz in Engel Form):**
Titu's Lemma states that for non-negative real numbers \(a_1, a_2, \ldots, a_n\) and \(b_1, b_2, \ldots, b_n\),
\[
\frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \cdots + \frac{a_n^2}{b_n} \geq \frac{(a_1 + a_2 + \cdots + a_n)^2}{b_1 + b_2 + \cdots + b_n}
\]
Applying Titu's Lemma to our problem, we set \(a_1 = \sqrt{x}\), \(a_2 = \sqrt{y}\), \(a_3 = \sqrt{z}\), \(b_1 = 2x + y\), \(b_2 = 2y + z\), \(b_3 = 2z + x\):
\[
\frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \geq \frac{(x + y + z)^2}{2x + y + 2y + z + 2z + x} = \frac{(x + y + z)^2}{3(x + y + z)} = \frac{x + y + z}{3}
\]
5. **Conclusion:**
Since \( \frac{x + y + z}{3} \leq 1 \) for all positive \(x, y, z\), we have:
\[
\frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq 1
\]
Therefore, the maximum value of \( M \) is \(1\).
The final answer is \(\boxed{1}\).
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Determine all positive integer $a$ such that the equation $2x^2 - 30x + a = 0$ has two prime roots, i.e. both roots are prime numbers.
|
1. Let \( p_1 \) and \( p_2 \) (with \( p_1 \leq p_2 \)) be the prime roots of the quadratic equation \( 2x^2 - 30x + a = 0 \). By Vieta's formulas, we know:
\[
p_1 + p_2 = \frac{30}{2} = 15
\]
and
\[
p_1 p_2 = \frac{a}{2}
\]
2. Since \( p_1 \) and \( p_2 \) are prime numbers and their sum is 15, we need to find two prime numbers that add up to 15. The only pair of prime numbers that satisfy this condition are \( p_1 = 2 \) and \( p_2 = 13 \) (since 2 is the only even prime number and 13 is the only prime number that, when added to 2, gives 15).
3. Substituting \( p_1 = 2 \) and \( p_2 = 13 \) into the product equation:
\[
p_1 p_2 = 2 \times 13 = 26
\]
Therefore,
\[
\frac{a}{2} = 26 \implies a = 52
\]
4. We verify that \( a = 52 \) satisfies the original equation:
\[
2x^2 - 30x + 52 = 0
\]
The roots of this equation are indeed \( x = 2 \) and \( x = 13 \), both of which are prime numbers.
Thus, the only positive integer \( a \) such that the equation \( 2x^2 - 30x + a = 0 \) has two prime roots is \( a = 52 \).
The final answer is \( \boxed{52} \).
|
52
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of integer $n$ from the set $\{2000,2001,...,2010\}$ such that $2^{2n} + 2^n + 5$ is divisible by $7$
(A): $0$, (B): $1$, (C): $2$, (D): $3$, (E) None of the above.
|
To solve the problem, we need to determine the number of integers \( n \) from the set \(\{2000, 2001, \ldots, 2010\}\) such that \(2^{2n} + 2^n + 5\) is divisible by 7.
1. **Identify the periodicity of \(2^n \mod 7\):**
\[
\begin{aligned}
2^1 &\equiv 2 \pmod{7}, \\
2^2 &\equiv 4 \pmod{7}, \\
2^3 &\equiv 8 \equiv 1 \pmod{7}.
\end{aligned}
\]
Since \(2^3 \equiv 1 \pmod{7}\), the powers of 2 modulo 7 repeat every 3 steps. Therefore, \(2^n \mod 7\) has a cycle of length 3: \(\{2, 4, 1\}\).
2. **Express \(n\) in terms of modulo 3:**
Since the powers of 2 repeat every 3 steps, we can write \(n\) as \(n = 3k + r\) where \(r \in \{0, 1, 2\}\).
3. **Evaluate \(2^{2n} + 2^n + 5 \mod 7\) for \(r = 0, 1, 2\):**
- For \(r = 0\):
\[
n = 3k \implies 2^n \equiv 1 \pmod{7}, \quad 2^{2n} \equiv 1 \pmod{7}
\]
\[
2^{2n} + 2^n + 5 \equiv 1 + 1 + 5 \equiv 7 \equiv 0 \pmod{7}
\]
- For \(r = 1\):
\[
n = 3k + 1 \implies 2^n \equiv 2 \pmod{7}, \quad 2^{2n} \equiv 2^2 \equiv 4 \pmod{7}
\]
\[
2^{2n} + 2^n + 5 \equiv 4 + 2 + 5 \equiv 11 \equiv 4 \pmod{7}
\]
- For \(r = 2\):
\[
n = 3k + 2 \implies 2^n \equiv 4 \pmod{7}, \quad 2^{2n} \equiv 4^2 \equiv 16 \equiv 2 \pmod{7}
\]
\[
2^{2n} + 2^n + 5 \equiv 2 + 4 + 5 \equiv 11 \equiv 4 \pmod{7}
\]
4. **Count the number of \(n\) in the set \(\{2000, 2001, \ldots, 2010\}\) that satisfy the condition:**
- From the above calculations, \(2^{2n} + 2^n + 5 \equiv 0 \pmod{7}\) only when \(n \equiv 0 \pmod{3}\).
- The set \(\{2000, 2001, \ldots, 2010\}\) contains 11 numbers. We need to find how many of these are divisible by 3.
- The sequence of numbers divisible by 3 in this range is \(\{2001, 2004, 2007, 2010\}\).
5. **Count the numbers:**
There are 4 numbers in the set \(\{2000, 2001, \ldots, 2010\}\) that are divisible by 3.
The final answer is \(\boxed{4}\).
|
4
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many real numbers $a \in (1,9)$ such that the corresponding number $a- \frac1a$ is an integer?
(A): $0$, (B): $1$, (C): $8$, (D): $9$, (E) None of the above.
|
1. Let \( k = a - \frac{1}{a} \). We need to find the values of \( a \) such that \( k \) is an integer and \( a \in (1, 9) \).
2. Rewrite the equation:
\[
k = a - \frac{1}{a} \implies k = \frac{a^2 - 1}{a}
\]
This implies:
\[
a^2 - ka - 1 = 0
\]
3. Solve the quadratic equation for \( a \):
\[
a = \frac{k \pm \sqrt{k^2 + 4}}{2}
\]
Since \( a > 0 \), we take the positive root:
\[
a = \frac{k + \sqrt{k^2 + 4}}{2}
\]
4. We need \( a \in (1, 9) \). Therefore:
\[
1 < \frac{k + \sqrt{k^2 + 4}}{2} < 9
\]
5. Multiply the inequality by 2:
\[
2 < k + \sqrt{k^2 + 4} < 18
\]
6. Subtract \( k \) from all parts of the inequality:
\[
2 - k < \sqrt{k^2 + 4} < 18 - k
\]
7. Consider the left part of the inequality:
\[
2 - k < \sqrt{k^2 + 4}
\]
Square both sides:
\[
(2 - k)^2 < k^2 + 4
\]
Simplify:
\[
4 - 4k + k^2 < k^2 + 4
\]
\[
4 - 4k < 4
\]
\[
-4k < 0
\]
\[
k > 0
\]
8. Consider the right part of the inequality:
\[
\sqrt{k^2 + 4} < 18 - k
\]
Square both sides:
\[
k^2 + 4 < (18 - k)^2
\]
Simplify:
\[
k^2 + 4 < 324 - 36k + k^2
\]
\[
4 < 324 - 36k
\]
\[
36k < 320
\]
\[
k < \frac{320}{36} \approx 8.89
\]
9. Since \( k \) must be an integer, we have:
\[
1 \leq k \leq 8
\]
10. For each integer value of \( k \) from 1 to 8, there is a corresponding \( a \) in the interval \( (1, 9) \). Therefore, there are 8 such values of \( a \).
The final answer is \( \boxed{8} \)
|
8
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many positive integers a less than $100$ such that $4a^2 + 3a + 5$ is divisible by $6$.
|
1. Define the function \( f(a) = 4a^2 + 3a + 5 \).
2. We need to find the values of \( a \) such that \( f(a) \) is divisible by \( 6 \). This means \( f(a) \equiv 0 \pmod{6} \).
3. We will check the values of \( f(a) \) for \( a \) modulo \( 6 \).
- If \( a \equiv 0 \pmod{6} \):
\[
f(a) = 4(0)^2 + 3(0) + 5 \equiv 5 \pmod{6}
\]
Thus, \( f(a) \not\equiv 0 \pmod{6} \).
- If \( a \equiv 1 \pmod{6} \):
\[
f(a) = 4(1)^2 + 3(1) + 5 = 4 + 3 + 5 = 12 \equiv 0 \pmod{6}
\]
Thus, \( f(a) \equiv 0 \pmod{6} \).
- If \( a \equiv 2 \pmod{6} \):
\[
f(a) = 4(2)^2 + 3(2) + 5 = 16 + 6 + 5 = 27 \equiv 3 \pmod{6}
\]
Thus, \( f(a) \not\equiv 0 \pmod{6} \).
- If \( a \equiv 3 \pmod{6} \):
\[
f(a) = 4(3)^2 + 3(3) + 5 = 36 + 9 + 5 = 50 \equiv 2 \pmod{6}
\]
Thus, \( f(a) \not\equiv 0 \pmod{6} \).
- If \( a \equiv 4 \pmod{6} \):
\[
f(a) = 4(4)^2 + 3(4) + 5 = 64 + 12 + 5 = 81 \equiv 3 \pmod{6}
\]
Thus, \( f(a) \not\equiv 0 \pmod{6} \).
- If \( a \equiv 5 \pmod{6} \):
\[
f(a) = 4(5)^2 + 3(5) + 5 = 100 + 15 + 5 = 120 \equiv 0 \pmod{6}
\]
Thus, \( f(a) \equiv 0 \pmod{6} \).
4. From the above calculations, \( f(a) \equiv 0 \pmod{6} \) when \( a \equiv 1 \pmod{6} \) or \( a \equiv 5 \pmod{6} \).
5. We need to count the number of positive integers \( a \) less than \( 100 \) that satisfy \( a \equiv 1 \pmod{6} \) or \( a \equiv 5 \pmod{6} \).
- For \( a \equiv 1 \pmod{6} \):
The sequence is \( 1, 7, 13, \ldots, 97 \). This is an arithmetic sequence with the first term \( a_1 = 1 \) and common difference \( d = 6 \).
The \( n \)-th term of the sequence is given by:
\[
a_n = 1 + (n-1) \cdot 6 = 6n - 5
\]
Setting \( 6n - 5 < 100 \):
\[
6n < 105 \implies n < 17.5
\]
Thus, \( n \) can be \( 1, 2, \ldots, 16 \), giving us \( 16 \) terms.
- For \( a \equiv 5 \pmod{6} \):
The sequence is \( 5, 11, 17, \ldots, 95 \). This is an arithmetic sequence with the first term \( a_1 = 5 \) and common difference \( d = 6 \).
The \( n \)-th term of the sequence is given by:
\[
a_n = 5 + (n-1) \cdot 6 = 6n - 1
\]
Setting \( 6n - 1 < 100 \):
\[
6n < 101 \implies n < 16.83
\]
Thus, \( n \) can be \( 1, 2, \ldots, 16 \), giving us \( 16 \) terms.
6. Adding the number of terms from both sequences, we get:
\[
16 + 16 = 32
\]
The final answer is \(\boxed{32}\).
|
32
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a, b, c$ be positive integers such that $a + 2b +3c = 100$.
Find the greatest value of $M = abc$
|
1. We start with the given equation:
\[
a + 2b + 3c = 100
\]
where \(a\), \(b\), and \(c\) are positive integers.
2. We aim to maximize the product \(M = abc\). To do this, we can use the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality), which states:
\[
\frac{a + 2b + 3c}{3} \geq \sqrt[3]{a \cdot 2b \cdot 3c}
\]
Substituting \(a + 2b + 3c = 100\) into the inequality, we get:
\[
\frac{100}{3} \geq \sqrt[3]{6abc}
\]
Simplifying, we find:
\[
\frac{100}{3} \geq \sqrt[3]{6abc} \implies \left(\frac{100}{3}\right)^3 \geq 6abc
\]
\[
\left(\frac{100}{3}\right)^3 = \frac{1000000}{27} \implies 6abc \leq \frac{1000000}{27}
\]
\[
abc \leq \frac{1000000}{162} \approx 6172.8395
\]
3. Since \(abc\) must be an integer, the maximum possible value of \(abc\) is less than or equal to 6172. We need to check if this value can be achieved with integer values of \(a\), \(b\), and \(c\).
4. We test the values \(a = 33\), \(b = 17\), and \(c = 11\):
\[
a + 2b + 3c = 33 + 2(17) + 3(11) = 33 + 34 + 33 = 100
\]
\[
abc = 33 \cdot 17 \cdot 11 = 6171
\]
5. We verify that 6171 is the maximum possible value by considering the prime factorization of 6172:
\[
6172 = 2^2 \cdot 1543
\]
Since 1543 is a prime number, it is not possible to achieve \(abc = 6172\) with positive integers \(a\), \(b\), and \(c\).
6. Therefore, the maximum possible value of \(abc\) is:
\[
\boxed{6171}
\]
|
6171
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
An integer is called "octal" if it is divisible by $8$ or if at least one of its digits is $8$.
How many integers between $1$ and $100$ are octal?
(A): $22$, (B): $24$, (C): $27$, (D): $30$, (E): $33$
|
1. **Identify numbers with a digit of $8$ between $1$ and $100$:**
- The numbers are: $8, 18, 28, 38, 48, 58, 68, 78, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 98$.
- There are $19$ such numbers.
2. **Identify numbers divisible by $8$ between $1$ and $100$:**
- These numbers are: $8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96$.
- There are $\left\lfloor \frac{100}{8} \right\rfloor = 12$ such numbers.
3. **Identify overcounted numbers (numbers that are both divisible by $8$ and have a digit $8$):**
- These numbers are: $8, 48, 88$.
- There are $3$ such numbers.
4. **Calculate the total number of octal numbers:**
- Total octal numbers = (Numbers with a digit $8$) + (Numbers divisible by $8$) - (Overcounted numbers)
- Total octal numbers = $19 + 12 - 3 = 28$.
The final answer is $\boxed{28}$.
|
28
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A cube with sides of length 3cm is painted red
and then cut into 3 x 3 x 3 = 27 cubes with sides of length 1cm.
If a denotes the number of small cubes (of 1cm x 1cm x 1cm) that
are not painted at all, b the number painted on one sides, c the
number painted on two sides, and d the number painted on three
sides, determine the value a-b-c+d.
|
1. **Determine the number of small cubes that are not painted at all ($a$):**
- The small cubes that are not painted at all are those that are completely inside the larger cube, not touching any face.
- For a cube of side length 3 cm, the inner cube that is not painted has side length \(3 - 2 = 1\) cm (since 1 cm on each side is painted).
- Therefore, there is only \(1 \times 1 \times 1 = 1\) small cube that is not painted at all.
- Hence, \(a = 1\).
2. **Determine the number of small cubes painted on one side ($b$):**
- The small cubes painted on one side are those on the faces of the cube but not on the edges or corners.
- Each face of the cube has \(3 \times 3 = 9\) small cubes.
- The cubes on the edges of each face (excluding the corners) are \(3 \times 4 = 12\) (since each edge has 3 cubes, and there are 4 edges per face, but each corner is counted twice).
- Therefore, the number of cubes on each face that are painted on one side is \(9 - 4 = 5\).
- Since there are 6 faces, the total number of cubes painted on one side is \(6 \times 5 = 30\).
- Hence, \(b = 6 \times (3-2)^2 = 6 \times 1 = 6\).
3. **Determine the number of small cubes painted on two sides ($c$):**
- The small cubes painted on two sides are those on the edges of the cube but not on the corners.
- Each edge of the cube has 3 small cubes, but the two end cubes are corners.
- Therefore, each edge has \(3 - 2 = 1\) cube painted on two sides.
- Since there are 12 edges, the total number of cubes painted on two sides is \(12 \times 1 = 12\).
- Hence, \(c = 12 \times (3-2) = 12\).
4. **Determine the number of small cubes painted on three sides ($d$):**
- The small cubes painted on three sides are those on the corners of the cube.
- A cube has 8 corners.
- Hence, \(d = 8\).
5. **Calculate \(a - b - c + d\):**
\[
a - b - c + d = 1 - 6 - 12 + 8 = -9
\]
The final answer is \(\boxed{-9}\)
|
-9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
What is the largest integer less than or equal to $4x^3 - 3x$, where $x=\frac{\sqrt[3]{2+\sqrt3}+\sqrt[3]{2-\sqrt3}}{2}$ ?
(A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) None of the above.
|
1. Given \( x = \frac{\sqrt[3]{2+\sqrt{3}} + \sqrt[3]{2-\sqrt{3}}}{2} \), we start by letting \( y = \sqrt[3]{2+\sqrt{3}} + \sqrt[3]{2-\sqrt{3}} \). Therefore, \( x = \frac{y}{2} \).
2. We need to find the value of \( 4x^3 - 3x \). First, we will find \( y^3 \):
\[
y = \sqrt[3]{2+\sqrt{3}} + \sqrt[3]{2-\sqrt{3}}
\]
Cubing both sides, we get:
\[
y^3 = \left( \sqrt[3]{2+\sqrt{3}} + \sqrt[3]{2-\sqrt{3}} \right)^3
\]
Using the binomial expansion:
\[
y^3 = \left( \sqrt[3]{2+\sqrt{3}} \right)^3 + \left( \sqrt[3]{2-\sqrt{3}} \right)^3 + 3 \cdot \sqrt[3]{2+\sqrt{3}} \cdot \sqrt[3]{2-\sqrt{3}} \cdot \left( \sqrt[3]{2+\sqrt{3}} + \sqrt[3]{2-\sqrt{3}} \right)
\]
Simplifying the terms:
\[
y^3 = (2+\sqrt{3}) + (2-\sqrt{3}) + 3 \cdot \sqrt[3]{(2+\sqrt{3})(2-\sqrt{3})} \cdot y
\]
Since \( (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1 \):
\[
y^3 = 4 + 3 \cdot \sqrt[3]{1} \cdot y
\]
\[
y^3 = 4 + 3y
\]
3. Now, substituting \( y = 2x \) into the equation \( y^3 = 4 + 3y \):
\[
(2x)^3 = 4 + 3(2x)
\]
\[
8x^3 = 4 + 6x
\]
4. Rearranging the equation:
\[
8x^3 - 6x = 4
\]
Dividing by 2:
\[
4x^3 - 3x = 2
\]
5. We need to find the largest integer less than or equal to \( 4x^3 - 3x \):
\[
4x^3 - 3x = 2
\]
6. The largest integer less than or equal to 2 is 1. Therefore, the answer is:
\[
\boxed{1}
\]
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $x=\frac{\sqrt{6+2\sqrt5}+\sqrt{6-2\sqrt5}}{\sqrt{20}}$. The value of $$H=(1+x^5-x^7)^{{2012}^{3^{11}}}$$
is
(A) $1$ (B) $11$ (C) $21$ (D) $101$ (E) None of the above
|
1. First, we need to simplify the expression for \( x \):
\[
x = \frac{\sqrt{6 + 2\sqrt{5}} + \sqrt{6 - 2\sqrt{5}}}{\sqrt{20}}
\]
2. We start by simplifying the terms inside the square roots. Notice that:
\[
\sqrt{6 + 2\sqrt{5}} = \sqrt{(\sqrt{5} + 1)^2} = \sqrt{5} + 1
\]
and
\[
\sqrt{6 - 2\sqrt{5}} = \sqrt{(\sqrt{5} - 1)^2} = \sqrt{5} - 1
\]
3. Substituting these back into the expression for \( x \):
\[
x = \frac{(\sqrt{5} + 1) + (\sqrt{5} - 1)}{\sqrt{20}}
\]
4. Simplify the numerator:
\[
x = \frac{2\sqrt{5}}{\sqrt{20}}
\]
5. Simplify the denominator:
\[
\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}
\]
6. Substitute back into the expression for \( x \):
\[
x = \frac{2\sqrt{5}}{2\sqrt{5}} = 1
\]
7. Now, we need to evaluate the expression \( H \):
\[
H = (1 + x^5 - x^7)^{2012^{3^{11}}}
\]
8. Since \( x = 1 \):
\[
x^5 = 1^5 = 1 \quad \text{and} \quad x^7 = 1^7 = 1
\]
9. Substitute these values back into the expression for \( H \):
\[
H = (1 + 1 - 1)^{2012^{3^{11}}} = 1^{2012^{3^{11}}}
\]
10. Any number raised to any power is still 1:
\[
1^{2012^{3^{11}}} = 1
\]
The final answer is \(\boxed{1}\).
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $f(x)$ be a function such that $f(x)+2f\left(\frac{x+2010}{x-1}\right)=4020 - x$ for all $x \ne 1$.
Then the value of $f(2012)$ is
(A) $2010$, (B) $2011$, (C) $2012$, (D) $2014$, (E) None of the above.
|
1. Given the functional equation:
\[
f(x) + 2f\left(\frac{x+2010}{x-1}\right) = 4020 - x
\]
for all \( x \neq 1 \).
2. Let us denote \( y = \frac{x+2010}{x-1} \). Then, we need to express \( x \) in terms of \( y \):
\[
y = \frac{x+2010}{x-1} \implies y(x-1) = x + 2010 \implies yx - y = x + 2010 \implies yx - x = y + 2010 \implies x(y-1) = y + 2010 \implies x = \frac{y + 2010}{y-1}
\]
3. Substitute \( y = \frac{x+2010}{x-1} \) back into the original equation:
\[
f(x) + 2f(y) = 4020 - x
\]
Now, substitute \( x = \frac{y + 2010}{y-1} \) into the equation:
\[
f\left(\frac{y + 2010}{y-1}\right) + 2f(y) = 4020 - \frac{y + 2010}{y-1}
\]
4. Simplify the right-hand side:
\[
4020 - \frac{y + 2010}{y-1} = 4020 - \left(\frac{y + 2010}{y-1}\right) = 4020 - \frac{y + 2010}{y-1} = 4020 - \frac{y + 2010}{y-1}
\]
5. To find \( f(2012) \), let \( x = 2012 \):
\[
f(2012) + 2f\left(\frac{2012 + 2010}{2012 - 1}\right) = 4020 - 2012
\]
Simplify the argument of the second function:
\[
\frac{2012 + 2010}{2012 - 1} = \frac{4022}{2011} = 2
\]
Thus, the equation becomes:
\[
f(2012) + 2f(2) = 2008
\]
6. Now, let \( x = 2 \) in the original equation:
\[
f(2) + 2f\left(\frac{2 + 2010}{2 - 1}\right) = 4020 - 2
\]
Simplify the argument of the second function:
\[
\frac{2 + 2010}{2 - 1} = 2012
\]
Thus, the equation becomes:
\[
f(2) + 2f(2012) = 4018
\]
7. We now have a system of linear equations:
\[
\begin{cases}
f(2012) + 2f(2) = 2008 \\
f(2) + 2f(2012) = 4018
\end{cases}
\]
8. Solve this system of equations. Multiply the first equation by 2:
\[
2f(2012) + 4f(2) = 4016
\]
Subtract the second equation from this result:
\[
(2f(2012) + 4f(2)) - (f(2) + 2f(2012)) = 4016 - 4018
\]
Simplify:
\[
2f(2) = -2 \implies f(2) = -1
\]
9. Substitute \( f(2) = -1 \) back into the first equation:
\[
f(2012) + 2(-1) = 2008 \implies f(2012) - 2 = 2008 \implies f(2012) = 2010
\]
The final answer is \( \boxed{2010} \).
|
2010
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
[b]Q4.[/b] A man travels from town $A$ to town $E$ through $B,C$ and $D$ with uniform speeds 3km/h, 2km/h, 6km/h and 3km/h on the horizontal, up slope, down slope and horizontal road, respectively. If the road between town $A$ and town $E$ can be classified as horizontal, up slope, down slope and horizontal and total length of each typr of road is the same, what is the average speed of his journey?
\[(A) \; 2 \text{km/h} \qquad (B) \; 2,5 \text{km/h} ; \qquad (C ) \; 3 \text{km/h} ; \qquad (D) \; 3,5 \text{km/h} ; \qquad (E) \; 4 \text{km/h}.\]
|
1. Let the length of each segment \( AB = BC = CD = DE \) be \( x \) km. Then the total distance of the journey from \( A \) to \( E \) is \( 4x \) km.
2. Calculate the time taken for each segment:
- For segment \( AB \) (horizontal road), the speed is \( 3 \) km/h. The time taken is:
\[
t_{AB} = \frac{x}{3} \text{ hours}
\]
- For segment \( BC \) (up slope), the speed is \( 2 \) km/h. The time taken is:
\[
t_{BC} = \frac{x}{2} \text{ hours}
\]
- For segment \( CD \) (down slope), the speed is \( 6 \) km/h. The time taken is:
\[
t_{CD} = \frac{x}{6} \text{ hours}
\]
- For segment \( DE \) (horizontal road), the speed is \( 3 \) km/h. The time taken is:
\[
t_{DE} = \frac{x}{3} \text{ hours}
\]
3. Sum the times for all segments to get the total time:
\[
t_{\text{total}} = t_{AB} + t_{BC} + t_{CD} + t_{DE} = \frac{x}{3} + \frac{x}{2} + \frac{x}{6} + \frac{x}{3}
\]
4. Simplify the total time:
\[
t_{\text{total}} = \frac{x}{3} + \frac{x}{2} + \frac{x}{6} + \frac{x}{3} = \frac{2x}{6} + \frac{3x}{6} + \frac{x}{6} + \frac{2x}{6} = \frac{8x}{6} = \frac{4x}{3} \text{ hours}
\]
5. Calculate the average speed using the formula \( r = \frac{d}{t} \):
\[
r = \frac{4x}{\frac{4x}{3}} = \frac{4x \cdot 3}{4x} = 3 \text{ km/h}
\]
Thus, the average speed of his journey is \( \boxed{3} \) km/h.
|
3
|
Calculus
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
[b]Q11.[/b] Let be given a sequense $a_1=5, \; a_2=8$ and $a_{n+1}=a_n+3a_{n-1}, \qquad n=1,2,3,...$ Calculate the greatest common divisor of $a_{2011}$ and $a_{2012}$.
|
1. **Initial Terms and Recurrence Relation:**
Given the sequence:
\[
a_1 = 5, \quad a_2 = 8
\]
and the recurrence relation:
\[
a_{n+1} = a_n + 3a_{n-1} \quad \text{for} \quad n = 1, 2, 3, \ldots
\]
2. **Modulo 3 Analysis:**
We start by examining the sequence modulo 3:
\[
a_1 \equiv 5 \equiv 2 \pmod{3}
\]
\[
a_2 \equiv 8 \equiv 2 \pmod{3}
\]
Using the recurrence relation modulo 3:
\[
a_{n+1} \equiv a_n + 3a_{n-1} \equiv a_n \pmod{3}
\]
This implies:
\[
a_{n+1} \equiv a_n \pmod{3}
\]
Since \(a_1 \equiv 2 \pmod{3}\) and \(a_2 \equiv 2 \pmod{3}\), it follows that:
\[
a_n \equiv 2 \pmod{3} \quad \text{for all} \quad n \geq 1
\]
3. **Greatest Common Divisor Analysis:**
We need to find \(\gcd(a_{2011}, a_{2012})\). From the recurrence relation:
\[
a_{2012} = a_{2011} + 3a_{2010}
\]
Therefore:
\[
\gcd(a_{2011}, a_{2012}) = \gcd(a_{2011}, a_{2011} + 3a_{2010})
\]
Using the property of gcd:
\[
\gcd(a, b) = \gcd(a, b - ka) \quad \text{for any integer} \quad k
\]
We get:
\[
\gcd(a_{2011}, a_{2011} + 3a_{2010}) = \gcd(a_{2011}, 3a_{2010})
\]
Since \(a_{2011}\) and \(a_{2010}\) are both congruent to 2 modulo 3, we have:
\[
\gcd(a_{2011}, 3a_{2010}) = \gcd(a_{2011}, 3)
\]
Because \(a_{2011} \equiv 2 \pmod{3}\), it is not divisible by 3. Therefore:
\[
\gcd(a_{2011}, 3) = 1
\]
Conclusion:
\[
\boxed{1}
\]
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[b]Q12.[/b] Find all positive integers $P$ such that the sum and product of all its divisors are $2P$ and $P^2$, respectively.
|
1. Let \( P \) be a positive integer. We are given that the sum of all divisors of \( P \) is \( 2P \) and the product of all divisors of \( P \) is \( P^2 \).
2. Let \( \tau(P) \) denote the number of divisors of \( P \). The product of all divisors of \( P \) is known to be \( P^{\tau(P)/2} \). This is because each divisor \( d \) of \( P \) can be paired with \( P/d \), and the product of each pair is \( P \). Since there are \( \tau(P) \) divisors, there are \( \tau(P)/2 \) such pairs.
3. Given that the product of all divisors is \( P^2 \), we have:
\[
P^{\tau(P)/2} = P^2
\]
Dividing both sides by \( P \) (assuming \( P \neq 0 \)):
\[
\tau(P)/2 = 2
\]
Therefore:
\[
\tau(P) = 4
\]
4. The number of divisors \( \tau(P) = 4 \) implies that \( P \) can be either of the form \( q^3 \) (where \( q \) is a prime number) or \( P = qr \) (where \( q \) and \( r \) are distinct prime numbers).
5. We need to check both forms to see which one satisfies the condition that the sum of the divisors is \( 2P \).
6. **Case 1: \( P = q^3 \)**
- The divisors of \( P \) are \( 1, q, q^2, q^3 \).
- The sum of the divisors is:
\[
1 + q + q^2 + q^3
\]
- We need this sum to be \( 2P = 2q^3 \):
\[
1 + q + q^2 + q^3 = 2q^3
\]
- Rearranging, we get:
\[
1 + q + q^2 = q^3
\]
- This equation does not hold for any prime \( q \) because \( q^3 \) grows much faster than \( q^2 \).
7. **Case 2: \( P = qr \)**
- The divisors of \( P \) are \( 1, q, r, qr \).
- The sum of the divisors is:
\[
1 + q + r + qr
\]
- We need this sum to be \( 2P = 2qr \):
\[
1 + q + r + qr = 2qr
\]
- Rearranging, we get:
\[
1 + q + r = qr
\]
- Solving for \( q \) and \( r \), we can try small prime values:
- Let \( q = 2 \) and \( r = 3 \):
\[
1 + 2 + 3 = 2 \cdot 3
\]
\[
6 = 6
\]
- This holds true, so \( P = 2 \cdot 3 = 6 \).
8. Therefore, the only positive integer \( P \) that satisfies the given conditions is \( P = 6 \).
The final answer is \( \boxed{6} \)
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[b]Q13.[/b] Determine the greatest value of the sum $M=11xy+3x+2012yz$, where $x,y,z$ are non negative integers satisfying condition $x+y+z=1000.$
|
1. We start with the given expression \( M = 11xy + 3x + 2012yz \) and the constraint \( x + y + z = 1000 \), where \( x, y, z \) are non-negative integers.
2. To find the maximum value of \( M \), we can express \( x \) in terms of \( y \) and \( z \) using the constraint:
\[
x = 1000 - y - z
\]
3. Substitute \( x = 1000 - y - z \) into \( M \):
\[
M = 11(1000 - y - z)y + 3(1000 - y - z) + 2012yz
\]
Simplify the expression:
\[
M = 11000y - 11y^2 - 11yz + 3000 - 3y - 3z + 2012yz
\]
Combine like terms:
\[
M = 11000y - 11y^2 + (2012 - 11)yz + 3000 - 3y - 3z
\]
\[
M = 11000y - 11y^2 + 2001yz + 3000 - 3y - 3z
\]
4. To maximize \( M \), we need to consider the values of \( y \) and \( z \). Notice that the term \( 2001yz \) will be maximized when both \( y \) and \( z \) are as large as possible.
5. By the AM-GM inequality, the product \( yz \) is maximized when \( y = z \). Given \( y + z = 1000 - x \), the maximum product occurs when \( y = z = 500 \) and \( x = 0 \).
6. Substitute \( y = 500 \) and \( z = 500 \) into \( M \):
\[
M = 11 \cdot 0 \cdot 500 + 3 \cdot 0 + 2012 \cdot 500 \cdot 500
\]
\[
M = 0 + 0 + 2012 \cdot 250000
\]
\[
M = 503000000
\]
7. Therefore, the maximum value of \( M \) is \( 503000000 \).
The final answer is \( \boxed{503000000} \).
|
503000000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[b]Q14.[/b] Let be given a trinagle $ABC$ with $\angle A=90^o$ and the bisectrices of angles $B$ and $C$ meet at $I$. Suppose that $IH$ is perpendicular to $BC$ ($H$ belongs to $BC$). If $HB=5 \text{cm}, \; HC=8 \text{cm}$, compute the area of $\triangle ABC$.
|
1. Given a right triangle \( \triangle ABC \) with \( \angle A = 90^\circ \), the incenter \( I \) is the intersection of the angle bisectors of \( \angle B \) and \( \angle C \). The perpendicular from \( I \) to \( BC \) meets \( BC \) at \( H \), with \( HB = 5 \) cm and \( HC = 8 \) cm.
2. The lengths \( HB \) and \( HC \) can be expressed in terms of the semi-perimeter \( s \) and the sides \( b \) and \( c \) of the triangle:
\[
HB = s - b \quad \text{and} \quad HC = s - c
\]
where \( s = \frac{a + b + c}{2} \).
3. Since \( HB = 5 \) cm and \( HC = 8 \) cm, we have:
\[
s - b = 5 \quad \text{and} \quad s - c = 8
\]
4. Adding these two equations, we get:
\[
(s - b) + (s - c) = 5 + 8 \implies 2s - (b + c) = 13
\]
Since \( s = \frac{a + b + c}{2} \), we can substitute \( 2s = a + b + c \):
\[
a + b + c - (b + c) = 13 \implies a = 13
\]
5. The area \( S \) of \( \triangle ABC \) can be calculated using the formula for the area of a right triangle:
\[
S = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2
\]
Here, the legs are \( b \) and \( c \), and the hypotenuse \( a = 13 \).
6. Using the Pythagorean theorem:
\[
a^2 = b^2 + c^2 \implies 13^2 = b^2 + c^2 \implies 169 = b^2 + c^2
\]
7. We also know:
\[
s = \frac{a + b + c}{2} = \frac{13 + b + c}{2}
\]
Substituting \( s - b = 5 \) and \( s - c = 8 \):
\[
\frac{13 + b + c}{2} - b = 5 \implies \frac{13 + b + c - 2b}{2} = 5 \implies 13 + c - b = 10 \implies c - b = -3
\]
\[
\frac{13 + b + c}{2} - c = 8 \implies \frac{13 + b + c - 2c}{2} = 8 \implies 13 + b - c = 16 \implies b - c = 3
\]
8. Solving the system of equations \( c - b = -3 \) and \( b - c = 3 \):
\[
b = c + 3
\]
Substituting into \( b^2 + c^2 = 169 \):
\[
(c + 3)^2 + c^2 = 169 \implies c^2 + 6c + 9 + c^2 = 169 \implies 2c^2 + 6c + 9 = 169 \implies 2c^2 + 6c - 160 = 0
\]
\[
c^2 + 3c - 80 = 0
\]
9. Solving the quadratic equation:
\[
c = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-80)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 320}}{2} = \frac{-3 \pm \sqrt{329}}{2}
\]
Since \( c \) must be positive:
\[
c = \frac{-3 + \sqrt{329}}{2}
\]
\[
b = c + 3 = \frac{-3 + \sqrt{329}}{2} + 3 = \frac{3 + \sqrt{329}}{2}
\]
10. The area \( S \) of \( \triangle ABC \) is:
\[
S = \frac{1}{2} \times b \times c = \frac{1}{2} \times \frac{3 + \sqrt{329}}{2} \times \frac{-3 + \sqrt{329}}{2} = \frac{1}{2} \times \frac{(3 + \sqrt{329})(-3 + \sqrt{329})}{4}
\]
\[
= \frac{1}{2} \times \frac{(3^2 - (\sqrt{329})^2)}{4} = \frac{1}{2} \times \frac{(9 - 329)}{4} = \frac{1}{2} \times \frac{-320}{4} = \frac{1}{2} \times -80 = -40
\]
11. Since the area cannot be negative, we take the absolute value:
\[
S = 40
\]
The final answer is \( \boxed{40} \) cm\(^2\).
|
40
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $A$ be an even number but not divisible by $10$. The last two digits of $A^{20}$ are:
(A): $46$, (B): $56$, (C): $66$, (D): $76$, (E): None of the above.
|
1. **Understanding the problem**:
We need to find the last two digits of \(A^{20}\) where \(A\) is an even number not divisible by 10. This means \(A\) is not divisible by 2 and 5 simultaneously.
2. **Using Euler's Totient Theorem**:
Euler's Totient Theorem states that \(A^{\phi(n)} \equiv 1 \pmod{n}\) where \(\gcd(A, n) = 1\). Here, we choose \(n = 25\) because we are interested in the last two digits, which can be found using modulo 100. Since \(A\) is even and not divisible by 10, \(\gcd(A, 25) = 1\).
3. **Calculating \(\phi(25)\)**:
\[
\phi(25) = 25 \left(1 - \frac{1}{5}\right) = 25 \cdot \frac{4}{5} = 20
\]
Therefore, \(A^{20} \equiv 1 \pmod{25}\).
4. **Considering modulo 4**:
Since \(A\) is even, \(A \equiv 0 \pmod{4}\). Therefore, \(A^{20} \equiv 0^{20} \equiv 0 \pmod{4}\).
5. **Using the Chinese Remainder Theorem**:
We now have the system of congruences:
\[
\begin{cases}
A^{20} \equiv 1 \pmod{25} \\
A^{20} \equiv 0 \pmod{4}
\end{cases}
\]
We need to find a number \(x\) such that:
\[
\begin{cases}
x \equiv 1 \pmod{25} \\
x \equiv 0 \pmod{4}
\end{cases}
\]
6. **Solving the system**:
We can solve this system by checking numbers that satisfy \(x \equiv 1 \pmod{25}\) and also \(x \equiv 0 \pmod{4}\). We start with \(x = 1\) and keep adding 25 until we find a number that is divisible by 4:
\[
\begin{aligned}
1 &\equiv 1 \pmod{25} \quad \text{but} \quad 1 \not\equiv 0 \pmod{4}, \\
26 &\equiv 1 \pmod{25} \quad \text{but} \quad 26 \not\equiv 0 \pmod{4}, \\
51 &\equiv 1 \pmod{25} \quad \text{but} \quad 51 \not\equiv 0 \pmod{4}, \\
76 &\equiv 1 \pmod{25} \quad \text{and} \quad 76 \equiv 0 \pmod{4}.
\end{aligned}
\]
Therefore, \(x = 76\) satisfies both congruences.
Conclusion:
\[
\boxed{76}
\]
|
76
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The number of integer solutions $x$ of the equation below
$(12x -1)(6x - 1)(4x -1)(3x - 1) = 330$ is
(A): $0$, (B): $1$, (C): $2$, (D): $3$, (E): None of the above.
|
1. Start with the given equation:
\[
(12x - 1)(6x - 1)(4x - 1)(3x - 1) = 330
\]
2. Notice that the equation can be rearranged in any order due to the commutative property of multiplication:
\[
(12x - 1)(3x - 1)(6x - 1)(4x - 1) = 330
\]
3. To simplify the problem, let's introduce a substitution. Let:
\[
y = x(12x - 5)
\]
4. Then, the equation becomes:
\[
(3y + 1)(2y + 1) = 330
\]
5. Factorize 330:
\[
330 = 2 \cdot 3 \cdot 5 \cdot 11
\]
6. Since \((3y + 1)\) and \((2y + 1)\) are factors of 330, we need to find pairs of factors that satisfy the equation. We test different pairs of factors:
- \( (3y + 1) = 2 \cdot 11 = 22 \) and \( (2y + 1) = 3 \cdot 5 = 15 \)
- \( (3y + 1) = 3 \cdot 5 = 15 \) and \( (2y + 1) = 2 \cdot 11 = 22 \)
7. Solve for \(y\) in each case:
- For \( (3y + 1) = 22 \):
\[
3y + 1 = 22 \implies 3y = 21 \implies y = 7
\]
- For \( (2y + 1) = 15 \):
\[
2y + 1 = 15 \implies 2y = 14 \implies y = 7
\]
8. Since both cases give \( y = 7 \), we substitute back to find \(x\):
\[
y = x(12x - 5) = 7
\]
9. Solve the quadratic equation:
\[
12x^2 - 5x - 7 = 0
\]
10. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 12, \quad b = -5, \quad c = -7
\]
\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 12 \cdot (-7)}}{2 \cdot 12}
\]
\[
x = \frac{5 \pm \sqrt{25 + 336}}{24}
\]
\[
x = \frac{5 \pm \sqrt{361}}{24}
\]
\[
x = \frac{5 \pm 19}{24}
\]
11. This gives two potential solutions:
\[
x = \frac{24}{24} = 1 \quad \text{and} \quad x = \frac{-14}{24} = -\frac{7}{12}
\]
12. Since we are looking for integer solutions, only \( x = 1 \) is valid.
The final answer is \( \boxed{1} \)
|
1
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The positive numbers $a, b, c,d,e$ are such that the following identity hold for all real number $x$:
$(x + a)(x + b)(x + c) = x^3 + 3dx^2 + 3x + e^3$. Find the smallest value of $d$.
|
1. Given the identity:
\[
(x + a)(x + b)(x + c) = x^3 + 3dx^2 + 3x + e^3
\]
we can expand the left-hand side and compare coefficients with the right-hand side.
2. Expanding the left-hand side:
\[
(x + a)(x + b)(x + c) = x^3 + (a+b+c)x^2 + (ab+bc+ca)x + abc
\]
3. By comparing coefficients with the right-hand side \(x^3 + 3dx^2 + 3x + e^3\), we get the following system of equations:
\[
a + b + c = 3d
\]
\[
ab + bc + ca = 3
\]
\[
abc = e^3
\]
4. To find the smallest value of \(d\), we use the well-known inequality for positive numbers \(a, b, c\):
\[
(a + b + c)^2 \geq 3(ab + bc + ca)
\]
5. Substituting the given expressions into the inequality:
\[
(3d)^2 \geq 3 \cdot 3
\]
\[
9d^2 \geq 9
\]
\[
d^2 \geq 1
\]
\[
d \geq 1
\]
6. Therefore, the smallest value of \(d\) that satisfies the inequality is:
\[
d = 1
\]
The final answer is \(\boxed{1}\)
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
What is the largest integer not exceeding $8x^3 +6x - 1$, where $x =\frac12 \left(\sqrt[3]{2+\sqrt5} + \sqrt[3]{2-\sqrt5}\right)$ ?
(A): $1$, (B): $2$, (C): $3$, (D): $4$, (E) None of the above.
|
1. Given \( x = \frac{1}{2} \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right) \), we need to evaluate \( 8x^3 + 6x - 1 \).
2. First, let's find \( 8x^3 \):
\[
8x^3 = 8 \left( \frac{1}{2} \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right) \right)^3
\]
Simplifying inside the cube:
\[
8x^3 = 8 \left( \frac{1}{8} \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right)^3 \right)
\]
\[
8x^3 = \left( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right)^3
\]
3. Using the identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\), let \( a = \sqrt[3]{2 + \sqrt{5}} \) and \( b = \sqrt[3]{2 - \sqrt{5}} \):
\[
a^3 = 2 + \sqrt{5}, \quad b^3 = 2 - \sqrt{5}
\]
\[
ab = \sqrt[3]{(2 + \sqrt{5})(2 - \sqrt{5})} = \sqrt[3]{4 - 5} = \sqrt[3]{-1} = -1
\]
Therefore:
\[
(a + b)^3 = (2 + \sqrt{5}) + (2 - \sqrt{5}) + 3(-1)(a + b)
\]
Simplifying:
\[
(a + b)^3 = 4 + 3(-1)(a + b)
\]
\[
(a + b)^3 = 4 - 3(a + b)
\]
4. Since \( x = \frac{1}{2}(a + b) \), we have \( a + b = 2x \):
\[
(2x)^3 = 4 - 3(2x)
\]
\[
8x^3 = 4 - 6x
\]
5. Now, we need to find \( 8x^3 + 6x - 1 \):
\[
8x^3 + 6x - 1 = (4 - 6x) + 6x - 1
\]
\[
8x^3 + 6x - 1 = 4 - 1
\]
\[
8x^3 + 6x - 1 = 3
\]
The final answer is \(\boxed{3}\).
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $x_0 = [a], x_1 = [2a] - [a], x_2 = [3a] - [2a], x_3 = [3a] - [4a],x_4 = [5a] - [4a],x_5 = [6a] - [5a], . . . , $ where $a=\frac{\sqrt{2013}}{\sqrt{2014}}$ .The value of $x_9$ is:
(A): $2$ (B): $3$ (C): $4$ (D): $5$ (E): None of the above.
|
1. First, we need to understand the notation $[x]$, which represents the floor function, i.e., the greatest integer less than or equal to $x$.
2. Given $a = \frac{\sqrt{2013}}{\sqrt{2014}} = \sqrt{\frac{2013}{2014}}$.
3. We need to find the value of $x_9 = [10a] - [9a]$.
Let's calculate $10a$ and $9a$ step by step:
4. Calculate $10a$:
\[
10a = 10 \cdot \sqrt{\frac{2013}{2014}} = \sqrt{100 \cdot \frac{2013}{2014}} = \sqrt{\frac{201300}{2014}}
\]
Since $\frac{201300}{2014} \approx 99.5$, we have:
\[
\sqrt{\frac{201300}{2014}} \approx \sqrt{99.5} \approx 9.975
\]
Therefore, $[10a] = [9.975] = 9$.
5. Calculate $9a$:
\[
9a = 9 \cdot \sqrt{\frac{2013}{2014}} = \sqrt{81 \cdot \frac{2013}{2014}} = \sqrt{\frac{162117}{2014}}
\]
Since $\frac{162117}{2014} \approx 80.5$, we have:
\[
\sqrt{\frac{162117}{2014}} \approx \sqrt{80.5} \approx 8.975
\]
Therefore, $[9a] = [8.975] = 8$.
6. Now, we can find $x_9$:
\[
x_9 = [10a] - [9a] = 9 - 8 = 1
\]
The final answer is $\boxed{1}$.
|
1
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The number $n$ is called a composite number if it can be written in the form $n = a\times b$, where $a, b$ are positive
integers greater than $1$. Write number $2013$ in a sum of $m$ composite numbers. What is the largest value of $m$?
(A): $500$, (B): $501$, (C): $502$, (D): $503$, (E): None of the above.
|
To solve the problem, we need to express the number 2013 as a sum of the maximum number of composite numbers.
1. **Understanding Composite Numbers**:
A composite number is a positive integer that has at least one positive divisor other than one or itself. In other words, it can be written as \( n = a \times b \) where \( a \) and \( b \) are integers greater than 1.
2. **Initial Decomposition**:
We start by noting that 4 is the smallest composite number. We can use 4 to decompose 2013 into a sum of composite numbers.
3. **Maximizing the Number of Terms**:
To maximize the number of composite numbers in the sum, we should use the smallest composite number (which is 4) as many times as possible.
4. **Calculation**:
\[
2013 = 4 \times m + r
\]
where \( m \) is the number of 4's we can use, and \( r \) is the remainder. We need \( r \) to be a composite number to ensure all parts of the sum are composite numbers.
5. **Finding \( m \) and \( r \)**:
\[
2013 \div 4 = 503 \text{ remainder } 1
\]
This means:
\[
2013 = 4 \times 503 + 1
\]
However, 1 is not a composite number. Therefore, we need to adjust the number of 4's to ensure the remainder is a composite number.
6. **Adjusting the Remainder**:
We reduce the number of 4's by 1 and check the new remainder:
\[
2013 = 4 \times 502 + 5
\]
Here, 5 is not a composite number. We continue adjusting:
\[
2013 = 4 \times 501 + 9
\]
Here, 9 is a composite number (since \( 9 = 3 \times 3 \)).
7. **Conclusion**:
The maximum number of composite numbers we can use is 501 terms of 4 and 1 term of 9, making a total of 502 composite numbers.
\[
\boxed{502}
\]
|
502
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let the numbers x and y satisfy the conditions $\begin{cases} x^2 + y^2 - xy = 2 \\
x^4 + y^4 + x^2y^2 = 8 \end{cases}$
The value of $P = x^8 + y^8 + x^{2014}y^{2014}$ is:
(A): $46$, (B): $48$, (C): $50$, (D): $52$, (E) None of the above.
|
1. We start with the given system of equations:
\[
\begin{cases}
x^2 + y^2 - xy = 2 \\
x^4 + y^4 + x^2y^2 = 8
\end{cases}
\]
2. From the first equation, we can express \(x^2 + y^2\) in terms of \(xy\):
\[
x^2 + y^2 = 2 + xy
\]
3. Next, we square both sides of the equation \(x^2 + y^2 = 2 + xy\):
\[
(x^2 + y^2)^2 = (2 + xy)^2
\]
Expanding both sides, we get:
\[
x^4 + y^4 + 2x^2y^2 = 4 + 4xy + x^2y^2
\]
4. We know from the second given equation that:
\[
x^4 + y^4 + x^2y^2 = 8
\]
5. Substituting \(x^4 + y^4 + x^2y^2 = 8\) into the expanded equation, we have:
\[
8 + x^2y^2 = 4 + 4xy + x^2y^2
\]
6. Simplifying this, we get:
\[
8 = 4 + 4xy
\]
\[
4 = 4xy
\]
\[
xy = 1
\]
7. Now, substituting \(xy = 1\) back into the expression for \(x^2 + y^2\):
\[
x^2 + y^2 = 2 + 1 = 3
\]
8. We now need to find \(x^8 + y^8 + x^{2014}y^{2014}\). First, we find \(x^8 + y^8\):
\[
x^8 + y^8 = (x^4 + y^4)^2 - 2(x^2y^2)^2
\]
9. From the second given equation, we know:
\[
x^4 + y^4 = 8 - x^2y^2 = 8 - 1 = 7
\]
10. Therefore:
\[
x^8 + y^8 = 7^2 - 2(1^2) = 49 - 2 = 47
\]
11. Since \(xy = 1\), we have:
\[
x^{2014}y^{2014} = (xy)^{2014} = 1^{2014} = 1
\]
12. Finally, we sum these results to find \(P\):
\[
P = x^8 + y^8 + x^{2014}y^{2014} = 47 + 1 = 48
\]
The final answer is \(\boxed{48}\).
|
48
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many zeros are there in the last digits of the following number $P = 11\times12\times ...\times 88\times 89$ ?
(A): $16$, (B): $17$, (C): $18$, (D): $19$, (E) None of the above.
|
To determine the number of zeros in the last digits of the product \( P = 11 \times 12 \times \cdots \times 88 \times 89 \), we need to count the number of factors of 10 in \( P \). A factor of 10 is composed of a factor of 2 and a factor of 5. Since there are generally more factors of 2 than factors of 5 in a factorial, we only need to count the number of factors of 5.
1. **Using Legendre's Formula**:
Legendre's formula for the exponent of a prime \( p \) in \( n! \) is given by:
\[
e_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots
\]
We need to apply this formula to both \( 89! \) and \( 10! \) to find the number of factors of 5 in each.
2. **Counting factors of 5 in \( 89! \)**:
\[
e_5(89!) = \left\lfloor \frac{89}{5} \right\rfloor + \left\lfloor \frac{89}{25} \right\rfloor + \left\lfloor \frac{89}{125} \right\rfloor + \cdots
\]
Calculating each term:
\[
\left\lfloor \frac{89}{5} \right\rfloor = 17, \quad \left\lfloor \frac{89}{25} \right\rfloor = 3, \quad \left\lfloor \frac{89}{125} \right\rfloor = 0
\]
Therefore:
\[
e_5(89!) = 17 + 3 = 20
\]
3. **Counting factors of 5 in \( 10! \)**:
\[
e_5(10!) = \left\lfloor \frac{10}{5} \right\rfloor + \left\lfloor \frac{10}{25} \right\rfloor + \left\lfloor \frac{10}{125} \right\rfloor + \cdots
\]
Calculating each term:
\[
\left\lfloor \frac{10}{5} \right\rfloor = 2, \quad \left\lfloor \frac{10}{25} \right\rfloor = 0
\]
Therefore:
\[
e_5(10!) = 2
\]
4. **Counting factors of 5 in \( P \)**:
Since \( P = \frac{89!}{10!} \), the number of factors of 5 in \( P \) is:
\[
e_5(P) = e_5(89!) - e_5(10!) = 20 - 2 = 18
\]
Conclusion:
The number of zeros in the last digits of \( P \) is \( \boxed{18} \).
|
18
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let be given $a < b < c$ and $f(x) =\frac{c(x - a)(x - b)}{(c - a)(c - b)}+\frac{a(x - b)(x - c)}{(a - b)(a -c)}+\frac{b(x -c)(x - a)}{(b - c)(b - a)}$.
Determine $f(2014)$.
|
1. **Identify the form of \( f(x) \)**:
The given function is:
\[
f(x) = \frac{c(x - a)(x - b)}{(c - a)(c - b)} + \frac{a(x - b)(x - c)}{(a - b)(a - c)} + \frac{b(x - c)(x - a)}{(b - c)(b - a)}
\]
This is a quadratic polynomial in \( x \).
2. **Check the roots of \( f(x) \)**:
Notice that \( f(a) \), \( f(b) \), and \( f(c) \) can be evaluated directly:
\[
f(a) = \frac{c(a - a)(a - b)}{(c - a)(c - b)} + \frac{a(a - b)(a - c)}{(a - b)(a - c)} + \frac{b(a - c)(a - a)}{(b - c)(b - a)} = 0 + a + 0 = a
\]
\[
f(b) = \frac{c(b - a)(b - b)}{(c - a)(c - b)} + \frac{a(b - b)(b - c)}{(a - b)(a - c)} + \frac{b(b - c)(b - a)}{(b - c)(b - a)} = 0 + 0 + b = b
\]
\[
f(c) = \frac{c(c - a)(c - b)}{(c - a)(c - b)} + \frac{a(c - b)(c - c)}{(a - b)(a - c)} + \frac{b(c - c)(c - a)}{(b - c)(b - a)} = c + 0 + 0 = c
\]
3. **Form of \( f(x) \)**:
Since \( f(a) = a \), \( f(b) = b \), and \( f(c) = c \), and \( f(x) \) is a quadratic polynomial, we can conclude that:
\[
f(x) = x
\]
This is because a quadratic polynomial that matches the identity function at three distinct points must be the identity function itself.
4. **Evaluate \( f(2014) \)**:
Given that \( f(x) = x \) for all \( x \), we have:
\[
f(2014) = 2014
\]
The final answer is \(\boxed{2014}\)
|
2014
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
How many integers are there in $\{0,1, 2,..., 2014\}$ such that $C^x_{2014} \ge C^{999}{2014}$ ?
(A): $15$, (B): $16$, (C): $17$, (D): $18$, (E) None of the above.
Note: $C^{m}_{n}$ stands for $\binom {m}{n}$
|
1. We start by understanding the properties of binomial coefficients. The binomial coefficient $\binom{n}{k}$ is defined as:
\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]
It is known that $\binom{n}{k}$ is symmetric, meaning:
\[
\binom{n}{k} = \binom{n}{n-k}
\]
Additionally, $\binom{n}{k}$ increases as $k$ goes from $0$ to $\left\lfloor \frac{n}{2} \right\rfloor$ and decreases as $k$ goes from $\left\lceil \frac{n}{2} \right\rceil$ to $n$.
2. Given the problem, we need to find the number of integers $x$ in the set $\{0, 1, 2, \ldots, 2014\}$ such that:
\[
\binom{2014}{x} \geq \binom{2014}{999}
\]
3. Using the properties of binomial coefficients, we know that $\binom{2014}{x}$ is maximized at $x = 1007$ (since $1007$ is the closest integer to $\frac{2014}{2}$). Therefore, $\binom{2014}{999}$ is less than or equal to $\binom{2014}{x}$ for $999 \leq x \leq 1015$.
4. To find the range of $x$ for which $\binom{2014}{x} \geq \binom{2014}{999}$, we need to consider the interval:
\[
999 \leq x \leq 2014 - 999 = 1015
\]
5. The number of integers in this interval is calculated as follows:
\[
1015 - 999 + 1 = 17
\]
Conclusion:
There are 17 integers in the set $\{0, 1, 2, \ldots, 2014\}$ such that $\binom{2014}{x} \geq \binom{2014}{999}$.
The final answer is $\boxed{17}$
|
17
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a$ and $b$ satisfy the conditions $\begin{cases}
a^3 - 6a^2 + 15a = 9 \\
b^3 - 3b^2 + 6b = -1 \end{cases}$ .
The value of $(a - b)^{2014}$ is:
(A): $1$, (B): $2$, (C): $3$, (D): $4$, (E) None of the above.
|
1. We start with the given system of equations:
\[
\begin{cases}
a^3 - 6a^2 + 15a = 9 \\
b^3 - 3b^2 + 6b = -1
\end{cases}
\]
2. Rewrite the first equation:
\[
a^3 - 6a^2 + 15a - 9 = 0
\]
We will try to express this in a form that might reveal a relationship with \(b\). Consider the transformation \(a = x + 1\):
\[
(x+1)^3 - 6(x+1)^2 + 15(x+1) - 9 = 0
\]
Expanding each term:
\[
(x+1)^3 = x^3 + 3x^2 + 3x + 1
\]
\[
6(x+1)^2 = 6(x^2 + 2x + 1) = 6x^2 + 12x + 6
\]
\[
15(x+1) = 15x + 15
\]
Substituting these into the equation:
\[
x^3 + 3x^2 + 3x + 1 - 6x^2 - 12x - 6 + 15x + 15 - 9 = 0
\]
Simplifying:
\[
x^3 + 3x^2 - 6x^2 + 3x - 12x + 15x + 1 - 6 + 15 - 9 = 0
\]
\[
x^3 - 3x^2 + 6x + 1 = 0
\]
This simplifies to:
\[
x^3 - 3x^2 + 6x + 1 = 0
\]
Notice that this is exactly the second equation with \(x = b\):
\[
b^3 - 3b^2 + 6b + 1 = 0
\]
Therefore, we have \(x = b\) and \(a = b + 1\).
3. Now, we need to find the value of \((a - b)^{2014}\):
\[
a - b = (b + 1) - b = 1
\]
Therefore:
\[
(a - b)^{2014} = 1^{2014} = 1
\]
The final answer is \(\boxed{1}\)
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Find the smallest positive integer $n$ such that the number $2^n + 2^8 + 2^{11}$ is a perfect square.
(A): $8$, (B): $9$, (C): $11$, (D): $12$, (E) None of the above.
|
1. We start with the expression \(2^n + 2^8 + 2^{11}\) and need to find the smallest positive integer \(n\) such that this expression is a perfect square.
2. Let's rewrite the expression in a more convenient form:
\[
2^n + 2^8 + 2^{11}
\]
Notice that \(2^8 = 256\) and \(2^{11} = 2048\). We can factor out the smallest power of 2 from the terms:
\[
2^n + 2^8 + 2^{11} = 2^8(2^{n-8} + 1 + 2^3)
\]
Simplifying further:
\[
2^8(2^{n-8} + 1 + 8)
\]
We need \(2^{n-8} + 9\) to be a perfect square. Let \(k^2 = 2^{n-8} + 9\), where \(k\) is an integer.
3. Rearrange the equation to solve for \(2^{n-8}\):
\[
2^{n-8} = k^2 - 9
\]
Factor the right-hand side:
\[
2^{n-8} = (k-3)(k+3)
\]
Since \(2^{n-8}\) is a power of 2, both \((k-3)\) and \((k+3)\) must be powers of 2. Let \(k-3 = 2^a\) and \(k+3 = 2^b\), where \(a < b\). Then:
\[
2^b - 2^a = 6
\]
4. We need to find powers of 2 that satisfy this equation. Let's check possible values for \(a\) and \(b\):
- If \(a = 1\) and \(b = 3\):
\[
2^3 - 2^1 = 8 - 2 = 6
\]
This works. So, \(a = 1\) and \(b = 3\).
5. Now, we have:
\[
k - 3 = 2^1 = 2 \quad \text{and} \quad k + 3 = 2^3 = 8
\]
Solving for \(k\):
\[
k = 2 + 3 = 5
\]
6. Substitute \(k = 5\) back into the equation \(2^{n-8} = k^2 - 9\):
\[
2^{n-8} = 5^2 - 9 = 25 - 9 = 16 = 2^4
\]
Therefore:
\[
n - 8 = 4 \implies n = 12
\]
The final answer is \(\boxed{12}\)
|
12
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a,b,c$ and $m$ ($0 \le m \le 26$) be integers such that $a + b + c = (a - b)(b- c)(c - a) = m$ (mod $27$) then $m$ is
(A): $0$, (B): $1$, (C): $25$, (D): $26$ (E): None of the above.
|
1. **Consider the given conditions:**
\[
a + b + c \equiv (a - b)(b - c)(c - a) \equiv m \pmod{27}
\]
We need to find the possible values of \(m\) under the constraint \(0 \le m \le 26\).
2. **Analyze the equation modulo 3:**
\[
a + b + c \equiv (a - b)(b - c)(c - a) \pmod{3}
\]
Since \(27 \equiv 0 \pmod{3}\), we can reduce the problem modulo 3.
3. **Consider all possible cases for \(a, b, c \mod 3\):**
- Case 1: \(a \equiv b \equiv c \pmod{3}\)
- Here, \(a + b + c \equiv 0 \pmod{3}\)
- \((a - b)(b - c)(c - a) \equiv 0 \pmod{3}\)
- Thus, \(m \equiv 0 \pmod{3}\)
- Case 2: Two of \(a, b, c\) are congruent modulo 3, and the third is different.
- Without loss of generality, assume \(a \equiv b \not\equiv c \pmod{3}\)
- Here, \(a + b + c \equiv 2a + c \pmod{3}\)
- \((a - b)(b - c)(c - a) \equiv 0 \cdot (b - c)(c - a) \equiv 0 \pmod{3}\)
- Thus, \(m \equiv 0 \pmod{3}\)
- Case 3: All three \(a, b, c\) are different modulo 3.
- Assume \(a \equiv 0, b \equiv 1, c \equiv 2 \pmod{3}\)
- Here, \(a + b + c \equiv 0 + 1 + 2 \equiv 3 \equiv 0 \pmod{3}\)
- \((a - b)(b - c)(c - a) \equiv (-1)(-1)(2) \equiv 2 \pmod{3}\)
- This case does not satisfy the condition \(a + b + c \equiv (a - b)(b - c)(c - a) \pmod{3}\)
4. **Conclusion from the cases:**
- From the above cases, the only consistent value for \(m\) modulo 3 is \(0\).
5. **Verify modulo 27:**
- Since \(m \equiv 0 \pmod{3}\) and \(0 \le m \le 26\), the only possible value for \(m\) that satisfies the given conditions is \(m = 0\).
The final answer is \(\boxed{0}\)
|
0
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let the numbers $a, b,c$ satisfy the relation $a^2+b^2+c^2 \le 8$.
Determine the maximum value of $M = 4(a^3 + b^3 + c^3) - (a^4 + b^4 + c^4)$
|
To determine the maximum value of \( M = 4(a^3 + b^3 + c^3) - (a^4 + b^4 + c^4) \) given the constraint \( a^2 + b^2 + c^2 \leq 8 \), we can proceed as follows:
1. **Rewrite the expression for \( M \):**
\[
M = 4(a^3 + b^3 + c^3) - (a^4 + b^4 + c^4)
\]
2. **Introduce the terms \( 4a^2, 4b^2, \) and \( 4c^2 \):**
\[
M = 4(a^3 + b^3 + c^3) - (a^4 + b^4 + c^4)
\]
We can rewrite \( M \) by adding and subtracting \( 4a^2, 4b^2, \) and \( 4c^2 \):
\[
M = 4(a^3 + b^3 + c^3) - (a^4 + b^4 + c^4) + 4(a^2 + b^2 + c^2) - 4(a^2 + b^2 + c^2)
\]
\[
M = 4(a^3 + b^3 + c^3 + a^2 + b^2 + c^2) - (a^4 + b^4 + c^4 + 4a^2 + 4b^2 + 4c^2)
\]
3. **Factor the expression:**
\[
M = 4(a^3 + b^3 + c^3 + a^2 + b^2 + c^2) - \left( (a^2 - 2)^2 + (b^2 - 2)^2 + (c^2 - 2)^2 \right)
\]
4. **Simplify the squared terms:**
\[
(a^2 - 2)^2 = a^4 - 4a^2 + 4
\]
\[
(b^2 - 2)^2 = b^4 - 4b^2 + 4
\]
\[
(c^2 - 2)^2 = c^4 - 4c^2 + 4
\]
Therefore,
\[
(a^2 - 2)^2 + (b^2 - 2)^2 + (c^2 - 2)^2 = a^4 + b^4 + c^4 - 4(a^2 + b^2 + c^2) + 12
\]
5. **Substitute back into the expression for \( M \):**
\[
M = 4(a^3 + b^3 + c^3 + a^2 + b^2 + c^2) - \left( a^4 + b^4 + c^4 - 4(a^2 + b^2 + c^2) + 12 \right)
\]
\[
M = 4(a^3 + b^3 + c^3 + a^2 + b^2 + c^2) - a^4 - b^4 - c^4 + 4(a^2 + b^2 + c^2) - 12
\]
\[
M = 4(a^3 + b^3 + c^3 + a^2 + b^2 + c^2) - a^4 - b^4 - c^4 + 4(a^2 + b^2 + c^2) - 12
\]
\[
M = 4(a^3 + b^3 + c^3) - a^4 - b^4 - c^4 + 12
\]
6. **Maximize the expression:**
To maximize \( M \), we need to consider the values of \( a, b, \) and \( c \) that satisfy \( a^2 + b^2 + c^2 \leq 8 \). By symmetry and testing boundary conditions, we find that the maximum value occurs when \( a = b = c = 2 \):
\[
a^2 + b^2 + c^2 = 2^2 + 2^2 + 2^2 = 4 + 4 + 4 = 12 \leq 8
\]
\[
M = 4(2^3 + 2^3 + 2^3) - (2^4 + 2^4 + 2^4)
\]
\[
M = 4(8 + 8 + 8) - (16 + 16 + 16)
\]
\[
M = 4 \cdot 24 - 48
\]
\[
M = 96 - 48
\]
\[
M = 48
\]
Therefore, the maximum value of \( M \) is \( \boxed{48} \).
|
48
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A monkey in Zoo becomes lucky if he eats three different fruits. What is the largest number of monkeys one can make lucky, by having $20$ oranges, $30$ bananas, $40$ peaches and $50$ tangerines? Justify your answer.
(A): $30$ (B): $35$ (C): $40$ (D): $45$ (E): None of the above.
|
To determine the largest number of monkeys that can be made lucky, we need to ensure that each monkey eats three different fruits. We have the following quantities of fruits:
- 20 oranges
- 30 bananas
- 40 peaches
- 50 tangerines
We need to maximize the number of monkeys that can eat three different fruits. Let's break down the steps:
1. **Initial Setup:**
We start with:
\[
\text{Oranges} = 20, \quad \text{Bananas} = 30, \quad \text{Peaches} = 40, \quad \text{Tangerines} = 50
\]
2. **First 10 Monkeys:**
Each of these monkeys can eat one orange, one banana, and one peach:
\[
\text{Oranges} = 20 - 10 = 10, \quad \text{Bananas} = 30 - 10 = 20, \quad \text{Peaches} = 40 - 10 = 30, \quad \text{Tangerines} = 50
\]
Number of lucky monkeys so far: 10
3. **Next 10 Monkeys:**
Each of these monkeys can eat one orange, one banana, and one tangerine:
\[
\text{Oranges} = 10 - 10 = 0, \quad \text{Bananas} = 20 - 10 = 10, \quad \text{Peaches} = 30, \quad \text{Tangerines} = 50 - 10 = 40
\]
Number of lucky monkeys so far: 20
4. **Next 10 Monkeys:**
Each of these monkeys can eat one banana, one peach, and one tangerine:
\[
\text{Oranges} = 0, \quad \text{Bananas} = 10 - 10 = 0, \quad \text{Peaches} = 30 - 10 = 20, \quad \text{Tangerines} = 40 - 10 = 30
\]
Number of lucky monkeys so far: 30
5. **Next 10 Monkeys:**
Each of these monkeys can eat one peach and two tangerines:
\[
\text{Oranges} = 0, \quad \text{Bananas} = 0, \quad \text{Peaches} = 20 - 10 = 10, \quad \text{Tangerines} = 30 - 10 = 20
\]
Number of lucky monkeys so far: 40
At this point, we have used up all the oranges and bananas, and we have 10 peaches and 20 tangerines left. However, we cannot make any more monkeys lucky because we need three different fruits for each monkey.
Therefore, the largest number of monkeys that can be made lucky is 40.
The final answer is $\boxed{40}$
|
40
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $x, y,z$ satisfy the following inequalities $\begin{cases} | x + 2y - 3z| \le 6 \\
| x - 2y + 3z| \le 6 \\
| x - 2y - 3z| \le 6 \\
| x + 2y + 3z| \le 6 \end{cases}$
Determine the greatest value of $M = |x| + |y| + |z|$.
|
To determine the greatest value of \( M = |x| + |y| + |z| \) given the inequalities:
\[
\begin{cases}
| x + 2y - 3z| \le 6 \\
| x - 2y + 3z| \le 6 \\
| x - 2y - 3z| \le 6 \\
| x + 2y + 3z| \le 6
\end{cases}
\]
1. **Understanding the inequalities**:
Each inequality represents a region in 3-dimensional space. The absolute value inequalities can be interpreted as planes that bound a region. For example, \( |x + 2y - 3z| \le 6 \) represents the region between the planes \( x + 2y - 3z = 6 \) and \( x + 2y - 3z = -6 \).
2. **Identifying the vertices**:
The solution to the system of inequalities is the intersection of these regions, which forms a convex polyhedron. The vertices of this polyhedron can be found by solving the system of equations formed by setting each inequality to equality. The vertices are:
\[
\{(\pm 6,0,0), (0,\pm 3,0), (0,0,\pm 2)\}
\]
3. **Analyzing the function \( M = |x| + |y| + |z| \)**:
For any fixed \( M > 0 \), the graph of \( M = |x| + |y| + |z| \) is a plane in the first octant. The goal is to find the maximum value of \( M \) such that the plane \( M = |x| + |y| + |z| \) intersects the convex polyhedron formed by the inequalities.
4. **Comparing the convex hulls**:
The convex hull of the vertices \(\{(\pm 6,0,0), (0,\pm 3,0), (0,0,\pm 2)\}\) is a polyhedron centered at the origin. The graph of \( M = |x| + |y| + |z| \) is a plane that intersects this polyhedron. Since the polyhedron is symmetric and centered at the origin, the maximum value of \( M \) occurs when the plane \( M = |x| + |y| + |z| \) just touches the farthest vertex of the polyhedron.
5. **Determining the maximum \( M \)**:
The farthest vertex from the origin in the set \(\{(\pm 6,0,0), (0,\pm 3,0), (0,0,\pm 2)\}\) is \((6,0,0)\). At this vertex, \( M = |6| + |0| + |0| = 6 \).
Therefore, the greatest value of \( M = |x| + |y| + |z| \) is \( 6 \).
The final answer is \(\boxed{6}\).
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose $x_1, x_2, x_3$ are the roots of polynomial $P(x) = x^3 - 6x^2 + 5x + 12$
The sum $|x_1| + |x_2| + |x_3|$ is
(A): $4$ (B): $6$ (C): $8$ (D): $14$ (E): None of the above.
|
1. **Identify the roots of the polynomial:**
Given the polynomial \( P(x) = x^3 - 6x^2 + 5x + 12 \), we need to find its roots. By the Rational Root Theorem, possible rational roots are the factors of the constant term (12) divided by the factors of the leading coefficient (1). Thus, the possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).
2. **Test potential roots:**
By testing \( x = -1 \):
\[
P(-1) = (-1)^3 - 6(-1)^2 + 5(-1) + 12 = -1 - 6 - 5 + 12 = 0
\]
Therefore, \( x = -1 \) is a root.
3. **Factor the polynomial:**
Since \( x = -1 \) is a root, we can factor \( P(x) \) as:
\[
P(x) = (x + 1)(x^2 - 7x + 12)
\]
4. **Factor the quadratic polynomial:**
Next, we factor \( x^2 - 7x + 12 \):
\[
x^2 - 7x + 12 = (x - 3)(x - 4)
\]
Therefore, the complete factorization of \( P(x) \) is:
\[
P(x) = (x + 1)(x - 3)(x - 4)
\]
5. **Identify all roots:**
The roots of the polynomial are \( x_1 = -1 \), \( x_2 = 3 \), and \( x_3 = 4 \).
6. **Calculate the sum of the absolute values of the roots:**
\[
|x_1| + |x_2| + |x_3| = |-1| + |3| + |4| = 1 + 3 + 4 = 8
\]
The final answer is \(\boxed{8}\).
|
8
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a, b, c$ be two-digit, three-digit, and four-digit numbers, respectively. Assume that the sum of all digits of number $a+b$, and the sum of all digits of $b + c$ are all equal to $2$. The largest value of $a + b + c$ is
(A): $1099$ (B): $2099$ (C): $1199$ (D): $2199$ (E): None of the above.
|
1. **Understanding the problem:**
- \(a\) is a two-digit number.
- \(b\) is a three-digit number.
- \(c\) is a four-digit number.
- The sum of the digits of \(a + b\) is 2.
- The sum of the digits of \(b + c\) is 2.
- We need to find the largest value of \(a + b + c\).
2. **Analyzing the constraints:**
- The sum of the digits of \(a + b\) is 2. This implies that \(a + b\) must be a number whose digits sum to 2.
- The sum of the digits of \(b + c\) is 2. This implies that \(b + c\) must be a number whose digits sum to 2.
3. **Maximizing \(a\):**
- Since \(a\) is a two-digit number, the maximum value of \(a\) is 99. However, we need to ensure that \(a + b\) has digits summing to 2.
4. **Finding \(b\) such that \(a + b\) has digits summing to 2:**
- If \(a = 99\), then \(b\) must be such that the digits of \(99 + b\) sum to 2.
- The smallest \(b\) that satisfies this is \(b = 911\) because \(99 + 911 = 1010\) and the digits of 1010 sum to 2.
5. **Maximizing \(c\):**
- We need to find \(c\) such that \(b + c\) has digits summing to 2.
- If \(b = 911\), then \(c\) must be such that the digits of \(911 + c\) sum to 2.
- The smallest \(c\) that satisfies this is \(c = 9189\) because \(911 + 9189 = 10100\) and the digits of 10100 sum to 2.
6. **Calculating \(a + b + c\):**
- \(a = 99\)
- \(b = 911\)
- \(c = 9189\)
- Therefore, \(a + b + c = 99 + 911 + 9189 = 10199\).
7. **Verifying the solution:**
- The sum of the digits of \(a + b = 1010\) is \(1 + 0 + 1 + 0 = 2\).
- The sum of the digits of \(b + c = 10100\) is \(1 + 0 + 1 + 0 + 0 = 2\).
- Both conditions are satisfied.
The final answer is \(\boxed{10199}\).
|
10199
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let a,b,c be three distinct positive numbers.
Consider the quadratic polynomial $P (x) =\frac{c(x - a)(x - b)}{(c -a)(c -b)}+\frac{a(x - b)(x - c)}{(a - b)(a - c)}+\frac{b(x -c)(x - a)}{(b - c)(b - a)}+ 1$.
The value of $P (2017)$ is
(A): $2015$ (B): $2016$ (C): $2017$ (D): $2018$ (E): None of the above.
|
1. Consider the given polynomial:
\[
P(x) = \frac{c(x - a)(x - b)}{(c - a)(c - b)} + \frac{a(x - b)(x - c)}{(a - b)(a - c)} + \frac{b(x - c)(x - a)}{(b - c)(b - a)} + 1
\]
2. We need to show that the expression:
\[
Q(x) = \frac{c(x - a)(x - b)}{(c - a)(c - b)} + \frac{a(x - b)(x - c)}{(a - b)(a - c)} + \frac{b(x - c)(x - a)}{(b - c)(b - a)}
\]
is equal to zero for all \( x \).
3. Notice that \( Q(x) \) is a quadratic polynomial in \( x \). Let's denote it as:
\[
Q(x) = A x^2 + B x + C
\]
4. To determine the coefficients \( A \), \( B \), and \( C \), we evaluate \( Q(x) \) at \( x = a \), \( x = b \), and \( x = c \):
\[
Q(a) = \frac{c(a - a)(a - b)}{(c - a)(c - b)} + \frac{a(a - b)(a - c)}{(a - b)(a - c)} + \frac{b(a - c)(a - a)}{(b - c)(b - a)} = 0 + 1 + 0 = 1
\]
\[
Q(b) = \frac{c(b - a)(b - b)}{(c - a)(c - b)} + \frac{a(b - b)(b - c)}{(a - b)(a - c)} + \frac{b(b - c)(b - a)}{(b - c)(b - a)} = 0 + 0 + 1 = 1
\]
\[
Q(c) = \frac{c(c - a)(c - b)}{(c - a)(c - b)} + \frac{a(c - b)(c - c)}{(a - b)(a - c)} + \frac{b(c - c)(c - a)}{(b - c)(b - a)} = 1 + 0 + 0 = 1
\]
5. Since \( Q(a) = Q(b) = Q(c) = 1 \), and \( Q(x) \) is a quadratic polynomial, the only way for \( Q(x) \) to be equal to 1 at three distinct points is if \( Q(x) \) is identically equal to 1. Therefore:
\[
Q(x) = 1 \quad \text{for all } x
\]
6. Substituting \( Q(x) \) back into the original polynomial \( P(x) \):
\[
P(x) = Q(x) + 1 = 1 + 1 = 2
\]
7. Therefore, \( P(x) = 2 \) for all \( x \), including \( x = 2017 \).
The final answer is \(\boxed{2}\)
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $T=\frac{1}{4}x^{2}-\frac{1}{5}y^{2}+\frac{1}{6}z^{2}$ where $x,y,z$ are real numbers such that $1 \leq x,y,z \leq 4$ and $x-y+z=4$.
Find the smallest value of $10 \times T$.
|
1. Given the function \( T = \frac{1}{4}x^2 - \frac{1}{5}y^2 + \frac{1}{6}z^2 \) and the constraints \( 1 \leq x, y, z \leq 4 \) and \( x - y + z = 4 \), we need to find the minimum value of \( 10 \times T \).
2. First, we rewrite \( T \) in a common denominator form:
\[
T = \frac{x^2}{4} - \frac{y^2}{5} + \frac{z^2}{6} = \frac{15x^2 - 12y^2 + 10z^2}{60}
\]
3. To find the minimum value of \( T \), we need to consider the constraints \( x - y + z = 4 \) and \( 1 \leq x, y, z \leq 4 \). We will test the boundary values of \( x, y, z \) to find the minimum value.
4. Let's start by setting \( x = 1 \):
\[
1 - y + z = 4 \implies z = y + 3
\]
Since \( 1 \leq z \leq 4 \), we have:
\[
1 \leq y + 3 \leq 4 \implies -2 \leq y \leq 1
\]
But \( y \geq 1 \), so \( y = 1 \) and \( z = 4 \).
5. Substitute \( x = 1 \), \( y = 1 \), and \( z = 4 \) into \( T \):
\[
T = \frac{1^2}{4} - \frac{1^2}{5} + \frac{4^2}{6} = \frac{1}{4} - \frac{1}{5} + \frac{16}{6}
\]
\[
T = \frac{1}{4} - \frac{1}{5} + \frac{8}{3} = \frac{15}{60} - \frac{12}{60} + \frac{160}{60} = \frac{163}{60}
\]
6. Now, let's test another boundary value, \( x = 4 \):
\[
4 - y + z = 4 \implies z = y
\]
Since \( 1 \leq z \leq 4 \), we have \( 1 \leq y \leq 4 \).
7. Substitute \( x = 4 \), \( y = 1 \), and \( z = 1 \) into \( T \):
\[
T = \frac{4^2}{4} - \frac{1^2}{5} + \frac{1^2}{6} = \frac{16}{4} - \frac{1}{5} + \frac{1}{6}
\]
\[
T = 4 - \frac{1}{5} + \frac{1}{6} = 4 - \frac{6}{30} + \frac{5}{30} = 4 - \frac{1}{30} = \frac{119}{30}
\]
8. Finally, let's test \( x = 2 \):
\[
2 - y + z = 4 \implies z = y + 2
\]
Since \( 1 \leq z \leq 4 \), we have:
\[
1 \leq y + 2 \leq 4 \implies -1 \leq y \leq 2
\]
But \( y \geq 1 \), so \( y = 1 \) and \( z = 3 \).
9. Substitute \( x = 2 \), \( y = 1 \), and \( z = 3 \) into \( T \):
\[
T = \frac{2^2}{4} - \frac{1^2}{5} + \frac{3^2}{6} = \frac{4}{4} - \frac{1}{5} + \frac{9}{6}
\]
\[
T = 1 - \frac{1}{5} + \frac{3}{2} = 1 - \frac{1}{5} + 1.5 = 1 - 0.2 + 1.5 = 2.3
\]
10. Therefore, the minimum value of \( T \) is \( 2.3 \).
11. The smallest value of \( 10 \times T \) is:
\[
10 \times 2.3 = 23
\]
The final answer is \( \boxed{23} \).
|
23
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of distinct real roots of the following equation $x^2 +\frac{9x^2}{(x + 3)^2} = 40$.
A. $0$ B. $1$ C. $2$ D. $3$ E. $4$
|
1. Start with the given equation:
\[
x^2 + \frac{9x^2}{(x + 3)^2} = 40
\]
2. Simplify the fraction:
\[
\frac{9x^2}{(x + 3)^2} = \frac{9x^2}{x^2 + 6x + 9}
\]
3. Substitute back into the equation:
\[
x^2 + \frac{9x^2}{x^2 + 6x + 9} = 40
\]
4. Let \( y = x + 3 \), then \( x = y - 3 \). Substitute \( x = y - 3 \) into the equation:
\[
(y - 3)^2 + \frac{9(y - 3)^2}{y^2} = 40
\]
5. Simplify the equation:
\[
(y - 3)^2 + \frac{9(y - 3)^2}{y^2} = 40
\]
\[
(y - 3)^2 \left(1 + \frac{9}{y^2}\right) = 40
\]
6. Let \( z = y - 3 \), then the equation becomes:
\[
z^2 \left(1 + \frac{9}{(z + 3)^2}\right) = 40
\]
7. Simplify further:
\[
z^2 + \frac{9z^2}{(z + 3)^2} = 40
\]
8. Multiply through by \((z + 3)^2\) to clear the fraction:
\[
z^2(z + 3)^2 + 9z^2 = 40(z + 3)^2
\]
\[
z^4 + 6z^3 + 9z^2 + 9z^2 = 40z^2 + 240z + 360
\]
\[
z^4 + 6z^3 + 18z^2 = 40z^2 + 240z + 360
\]
\[
z^4 + 6z^3 - 22z^2 - 240z - 360 = 0
\]
9. This is a quartic equation, which can have up to 4 real or complex roots. To determine the number of real roots, we can use the Descartes' Rule of Signs or numerical methods.
10. By analyzing the polynomial or using a graphing tool, we find that the quartic equation has 2 real roots.
The final answer is \(\boxed{2}\).
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose that $ABCDE$ is a convex pentagon with $\angle A = 90^o,\angle B = 105^o,\angle C = 90^o$ and $AB = 2,BC = CD = DE =\sqrt2$. If the length of $AE$ is $\sqrt{a }- b$ where $a, b$ are integers, what is the value of $a + b$?
|
1. **Identify the given information and draw the pentagon:**
- $\angle A = 90^\circ$
- $\angle B = 105^\circ$
- $\angle C = 90^\circ$
- $AB = 2$
- $BC = CD = DE = \sqrt{2}$
2. **Analyze the triangle $\triangle ABD$:**
- Since $\angle A = 90^\circ$ and $\angle B = 105^\circ$, the remaining angle $\angle DAB$ in $\triangle ABD$ is:
\[
\angle DAB = 180^\circ - \angle A - \angle B = 180^\circ - 90^\circ - 105^\circ = -15^\circ
\]
This is incorrect. Let's re-evaluate the angles.
3. **Re-evaluate the angles:**
- Since $\angle A = 90^\circ$ and $\angle C = 90^\circ$, $\angle B$ must be split between $\angle ABD$ and $\angle DBC$.
- Given $\angle B = 105^\circ$, we can split it as follows:
\[
\angle ABD = 60^\circ \quad \text{and} \quad \angle DBC = 45^\circ
\]
4. **Determine the properties of $\triangle ABD$:**
- Since $\angle ABD = 60^\circ$ and $AB = 2$, $\triangle ABD$ is an isosceles triangle with $AB = BD$.
- Therefore, $\triangle ABD$ is an equilateral triangle, and $AD = AB = BD = 2$.
5. **Determine the properties of $\triangle ADE$:**
- $\angle DAE = 90^\circ - \angle BAD = 90^\circ - 30^\circ = 60^\circ$.
- Using the Law of Cosines in $\triangle ADE$:
\[
AE^2 = AD^2 + DE^2 - 2 \cdot AD \cdot DE \cdot \cos(\angle DAE)
\]
\[
AE^2 = 2^2 + (\sqrt{2})^2 - 2 \cdot 2 \cdot \sqrt{2} \cdot \cos(60^\circ)
\]
\[
AE^2 = 4 + 2 - 2 \cdot 2 \cdot \sqrt{2} \cdot \frac{1}{2}
\]
\[
AE^2 = 6 - 2\sqrt{2}
\]
\[
AE = \sqrt{6 - 2\sqrt{2}}
\]
6. **Simplify the expression for $AE$:**
- We need to express $AE$ in the form $\sqrt{a} - b$.
- Notice that $6 - 2\sqrt{2}$ can be rewritten as $(\sqrt{3} - 1)^2$:
\[
(\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}
\]
- Therefore, $AE = \sqrt{3} - 1$.
7. **Identify $a$ and $b$:**
- From $AE = \sqrt{3} - 1$, we have $a = 3$ and $b = 1$.
- Thus, $a + b = 3 + 1 = 4$.
The final answer is $\boxed{4}$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $k$ be a positive integer such that $1 +\frac12+\frac13+ ... +\frac{1}{13}=\frac{k}{13!}$. Find the remainder when $k$ is divided by $7$.
|
1. We start with the given equation:
\[
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{13} = \frac{k}{13!}
\]
To find \( k \), we multiply both sides by \( 13! \):
\[
13! \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{13}\right) = k
\]
This can be rewritten as:
\[
k = 13! + \frac{13!}{2} + \frac{13!}{3} + \cdots + \frac{13!}{13}
\]
2. We need to find the remainder when \( k \) is divided by 7. Therefore, we compute:
\[
k \pmod{7}
\]
This is equivalent to:
\[
13! + \frac{13!}{2} + \frac{13!}{3} + \cdots + \frac{13!}{13} \pmod{7}
\]
3. First, we simplify \( 13! \pmod{7} \). Note that:
\[
13! = 1 \cdot 2 \cdot 3 \cdot \cdots \cdot 13
\]
We can reduce each term modulo 7:
\[
13! \equiv 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 0 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \pmod{7}
\]
Since \( 7 \equiv 0 \pmod{7} \), any product involving 7 or its multiples will be 0 modulo 7. Therefore:
\[
13! \equiv 0 \pmod{7}
\]
4. Next, we consider the terms \( \frac{13!}{2}, \frac{13!}{3}, \ldots, \frac{13!}{13} \) modulo 7. Since \( 13! \equiv 0 \pmod{7} \), each of these terms will also be 0 modulo 7:
\[
\frac{13!}{2} \equiv 0 \pmod{7}, \quad \frac{13!}{3} \equiv 0 \pmod{7}, \quad \ldots, \quad \frac{13!}{13} \equiv 0 \pmod{7}
\]
5. Summing these results, we get:
\[
k \equiv 0 + 0 + 0 + \cdots + 0 \pmod{7}
\]
Therefore:
\[
k \equiv 0 \pmod{7}
\]
The final answer is \(\boxed{0}\)
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find all $3$-digit numbers $\overline{abc}$ ($a,b \ne 0$) such that $\overline{bcd} \times a = \overline{1a4d}$ for some integer $d$ from $1$ to $9$
|
We are given the problem of finding all 3-digit numbers $\overline{abc}$ (where $a, b \neq 0$) such that $\overline{bcd} \times a = \overline{1a4d}$ for some integer $d$ from $1$ to $9$.
1. **Express the numbers in terms of their digits:**
- Let $\overline{abc} = 100a + 10b + c$
- Let $\overline{bcd} = 100b + 10c + d$
- Let $\overline{1a4d} = 1000 + 100a + 40 + d = 1040 + 100a + d$
2. **Set up the equation:**
\[
a \times (100b + 10c + d) = 1040 + 100a + d
\]
3. **Simplify the equation:**
\[
100ab + 10ac + ad = 1040 + 100a + d
\]
Subtract $d$ from both sides:
\[
100ab + 10ac + ad - d = 1040 + 100a
\]
Factor out $d$ on the left side:
\[
100ab + 10ac + d(a - 1) = 1040 + 100a
\]
4. **Consider the divisibility condition:**
Since $d(a - 1)$ must be divisible by $10$, we have:
\[
10 \mid d(a - 1)
\]
This implies that $d(a - 1)$ must be a multiple of $10$. Given that $d$ ranges from $1$ to $9$, $a - 1$ must be a multiple of $10$.
5. **Case analysis:**
- **Case 1: $a = 1$**
\[
100b + 10c + d = 1040 + 100 + d
\]
Simplifying:
\[
100b + 10c = 1140
\]
\[
10b + c = 114
\]
This is impossible since $b$ and $c$ are digits (0-9).
- **Case 2: $a = 6$**
\[
100b + 10c + d = 1040 + 600 + d
\]
Simplifying:
\[
100b + 10c = 1640
\]
\[
10b + c = 164
\]
This is impossible since $b$ and $c$ are digits (0-9).
- **Case 3: $a = 6$ and $d$ must be even (let $d = 2m$)**
\[
600b + 60c + 12m = 1040 + 600 + 2m
\]
Simplifying:
\[
600b + 60c + 10m = 1640
\]
\[
60b + 6c + m = 164
\]
Solving for $b, c, m$:
\[
b = 2, c = 7, m = 2
\]
So, $\overline{abc} = 627$.
- **Case 4: $d = 5$ and $a$ must be odd (3, 5, 7, 9)**
\[
100ab + 10ac + 5a = 1040 + 100a + 5
\]
Simplifying:
\[
a(20b + 2c - 19) = 209
\]
$a$ must divide $209$, but none of $3, 5, 7, 9$ divides $209$. So, this case is impossible.
Hence, the only value $\overline{abc}$ is $627$.
The final answer is $\boxed{627}$
|
627
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
In the below figure, there is a regular hexagon and three squares whose sides are equal to $4$ cm. Let $M,N$, and $P$ be the centers of the squares. The perimeter of the triangle $MNP$ can be written in the form $a + b\sqrt3$ (cm), where $a, b$ are integers. Compute the value of $a + b$.
[img]https://cdn.artofproblemsolving.com/attachments/e/8/5996e994d4bbed8d3b3269d3e38fc2ec5d2f0b.png[/img]
|
1. **Identify the properties of the shapes involved:**
- The hexagon is regular, meaning all its sides and angles are equal.
- Each square has a side length of 4 cm.
- The centers of the squares, \(M\), \(N\), and \(P\), form the vertices of triangle \(MNP\).
2. **Determine the side length of the equilateral triangle \(MNP\):**
- Since the squares are regular and their sides are parallel to the sides of the hexagon, the centers \(M\), \(N\), and \(P\) form an equilateral triangle.
- The distance between the centers of two adjacent squares can be calculated by considering the geometry of the squares and the hexagon.
3. **Calculate the distance between the centers of two adjacent squares:**
- Each square has a side length of 4 cm.
- The distance between the centers of two adjacent squares is the sum of the side of the square and the height of the equilateral triangle formed by the centers of the squares.
- The height of an equilateral triangle with side length \(s\) is given by \(\frac{\sqrt{3}}{2} s\).
4. **Calculate the height of the mini equilateral triangles inside the squares:**
- The height of one of these mini triangles is 2 cm (half the side length of the square).
- The side length of these mini triangles is \(\frac{4}{\sqrt{3}}\).
5. **Calculate the side length of \(\triangle{MNP}\):**
- The side length of \(\triangle{MNP}\) is the sum of the side length of the square and twice the height of the mini equilateral triangles.
- Therefore, the side length of \(\triangle{MNP}\) is \(4 + 2 \times \frac{4}{\sqrt{3}} = 4 + \frac{8}{\sqrt{3}} = 4 + \frac{8\sqrt{3}}{3}\).
6. **Calculate the perimeter of \(\triangle{MNP}\):**
- The perimeter of \(\triangle{MNP}\) is three times the side length.
- Thus, the perimeter is \(3 \left(4 + \frac{8\sqrt{3}}{3}\right) = 12 + 8\sqrt{3}\).
7. **Express the perimeter in the form \(a + b\sqrt{3}\):**
- Here, \(a = 12\) and \(b = 8\).
8. **Compute the value of \(a + b\):**
- \(a + b = 12 + 8 = 20\).
The final answer is \(\boxed{20}\).
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For a special event, the five Vietnamese famous dishes including Phở, (Vietnamese noodle), Nem (spring roll), Bún Chả (grilled pork noodle), Bánh cuốn (stuffed pancake), and Xôi gà (chicken sticky rice) are the options for the main courses for the dinner of Monday, Tuesday, and Wednesday. Every dish must be used exactly one time. How many choices do we have?
|
To solve this problem, we need to determine the number of ways to distribute 5 dishes over 3 days such that each dish is used exactly once. We will use the principle of inclusion-exclusion (PIE) to count the valid distributions.
1. **Total number of distributions without any restrictions:**
Each of the 5 dishes can be assigned to any of the 3 days. Therefore, the total number of ways to distribute the dishes is:
\[
3^5 = 243
\]
2. **Subtract the invalid distributions where fewer than 3 days are used:**
- **Distributions using exactly 1 day:**
There are 3 ways to choose which day to use, and all 5 dishes must be assigned to that day. Thus, there are:
\[
3 \times 1 = 3
\]
- **Distributions using exactly 2 days:**
There are \(\binom{3}{2} = 3\) ways to choose which 2 days to use. For each pair of days, each dish can be assigned to either of the 2 days, giving \(2^5\) ways. However, we must subtract the 2 cases where all dishes are assigned to only one of the two days (since we want exactly 2 days to be used). Thus, there are:
\[
3 \times (2^5 - 2) = 3 \times (32 - 2) = 3 \times 30 = 90
\]
3. **Apply the principle of inclusion-exclusion:**
The number of valid distributions is given by:
\[
3^5 - \left( \text{number of ways using exactly 1 day} \right) - \left( \text{number of ways using exactly 2 days} \right)
\]
Substituting the values, we get:
\[
243 - 3 - 90 = 150
\]
The final answer is \(\boxed{150}\).
|
150
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[THE PROBLEM OF PAINTING THE THÁP RÙA (THE CENTRAL TOWER) MODEL]
The following picture illustrates the model of the Tháp Rùa (the Central Tower) in Hanoi, which consists of $3$ levels. For the first and second levels, each has $10$ doorways among which $3$ doorways are located at the front, $3$ at the back, $2$ on the right side and $2$ on the left side. The top level of the tower model has no doorways. The front of the tower model is signified by a disk symbol on the top level.
We paint the tower model with three colors: Blue, Yellow and Brown by fulfilling the following requirements:
1. The top level is painted with only one color.
2. In the second level, the $3$ doorways at the front are painted with the same color which is different from the one used for the center doorway at the back. Besides, any two adjacent doorways, including the pairs at the same corners, are painted with different colors.
3. For the first level, we apply the same rules as for the second level.
[img]https://cdn.artofproblemsolving.com/attachments/2/3/18ee062b79693c4ccc26bf922a7f54e9f352ee.png[/img]
(a) In how many ways the first level can be painted?
(b) In how many ways the whole tower model can be painted?
|
### Part (a): In how many ways can the first level be painted?
1. **Choosing the color for the three doorways at the front:**
- There are 3 colors available: Blue, Yellow, and Brown.
- Therefore, there are \(3\) ways to choose the color for the three doorways at the front.
2. **Choosing the color for the center doorway at the back:**
- The color of the center doorway at the back must be different from the color of the three doorways at the front.
- Therefore, there are \(3 - 1 = 2\) ways to choose the color for the center doorway at the back.
3. **Coloring the remaining doorways:**
- The remaining doorways must be colored such that any two adjacent doorways have different colors.
- We need to consider the doorways on the left and right sides separately.
4. **Coloring the doorways on the left side:**
- There are 2 doorways on the left side.
- The first doorway on the left side must be a different color from the adjacent front doorway.
- The second doorway on the left side must be a different color from both the first doorway on the left side and the center doorway at the back.
- There are \(2\) choices for the first doorway on the left side and \(2\) choices for the second doorway on the left side.
- Therefore, there are \(2 \times 2 = 4\) ways to color the doorways on the left side.
5. **Coloring the doorways on the right side:**
- Similarly, there are \(2 \times 2 = 4\) ways to color the doorways on the right side.
6. **Combining all choices:**
- The total number of ways to paint the first level is:
\[
3 \text{ (front)} \times 2 \text{ (back)} \times 4 \text{ (left)} \times 4 \text{ (right)} = 3 \times 2 \times 4 \times 4 = 96
\]
### Part (b): In how many ways can the whole tower model be painted?
1. **Painting the top level:**
- The top level is painted with only one color.
- There are \(3\) ways to choose the color for the top level.
2. **Painting the second level:**
- The second level follows the same rules as the first level.
- From part (a), we know there are \(96\) ways to paint the second level.
3. **Painting the first level:**
- The first level follows the same rules as the second level.
- Therefore, there are \(96\) ways to paint the first level.
4. **Combining all choices:**
- The total number of ways to paint the entire Tháp Rùa tower is:
\[
3 \text{ (top level)} \times 96 \text{ (second level)} \times 96 \text{ (first level)} = 3 \times 96 \times 96 = 27648
\]
The final answer is \(\boxed{27648}\)
|
27648
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
How many rectangles can be formed by the vertices of a cube? (Note: square is also a special rectangle).
A. $6$ B. $8$ C. $12$ D. $18$ E. $16$
|
1. **Identify the number of square faces in the cube:**
A cube has 6 faces, and each face is a square. Therefore, there are 6 square faces.
2. **Identify the number of rectangles that bisect the cube:**
These rectangles are formed by selecting two opposite edges on one face and two opposite edges on the opposite face. Each of these rectangles includes two face diagonals. There are 6 such rectangles, one for each pair of opposite faces.
3. **Count the total number of rectangles:**
- Number of square faces: 6
- Number of bisecting rectangles: 6
Therefore, the total number of rectangles is:
\[
6 + 6 = 12
\]
Conclusion:
The total number of rectangles that can be formed by the vertices of a cube is $\boxed{12}$.
|
12
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many integers $n$ are there those satisfy the following inequality $n^4 - n^3 - 3n^2 - 3n - 17 < 0$?
A. $4$ B. $6$ C. $8$ D. $10$ E. $12$
|
1. We start with the inequality \( n^4 - n^3 - 3n^2 - 3n - 17 < 0 \).
2. We rewrite the inequality as \( n^4 - n^3 - 3n^2 - 3n - 18 + 1 < 0 \), which simplifies to \( n^4 - n^3 - 3n^2 - 3n - 18 < -1 \).
3. We factorize the polynomial \( n^4 - n^3 - 3n^2 - 3n - 18 \). We find that \( n = -2 \) and \( n = 3 \) are roots of the polynomial:
\[
P(n) = n^4 - n^3 - 3n^2 - 3n - 18
\]
Substituting \( n = -2 \):
\[
P(-2) = (-2)^4 - (-2)^3 - 3(-2)^2 - 3(-2) - 18 = 16 + 8 - 12 + 6 - 18 = 0
\]
Substituting \( n = 3 \):
\[
P(3) = 3^4 - 3^3 - 3 \cdot 3^2 - 3 \cdot 3 - 18 = 81 - 27 - 27 - 9 - 18 = 0
\]
4. We factorize \( P(n) \) as:
\[
P(n) = (n - 3)(n + 2)(n^2 + 3)
\]
Therefore, the inequality becomes:
\[
(n - 3)(n + 2)(n^2 + 3) + 1 < 0
\]
5. We analyze the sign of \( (n - 3)(n + 2)(n^2 + 3) + 1 \) for different values of \( n \):
- For \( n > 3 \):
\[
(n - 3)(n + 2) > 0 \implies P(n) + 1 > 0
\]
- For \( n = 3 \):
\[
P(3) + 1 = 0 + 1 > 0
\]
- For \( n = 2 \):
\[
P(2) + 1 = (-1)(4)(7) + 1 < 0
\]
- For \( n = 1 \):
\[
P(1) + 1 = (-2)(3)(4) + 1 < 0
\]
- For \( n = 0 \):
\[
P(0) + 1 = (-3)(2)(3) + 1 < 0
\]
- For \( n = -1 \):
\[
P(-1) + 1 = (-4)(1)(4) + 1 < 0
\]
- For \( n = -2 \):
\[
P(-2) + 1 = 0 + 1 > 0
\]
- For \( n < -2 \):
\[
(n - 3)(n + 2) > 0 \implies P(n) + 1 > 0
\]
6. From the analysis, the values of \( n \) that satisfy the inequality are \( n = 2, 1, 0, -1 \).
The final answer is \(\boxed{4}\)
|
4
|
Inequalities
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many ways of choosing four edges in a cube such that any two among those four choosen edges have no common point.
|
To solve the problem of choosing four edges in a cube such that any two among those four chosen edges have no common point, we need to consider the geometric properties of the cube and the constraints given.
1. **Case 1: The four edges are pairwise parallel.**
- A cube has 12 edges, and these edges can be grouped into 3 sets of 4 parallel edges each.
- For each set of parallel edges, we can choose 4 edges such that no two edges share a common vertex.
- Since there are 3 sets of parallel edges, there are \(3\) ways to choose this set of edges.
2. **Case 2: Two pairs of parallel edges from opposite faces, and any two edges from different faces are skew.**
- A cube has 6 faces, and these faces can be grouped into 3 pairs of opposite faces.
- For each pair of opposite faces, we can choose 2 parallel edges from one face and 2 parallel edges from the opposite face.
- Each face has 2 sets of parallel edges, so for each pair of opposite faces, there are \(2 \times 2 = 4\) ways to choose the edges.
- Since there are 3 pairs of opposite faces, there are \(3 \times 4 = 12\) ways to choose this set of edges.
Combining both cases, we get:
- Case 1: \(3\) ways
- Case 2: \(12\) ways
Therefore, the total number of ways to choose four edges such that any two among those four chosen edges have no common point is:
\[ 3 + 12 = 15 \]
The final answer is \(\boxed{15}\)
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
There are $100$ school students from two clubs $A$ and $B$ standing in circle. Among them $62$ students stand next to at least one student from club $A$, and $54$ students stand next to at least one student from club $B$.
1) How many students stand side-by-side with one friend from club $A$ and one friend from club $B$?
2) What is the number of students from club $A$?
|
1. Let \( p \) be the number of students who stand next to exactly one student from club \( A \) and one student from club \( B \).
2. Let \( q \) be the number of students who stand between two students from club \( A \).
3. Let \( r \) be the number of students who stand between two students from club \( B \).
From the problem, we have the following equations:
\[ p + q + r = 100 \]
\[ p + q = 62 \]
\[ p + r = 54 \]
We need to solve these equations to find the values of \( p \), \( q \), and \( r \).
First, subtract the second equation from the third equation:
\[ (p + r) - (p + q) = 54 - 62 \]
\[ r - q = -8 \]
\[ r = q - 8 \]
Now, substitute \( r = q - 8 \) into the first equation:
\[ p + q + (q - 8) = 100 \]
\[ p + 2q - 8 = 100 \]
\[ p + 2q = 108 \]
Next, subtract the second equation from this result:
\[ (p + 2q) - (p + q) = 108 - 62 \]
\[ q = 46 \]
Now, substitute \( q = 46 \) back into the second equation:
\[ p + 46 = 62 \]
\[ p = 16 \]
Finally, substitute \( q = 46 \) into \( r = q - 8 \):
\[ r = 46 - 8 \]
\[ r = 38 \]
Thus, the number of students who stand next to exactly one student from club \( A \) and one student from club \( B \) is \( p = 16 \).
The final answer is \( \boxed{16} \).
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a,b, c$ denote the real numbers such that $1 \le a, b, c\le 2$.
Consider $T = (a - b)^{2018} + (b - c)^{2018} + (c - a)^{2018}$.
Determine the largest possible value of $T$.
|
1. Given the conditions \(1 \le a, b, c \le 2\), we need to determine the largest possible value of \(T = (a - b)^{2018} + (b - c)^{2018} + (c - a)^{2018}\).
2. First, note that \(|a - b| \le 1\), \(|b - c| \le 1\), and \(|c - a| \le 1\) because \(a, b, c\) are all within the interval \([1, 2]\).
3. Since \(2018\) is an even number, \((x)^{2018} = |x|^{2018}\). Therefore, we can rewrite \(T\) as:
\[
T = |a - b|^{2018} + |b - c|^{2018} + |c - a|^{2018}
\]
4. To maximize \(T\), we need to consider the maximum values of \(|a - b|\), \(|b - c|\), and \(|c - a|\). Since \(|a - b|, |b - c|, |c - a| \le 1\), the maximum value of \(|x|^{2018}\) for any \(x\) in \([-1, 1]\) is \(1\).
5. We need to check if it is possible for \(T\) to achieve the value of \(2\). Assume \(a, b, c\) are such that the differences \(|a - b|\), \(|b - c|\), and \(|c - a|\) are maximized. Without loss of generality, assume \(a \ge b \ge c\).
6. If \(a = 2\), \(b = 1\), and \(c = 1\), then:
\[
|a - b| = |2 - 1| = 1, \quad |b - c| = |1 - 1| = 0, \quad |c - a| = |1 - 2| = 1
\]
7. Substituting these values into \(T\), we get:
\[
T = 1^{2018} + 0^{2018} + 1^{2018} = 1 + 0 + 1 = 2
\]
8. Therefore, the maximum value of \(T\) is indeed \(2\).
The final answer is \(\boxed{2}\).
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $f$ be a polynomial such that, for all real number $x$, $f(-x^2-x-1) = x^4 + 2x^3 + 2022x^2 + 2021x + 2019$.
Compute $f(2018)$.
|
1. We start with the given functional equation for the polynomial \( f \):
\[
f(-x^2 - x - 1) = x^4 + 2x^3 + 2022x^2 + 2021x + 2019
\]
2. To find \( f \), we assume that \( f \) is a polynomial of the form \( f(t) \). We need to express \( f(t) \) in terms of \( t \) such that \( t = -x^2 - x - 1 \).
3. Let \( t = -x^2 - x - 1 \). Then we need to express \( x^4 + 2x^3 + 2022x^2 + 2021x + 2019 \) in terms of \( t \).
4. Notice that:
\[
t = -x^2 - x - 1 \implies x^2 + x + t + 1 = 0
\]
Solving for \( x \) in terms of \( t \), we get:
\[
x^2 + x + t + 1 = 0
\]
This is a quadratic equation in \( x \). The roots of this equation are:
\[
x = \frac{-1 \pm \sqrt{1 - 4(t + 1)}}{2} = \frac{-1 \pm \sqrt{-3 - 4t}}{2}
\]
5. We need to express \( x^4 + 2x^3 + 2022x^2 + 2021x + 2019 \) in terms of \( t \). However, a simpler approach is to recognize that the polynomial \( f(t) \) must match the given polynomial for all \( x \).
6. Given that \( f(-x^2 - x - 1) = x^4 + 2x^3 + 2022x^2 + 2021x + 2019 \), we can try to find a polynomial \( f(t) \) that fits this form. By inspection, we can see that:
\[
f(t) = t^2 - 2019t - 1
\]
satisfies the given equation because:
\[
f(-x^2 - x - 1) = (-x^2 - x - 1)^2 - 2019(-x^2 - x - 1) - 1
\]
Simplifying the right-hand side:
\[
(-x^2 - x - 1)^2 = x^4 + 2x^3 + x^2 + 2x^3 + 2x^2 + 2 + x^2 + 2x + 1 = x^4 + 4x^3 + 5x^2 + 2x + 1
\]
\[
-2019(-x^2 - x - 1) = 2019x^2 + 2019x + 2019
\]
\[
f(-x^2 - x - 1) = x^4 + 4x^3 + 5x^2 + 2x + 1 + 2019x^2 + 2019x + 2019 - 1 = x^4 + 4x^3 + 2024x^2 + 2021x + 2019
\]
This matches the given polynomial, confirming that \( f(t) = t^2 - 2019t - 1 \).
7. Now, we need to compute \( f(2018) \):
\[
f(2018) = 2018^2 - 2019 \cdot 2018 - 1
\]
\[
= 2018^2 - 2018 \cdot 2019 - 1
\]
\[
= 2018(2018 - 2019) - 1
\]
\[
= 2018(-1) - 1
\]
\[
= -2018 - 1
\]
\[
= -2019
\]
The final answer is \(\boxed{-2019}\).
|
-2019
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Given an arbitrary angle $\alpha$, compute
$cos \alpha + cos \big( \alpha +\frac{2\pi }{3 }\big) + cos \big( \alpha +\frac{4\pi }{3 }\big)$ and $sin \alpha + sin \big( \alpha +\frac{2\pi }{3 } \big) + sin \big( \alpha +\frac{4\pi }{3 } \big)$ .
Generalize this result and justify your answer.
|
1. **Given Problem:**
We need to compute the following sums:
\[
\cos \alpha + \cos \left( \alpha + \frac{2\pi}{3} \right) + \cos \left( \alpha + \frac{4\pi}{3} \right)
\]
and
\[
\sin \alpha + \sin \left( \alpha + \frac{2\pi}{3} \right) + \sin \left( \alpha + \frac{4\pi}{3} \right).
\]
2. **Generalization:**
We generalize the problem to:
\[
\sum_{k=0}^{n-1} \cos \left( \alpha + \frac{2k\pi}{n} \right)
\]
and
\[
\sum_{k=0}^{n-1} \sin \left( \alpha + \frac{2k\pi}{n} \right).
\]
3. **Using Complex Exponentials:**
Let \( x = e^{i\alpha} \) and \( \omega = e^{i \frac{2\pi}{n}} \). Note that \( \omega \) is a primitive \( n \)-th root of unity.
4. **Sum of Cosines:**
\[
\sum_{k=0}^{n-1} \cos \left( \alpha + \frac{2k\pi}{n} \right) = \sum_{k=0}^{n-1} \Re \left( x \omega^k \right)
\]
where \( \Re \) denotes the real part.
5. **Sum of Complex Exponentials:**
\[
\sum_{k=0}^{n-1} x \omega^k = x \sum_{k=0}^{n-1} \omega^k
\]
Since \( \omega \) is a primitive \( n \)-th root of unity, we know that:
\[
\sum_{k=0}^{n-1} \omega^k = 0
\]
(This follows from the geometric series sum formula for roots of unity.)
6. **Real Part of the Sum:**
\[
\Re \left( x \sum_{k=0}^{n-1} \omega^k \right) = \Re (x \cdot 0) = \Re (0) = 0
\]
Therefore,
\[
\sum_{k=0}^{n-1} \cos \left( \alpha + \frac{2k\pi}{n} \right) = 0.
\]
7. **Sum of Sines:**
Similarly, for the sines, we have:
\[
\sum_{k=0}^{n-1} \sin \left( \alpha + \frac{2k\pi}{n} \right) = \sum_{k=0}^{n-1} \Im \left( x \omega^k \right)
\]
where \( \Im \) denotes the imaginary part.
8. **Imaginary Part of the Sum:**
\[
\Im \left( x \sum_{k=0}^{n-1} \omega^k \right) = \Im (x \cdot 0) = \Im (0) = 0
\]
Therefore,
\[
\sum_{k=0}^{n-1} \sin \left( \alpha + \frac{2k\pi}{n} \right) = 0.
\]
Conclusion:
\[
\cos \alpha + \cos \left( \alpha + \frac{2\pi}{3} \right) + \cos \left( \alpha + \frac{4\pi}{3} \right) = 0
\]
and
\[
\sin \alpha + \sin \left( \alpha + \frac{2\pi}{3} \right) + \sin \left( \alpha + \frac{4\pi}{3} \right) = 0.
\]
The final answer is \( \boxed{ 0 } \).
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
$(a)$ Let $x, y$ be integers, not both zero. Find the minimum possible value of $|5x^2 + 11xy - 5y^2|$.
$(b)$ Find all positive real numbers $t$ such that $\frac{9t}{10}=\frac{[t]}{t - [t]}$.
|
1. Let \( f(x, y) = 5x^2 + 11xy - 5y^2 \). We need to find the minimum possible value of \( |f(x, y)| \) for integers \( x \) and \( y \) not both zero.
2. Note that \( -f(x, y) = f(y, -x) \). Therefore, it suffices to prove that \( f(x, y) = k \) for \( k \in \{1, 2, 3, 4\} \) has no integral solution.
3. First, consider \( f(x, y) \) modulo 2. If \( f(x, y) \) is even, then both \( x \) and \( y \) must be even. This implies \( 4 \mid f(x, y) \), ruling out \( k = 2 \). Additionally, if \( f(x, y) = 4 \) has a solution, then \( f(x, y) = 1 \) must also have a solution.
4. Next, consider \( f(x, y) \) modulo 3. If \( f(x, y) \equiv 0 \pmod{3} \), then \( x^2 + xy - y^2 \equiv 0 \pmod{3} \). If neither \( x \) nor \( y \) is divisible by 3, then \( x^2 - y^2 \equiv 0 \pmod{3} \) implies \( xy \equiv 0 \pmod{3} \), a contradiction. Therefore, if 3 divides one of \( x \) or \( y \), it must divide both, implying \( 9 \mid f(x, y) \), ruling out \( k = 3 \).
5. Now, consider \( f(x, y) = 1 \). This equation is equivalent to:
\[
(10x + 11y)^2 - 221y^2 = 20
\]
Let \( A = 10x + 11y \). Then:
\[
A^2 \equiv 20 \pmod{13}
\]
However, \( 20 \) is a quadratic nonresidue modulo 13, as shown by:
\[
\left( \frac{20}{13} \right) = \left( \frac{5}{13} \right) = \left( \frac{13}{5} \right) = \left( \frac{3}{5} \right) = -1
\]
Therefore, \( 20 \) is not a quadratic residue modulo 13, and no solution exists for \( f(x, y) = 1 \).
6. Consequently, \( |f(x, y)| \geq 5 \). The minimum value is achieved when \( x = 1 \) and \( y = 0 \), giving \( f(1, 0) = 5 \).
The final answer is \( \boxed{5} \).
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ u_1$, $ u_2$, $ \ldots$, $ u_{1987}$ be an arithmetic progression with $ u_1 \equal{} \frac {\pi}{1987}$ and the common difference $ \frac {\pi}{3974}$. Evaluate
\[ S \equal{} \sum_{\epsilon_i\in\left\{ \minus{} 1, 1\right\}}\cos\left(\epsilon_1 u_1 \plus{} \epsilon_2 u_2 \plus{} \cdots \plus{} \epsilon_{1987} u_{1987}\right)
\]
|
1. Given the arithmetic progression \( u_1, u_2, \ldots, u_{1987} \) with \( u_1 = \frac{\pi}{1987} \) and common difference \( \frac{\pi}{3974} \), we can express the general term \( u_n \) as:
\[
u_n = u_1 + (n-1) \cdot \frac{\pi}{3974} = \frac{\pi}{1987} + (n-1) \cdot \frac{\pi}{3974}
\]
2. We need to evaluate the sum:
\[
S = \sum_{\epsilon_i \in \{-1, 1\}} \cos\left(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987}\right)
\]
3. Notice that for each term in the sum, \( \cos\left(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987}\right) \), there exists a corresponding term \( \cos\left(-\epsilon_1 u_1 - \epsilon_2 u_2 - \cdots - \epsilon_{1987} u_{1987}\right) \).
4. Using the property of the cosine function, \( \cos(x) = \cos(-x) \), we can pair each term with its corresponding negative:
\[
\cos\left(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987}\right) = \cos\left(-(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987})\right)
\]
5. Since \( \epsilon_i \) can be either \( 1 \) or \( -1 \), for every combination of \( \epsilon_i \), there is an equal and opposite combination. Therefore, each term in the sum \( S \) is paired with its negative counterpart.
6. The sum of each pair is zero:
\[
\cos\left(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987}\right) + \cos\left(-(\epsilon_1 u_1 + \epsilon_2 u_2 + \cdots + \epsilon_{1987} u_{1987})\right) = 0
\]
7. Since every term in the sum \( S \) cancels out with its corresponding negative term, the total sum \( S \) is zero.
\[
S = 0
\]
The final answer is \(\boxed{0}\)
|
0
|
Combinatorics
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
$1991$ students sit around a circle and play the following game. Starting from some student $A$ and counting clockwise, each student on turn says a number. The numbers are $1,2,3,1,2,3,...$ A student who says $2$ or $3$ must leave the circle. The game is over when there is only one student left. What position was the remaining student sitting at the beginning of the game?
|
1. **Initial Setup**: We start with 1991 students sitting in a circle, each assigned a number from 1 to 1991. The students count off in sequence, saying the numbers 1, 2, 3 repeatedly. Students who say 2 or 3 leave the circle.
2. **First Round**: In the first round, every third student remains. Therefore, the positions of the remaining students are:
\[
1, 4, 7, 10, \ldots, 1990
\]
This sequence can be described by the arithmetic progression:
\[
a_n = 1 + 3(n-1) = 3n - 2
\]
where \( n \) ranges from 1 to 664 (since \( 3 \times 664 - 2 = 1990 \)).
3. **Second Round**: In the second round, we again count off in threes. The new sequence of remaining students is:
\[
4, 13, 22, \ldots, 1984
\]
This sequence can be described by:
\[
b_n = 4 + 9(n-1) = 9n - 5
\]
where \( n \) ranges from 1 to 221 (since \( 9 \times 221 - 5 = 1984 \)).
4. **Third Round**: Continuing this process, the sequence of remaining students is:
\[
13, 40, 67, \ldots, 1975
\]
This sequence can be described by:
\[
c_n = 13 + 27(n-1) = 27n - 14
\]
where \( n \) ranges from 1 to 74 (since \( 27 \times 74 - 14 = 1975 \)).
5. **Fourth Round**: The sequence of remaining students is:
\[
40, 121, 202, \ldots, 1963
\]
This sequence can be described by:
\[
d_n = 40 + 81(n-1) = 81n - 41
\]
where \( n \) ranges from 1 to 25 (since \( 81 \times 25 - 41 = 1963 \)).
6. **Fifth Round**: The sequence of remaining students is:
\[
121, 364, 607, \ldots, 1894
\]
This sequence can be described by:
\[
e_n = 121 + 243(n-1) = 243n - 122
\]
where \( n \) ranges from 1 to 8 (since \( 243 \times 8 - 122 = 1894 \)).
7. **Sixth Round**: The sequence of remaining students is:
\[
364, 1093, 1822
\]
This sequence can be described by:
\[
f_n = 364 + 729(n-1) = 729n - 365
\]
where \( n \) ranges from 1 to 3 (since \( 729 \times 3 - 365 = 1822 \)).
8. **Seventh Round**: The sequence of remaining students is:
\[
1093
\]
This sequence can be described by:
\[
g_n = 1093 + 2187(n-1) = 2187n - 1094
\]
where \( n = 1 \).
Therefore, the last remaining student is at position 1093.
The final answer is \(\boxed{1093}\).
|
1093
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Define the sequences $a_{0}, a_{1}, a_{2}, ...$ and $b_{0}, b_{1}, b_{2}, ...$ by $a_{0}= 2, b_{0}= 1, a_{n+1}= 2a_{n}b_{n}/(a_{n}+b_{n}), b_{n+1}= \sqrt{a_{n+1}b_{n}}$. Show that the two sequences converge to the same limit, and find the limit.
|
1. Define the sequences \(a_n\) and \(b_n\) as given:
\[
a_0 = 2, \quad b_0 = 1
\]
\[
a_{n+1} = \frac{2a_n b_n}{a_n + b_n}, \quad b_{n+1} = \sqrt{a_{n+1} b_n}
\]
2. Define the ratio sequence \(c_n = \frac{b_n}{a_n}\). Initially, we have:
\[
c_0 = \frac{b_0}{a_0} = \frac{1}{2}
\]
3. Express \(a_{n+1}\) and \(b_{n+1}\) in terms of \(c_n\):
\[
a_{n+1} = \frac{2a_n b_n}{a_n + b_n} = \frac{2a_n \cdot c_n a_n}{a_n + c_n a_n} = \frac{2a_n^2 c_n}{a_n(1 + c_n)} = \frac{2a_n c_n}{1 + c_n}
\]
\[
b_{n+1} = \sqrt{a_{n+1} b_n} = \sqrt{\left(\frac{2a_n c_n}{1 + c_n}\right) b_n} = \sqrt{\frac{2a_n c_n \cdot c_n a_n}{1 + c_n}} = \sqrt{\frac{2a_n^2 c_n^2}{1 + c_n}} = a_n \sqrt{\frac{2c_n^2}{1 + c_n}}
\]
4. Now, compute \(c_{n+1}\):
\[
c_{n+1} = \frac{b_{n+1}}{a_{n+1}} = \frac{a_n \sqrt{\frac{2c_n^2}{1 + c_n}}}{\frac{2a_n c_n}{1 + c_n}} = \frac{\sqrt{\frac{2c_n^2}{1 + c_n}}}{\frac{2c_n}{1 + c_n}} = \frac{\sqrt{2c_n^2}}{2c_n} \cdot \sqrt{1 + c_n} = \sqrt{\frac{1 + c_n}{2}}
\]
5. We now have the recursive relation for \(c_n\):
\[
c_{n+1} = \sqrt{\frac{1 + c_n}{2}}
\]
6. To find the limit of \(c_n\), let \(L\) be the limit of \(c_n\) as \(n \to \infty\). Then:
\[
L = \sqrt{\frac{1 + L}{2}}
\]
7. Square both sides to solve for \(L\):
\[
L^2 = \frac{1 + L}{2}
\]
\[
2L^2 = 1 + L
\]
\[
2L^2 - L - 1 = 0
\]
8. Solve the quadratic equation:
\[
L = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}
\]
\[
L = \frac{4}{4} = 1 \quad \text{or} \quad L = \frac{-2}{4} = -\frac{1}{2}
\]
9. Since \(c_n = \frac{b_n}{a_n}\) and both \(a_n\) and \(b_n\) are positive, \(c_n\) must be positive. Therefore, the limit \(L\) must be \(1\).
10. Since \(c_n \to 1\), we have:
\[
\frac{b_n}{a_n} \to 1 \implies b_n \to a_n
\]
11. Let \(a_n \to L\) and \(b_n \to L\). Then:
\[
a_{n+1} = \frac{2a_n b_n}{a_n + b_n} \to \frac{2L \cdot L}{L + L} = \frac{2L^2}{2L} = L
\]
\[
b_{n+1} = \sqrt{a_{n+1} b_n} \to \sqrt{L \cdot L} = L
\]
Thus, both sequences \(a_n\) and \(b_n\) converge to the same limit \(L\).
The final answer is \(\boxed{1}\).
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of functions $ f: \mathbb N\rightarrow\mathbb N$ which satisfying:
(i) $ f(1) \equal{} 1$
(ii) $ f(n)f(n \plus{} 2) \equal{} f^2(n \plus{} 1) \plus{} 1997$ for every natural numbers n.
|
1. **Initial Setup and Substitution**:
Given the function \( f: \mathbb{N} \rightarrow \mathbb{N} \) satisfying:
\[
f(1) = 1
\]
and
\[
f(n)f(n+2) = f^2(n+1) + 1997 \quad \text{for all natural numbers } n,
\]
we can replace 1997 with a prime number \( p \). Let \( f(2) = m \), then:
\[
f(3) = m^2 + p.
\]
2. **Finding \( f(4) \)**:
Using the given functional equation for \( n = 2 \):
\[
f(2)f(4) = f^2(3) + p,
\]
substituting \( f(2) = m \) and \( f(3) = m^2 + p \):
\[
mf(4) = (m^2 + p)^2 + p.
\]
Solving for \( f(4) \):
\[
f(4) = \frac{(m^2 + p)^2 + p}{m} = m^3 + 2mp + \frac{p(p+1)}{m}.
\]
For \( f(4) \) to be an integer, \( m \) must divide \( p(p+1) \).
3. **General Form and Recurrence**:
We have:
\[
f(n)f(n+2) = f^2(n+1) + p,
\]
and
\[
f(n+1)f(n+3) = f^2(n+2) + p.
\]
Subtracting these equations:
\[
f(n)f(n+2) - f(n+1)f(n+3) = f^2(n+1) + p - (f^2(n+2) + p),
\]
simplifying:
\[
f(n)f(n+2) - f(n+1)f(n+3) = f^2(n+1) - f^2(n+2).
\]
Dividing both sides by \( f(n+2)f(n+3) \):
\[
\frac{f(n)}{f(n+3)} = \frac{f(n+1)}{f(n+2)}.
\]
This implies:
\[
\frac{f(n+1)}{f(n+2)} = \frac{f(n)}{f(n+1)}.
\]
4. **Finding the General Solution**:
From the above, we get:
\[
\frac{f(2)}{f(n)} = \frac{f(2)}{f(3)} \cdot \frac{f(3)}{f(4)} \cdots \frac{f(n-1)}{f(n)} = \frac{f(1) + f(3)}{f(2) + f(4)} \cdot \frac{f(2) + f(4)}{f(3) + f(5)} \cdots \frac{f(n-2) + f(n)}{f(n-1) + f(n+1)}.
\]
Simplifying:
\[
\frac{f(2)}{f(n)} = \frac{f(1) + f(3)}{f(n-1) + f(n+1)}.
\]
This implies:
\[
f(n-1) + f(n+1) = \frac{f(1) + f(3)}{f(2)} f(n) = \frac{p+1}{m} f(n) + mf(n).
\]
5. **Considering \( m \)**:
If \( m = 1 \), then:
\[
f(n+1) = f(n) + 1.
\]
This sequence is valid and integer-valued.
If \( m > 1 \), since \( p \) and \( p+1 \) are relatively prime, \( m \) must divide either \( p \) or \( p+1 \) but not both. If \( m \) divides \( p+1 \), it works. If \( m \) divides \( p \), then \( m = p \).
6. **Checking \( m = p \)**:
For \( m = p \):
\[
f(1) = 1, \quad f(2) = p, \quad f(3) = p^2 + p = p(p+1),
\]
\[
f(4) = p^3 + 2p^2 + p + 1.
\]
For \( f(5) \):
\[
f(5) = \frac{(p^3 + 2p^2 + p + 1)^2 + p}{p(p+1)}.
\]
The numerator is not divisible by \( p \), so \( f(5) \) is not an integer.
7. **Conclusion**:
The only valid \( m \) is 1, leading to the sequence \( f(n) = n \).
The final answer is \( \boxed{1} \).
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a\geq 1$ be a real number. Put $x_{1}=a,x_{n+1}=1+\ln{(\frac{x_{n}^{2}}{1+\ln{x_{n}}})}(n=1,2,...)$. Prove that the sequence $\{x_{n}\}$ converges and find its limit.
|
1. **Define the function and sequence:**
Let \( f(x) = 1 + \ln\left(\frac{x^2}{1 + \ln x}\right) \). The sequence is defined as \( x_1 = a \) and \( x_{n+1} = f(x_n) \) for \( n \geq 1 \).
2. **Check the derivative of \( f(x) \):**
Compute the derivative \( f'(x) \):
\[
f'(x) = \frac{d}{dx} \left( 1 + \ln\left(\frac{x^2}{1 + \ln x}\right) \right)
\]
Using the chain rule and quotient rule:
\[
f'(x) = \frac{d}{dx} \left( \ln\left(\frac{x^2}{1 + \ln x}\right) \right) = \frac{1}{\frac{x^2}{1 + \ln x}} \cdot \frac{d}{dx} \left( \frac{x^2}{1 + \ln x} \right)
\]
Simplify the derivative inside:
\[
\frac{d}{dx} \left( \frac{x^2}{1 + \ln x} \right) = \frac{(1 + \ln x) \cdot 2x - x^2 \cdot \frac{1}{x}}{(1 + \ln x)^2} = \frac{2x(1 + \ln x) - x}{(1 + \ln x)^2} = \frac{x(2 + 2\ln x - 1)}{(1 + \ln x)^2} = \frac{x(1 + 2\ln x)}{(1 + \ln x)^2}
\]
Therefore:
\[
f'(x) = \frac{1 + 2\ln x}{x(1 + \ln x)}
\]
3. **Verify the bounds of \( f'(x) \):**
We need to show \( 0 < f'(x) \leq 1 \) for \( x \geq 1 \):
\[
1 + 2\ln x \leq x(1 + \ln x)
\]
Define \( g(x) = x(1 + \ln x) - (1 + 2\ln x) \). Compute \( g'(x) \):
\[
g'(x) = (1 + \ln x) + x \cdot \frac{1}{x} - \frac{2}{x} = 1 + \ln x + 1 - \frac{2}{x} = 2 + \ln x - \frac{2}{x}
\]
For \( x \geq 1 \), \( g'(x) \geq 0 \) because \( \ln x \geq 0 \) and \( -\frac{2}{x} \geq -2 \). Since \( g(1) = 0 \), \( g(x) \geq 0 \) for \( x \geq 1 \). Thus, \( 1 + 2\ln x \leq x(1 + \ln x) \).
4. **Show the sequence is decreasing and bounded below:**
\[
x_{k+1} = f(x_k) = 1 + \ln\left(\frac{x_k^2}{1 + \ln x_k}\right)
\]
Since \( f'(x) \leq 1 \), \( f(x) \) is non-increasing. Also, \( x_{k+1} \leq x_k \) implies the sequence is decreasing. Since \( x_k \geq 1 \), the sequence is bounded below by 1.
5. **Convergence and limit:**
Since \( \{x_n\} \) is decreasing and bounded below, it converges to some limit \( L \). At the limit:
\[
L = 1 + \ln\left(\frac{L^2}{1 + \ln L}\right)
\]
Solve for \( L \):
\[
L = 1 + \ln\left(\frac{L^2}{1 + \ln L}\right)
\]
Assume \( L = 1 \):
\[
1 = 1 + \ln\left(\frac{1^2}{1 + \ln 1}\right) = 1 + \ln\left(\frac{1}{1}\right) = 1 + \ln 1 = 1
\]
Thus, \( L = 1 \) is a solution. Since the function \( f(x) \) is strictly decreasing for \( x \geq 1 \), the limit must be unique.
\(\blacksquare\)
The final answer is \( \boxed{ 1 } \)
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The sequence $\{a_{n}\}_{n\geq 0}$ is defined by $a_{0}=20,a_{1}=100,a_{n+2}=4a_{n+1}+5a_{n}+20(n=0,1,2,...)$. Find the smallest positive integer $h$ satisfying $1998|a_{n+h}-a_{n}\forall n=0,1,2,...$
|
1. **Initial Conditions and Recurrence Relation:**
The sequence $\{a_n\}_{n \geq 0}$ is defined by:
\[
a_0 = 20, \quad a_1 = 100, \quad a_{n+2} = 4a_{n+1} + 5a_n + 20 \quad \text{for} \quad n \geq 0
\]
2. **Modulo 3 Analysis:**
We start by analyzing the sequence modulo 3:
\[
a_0 \equiv 20 \equiv 2 \pmod{3}
\]
\[
a_1 \equiv 100 \equiv 1 \pmod{3}
\]
Using the recurrence relation modulo 3:
\[
a_{n+2} \equiv 4a_{n+1} + 5a_n + 20 \pmod{3}
\]
Since $4 \equiv 1 \pmod{3}$ and $5 \equiv 2 \pmod{3}$, we have:
\[
a_{n+2} \equiv a_{n+1} + 2a_n + 2 \pmod{3}
\]
3. **Computing Initial Terms Modulo 3:**
\[
a_2 \equiv a_1 + 2a_0 + 2 \equiv 1 + 2 \cdot 2 + 2 \equiv 1 + 4 + 2 \equiv 7 \equiv 1 \pmod{3}
\]
\[
a_3 \equiv a_2 + 2a_1 + 2 \equiv 1 + 2 \cdot 1 + 2 \equiv 1 + 2 + 2 \equiv 5 \equiv 2 \pmod{3}
\]
\[
a_4 \equiv a_3 + 2a_2 + 2 \equiv 2 + 2 \cdot 1 + 2 \equiv 2 + 2 + 2 \equiv 6 \equiv 0 \pmod{3}
\]
\[
a_5 \equiv a_4 + 2a_3 + 2 \equiv 0 + 2 \cdot 2 + 2 \equiv 0 + 4 + 2 \equiv 6 \equiv 0 \pmod{3}
\]
\[
a_6 \equiv a_5 + 2a_4 + 2 \equiv 0 + 2 \cdot 0 + 2 \equiv 0 + 0 + 2 \equiv 2 \pmod{3}
\]
\[
a_7 \equiv a_6 + 2a_5 + 2 \equiv 2 + 2 \cdot 0 + 2 \equiv 2 + 0 + 2 \equiv 4 \equiv 1 \pmod{3}
\]
4. **Pattern Identification:**
The sequence modulo 3 is:
\[
\{2, 1, 1, 2, 0, 0, 2, 1, \ldots\}
\]
We observe that the sequence repeats every 6 terms. Therefore, $3 \mid a_{n+h} - a_n$ for all $n \geq 0$ forces $h$ to be a multiple of 6.
5. **Modulo 1998 Analysis:**
We need to find the smallest $h$ such that $1998 \mid a_{n+h} - a_n$ for all $n \geq 0$. Note that $1998 = 2 \cdot 3^3 \cdot 37$.
6. **General Form of the Sequence:**
Using the recurrence relation, we can express $a_n$ in terms of its characteristic equation. The characteristic equation of the recurrence relation is:
\[
x^2 - 4x - 5 = 0
\]
Solving for $x$, we get:
\[
x = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}
\]
Thus, the roots are $x = 5$ and $x = -1$. Therefore, the general solution is:
\[
a_n = A \cdot 5^n + B \cdot (-1)^n + C
\]
where $A$, $B$, and $C$ are constants determined by initial conditions.
7. **Finding Constants:**
Using initial conditions:
\[
a_0 = 20 \implies A + B + C = 20
\]
\[
a_1 = 100 \implies 5A - B + C = 100
\]
Solving these equations, we find $A$, $B$, and $C$.
8. **Conclusion:**
Since $1998 = 2 \cdot 3^3 \cdot 37$, and we already know $h$ must be a multiple of 6, we need to check the smallest $h$ that satisfies the divisibility by $1998$. The smallest such $h$ is $6$.
The final answer is $\boxed{6}$.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ P(x)$ be a nonzero polynomial such that, for all real numbers $ x$, $ P(x^2 \minus{} 1) \equal{} P(x)P(\minus{}x)$. Determine the maximum possible number of real roots of $ P(x)$.
|
1. **Define the polynomial and the function:**
Let \( P(x) \) be a nonzero polynomial such that for all real numbers \( x \), \( P(x^2 - 1) = P(x)P(-x) \). Define the function \( f(x) = x^2 - 1 \).
2. **Factorization of \( P(x) \):**
By the Fundamental Theorem of Algebra, \( P(x) \) can be factored as:
\[
P(x) = \prod_{i=1}^{n} (x - c_i)
\]
where \( c_i \in \mathbb{C} \) and \( n = \deg P \).
3. **Substitute \( x^2 - 1 \) into \( P(x) \):**
Given the hypothesis \( P(x^2 - 1) = P(x)P(-x) \), we substitute the factorized form:
\[
\prod_{i=1}^{n} (x^2 - 1 - c_i) = P(x)P(-x) = \prod_{i=1}^{n} (x - c_i) \prod_{i=1}^{n} (-x - c_i) = (-1)^n \prod_{i=1}^{n} (x^2 - c_i)
\]
4. **Equate the polynomials:**
Since the polynomials on both sides must be equal, we have:
\[
\prod_{i=1}^{n} (x^2 - 1 - c_i) = (-1)^n \prod_{i=1}^{n} (x^2 - c_i)
\]
5. **Analyze the degrees and roots:**
The degrees of both sides are equal, so \( n \) must be even. Let \( n = 2m \). Let \( y = x^2 \). Then the equation becomes:
\[
\prod_{i=1}^{n} (y - 1 - c_i) = (-1)^n \prod_{i=1}^{n} (y - c_i)
\]
6. **Unique Factorization Theorem:**
By the Unique Factorization Theorem, the complex numbers \( c_i \) must satisfy that \( c_i^2 \) are a permutation of the numbers \( 1 - c_i \). This implies that the roots \( c_i \) can be grouped into sets where \( c_{i,j+1}^2 = 1 + c_{i,j} \).
7. **Roots of \( f^k(x) = 0 \):**
The roots of \( P(x) \) must be roots of \( f^k(x) = 0 \). We need to show that \( f^k(x) \) has at most 4 real roots.
8. **Iterative function analysis:**
Consider the function \( f(x) = x^2 - 1 \). The fixed points of \( f(x) \) are the solutions to \( x^2 - 1 = x \), which are \( x = \frac{1 \pm \sqrt{5}}{2} \). These are the only real fixed points.
9. **Behavior of \( f^k(x) \):**
For \( k \geq 2 \), the function \( f^k(x) \) will have at most 4 real roots. This is because each iteration of \( f(x) \) maps real numbers to real numbers, and the number of real roots does not increase.
10. **Conclusion:**
Therefore, the polynomial \( P(x) \) can have at most 4 real roots.
The final answer is \(\boxed{4}\)
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ S(n)$ be the sum of decimal digits of a natural number $ n$. Find the least value of $ S(m)$ if $ m$ is an integral multiple of $ 2003$.
|
1. **Claim 1:**
- We need to show that \(1001\) is the order of \(10\) modulo \(2003\).
- The order of an integer \(a\) modulo \(n\) is the smallest positive integer \(k\) such that \(a^k \equiv 1 \pmod{n}\).
- To prove this, we need to show that \(10^{1001} \equiv 1 \pmod{2003}\) and that no smaller positive integer \(k\) satisfies this condition.
- **Proof:**
- Consider the set of all positive divisors of \(1001\). The divisors are \(1, 7, 11, 13, 77, 91, 143, 1001\).
- We need to check if any of these divisors (other than \(1001\)) is the order of \(10\) modulo \(2003\).
- After calculations, we find that none of these numbers (except \(1001\)) is an order modulo \(2003\).
- We verify that \(10^{1001} \equiv 1 \pmod{2003}\) through detailed calculations:
\[
10^4 \equiv -15 \pmod{2003}
\]
\[
15^4 \equiv 550 \pmod{2003}
\]
\[
10^{16} \equiv 550 \pmod{2003}
\]
\[
550^2 \equiv 47 \pmod{2003}
\]
\[
10^{32} \equiv 47 \pmod{2003}
\]
\[
10^{64} \equiv 206 \pmod{2003}
\]
\[
10^{96} \equiv 206 \cdot 47 \equiv 1670 \pmod{2003}
\]
\[
10^{100} \equiv 1670 \cdot (-15) \equiv 989 \pmod{2003}
\]
\[
10^{200} \equiv 989^2 \equiv 657 \pmod{2003}
\]
\[
10^{400} \equiv 1004 \pmod{2003}
\]
\[
10^{800} \equiv 507 \pmod{2003}
\]
\[
10^{1000} \equiv 657 \cdot 507 \equiv 601 \pmod{2003}
\]
\[
10^{1001} \equiv 6010 \equiv 1 \pmod{2003}
\]
- Therefore, \(1001\) is indeed the order of \(10\) modulo \(2003\).
2. **Claim 2:**
- There is no number \(n\) with \(S(n) = 2\) that is divisible by \(2003\).
- **Proof:**
- Assume the contrary, that there exists \(n\) of the form \(2 \cdot 10^k\) or \(10^k + 1\) that is divisible by \(2003\).
- The first case \(2 \cdot 10^k\) is clearly impossible because \(2003\) is not divisible by \(2\).
- For the second case, if \(10^k \equiv -1 \pmod{2003}\), then \(2k\) must be divisible by \(1001\). Thus, \(k\) must be divisible by \(1001\), but then \(10^k \equiv 1 \pmod{2003}\), which is a contradiction.
3. **Claim 3:**
- There exist natural numbers \(a, b\) such that \(10^a + 10^b + 1\) is divisible by \(2003\).
- **Proof:**
- Assume the contrary. Consider the sets of remainders modulo \(2003\):
\[
A = \{10^a \mid a = 1, \dots, 1001\}
\]
\[
B = \{-10^b - 1 \mid b = 1, \dots, 1001\}
\]
- According to Claim 1, neither of the elements from one set gives the same remainder.
- Claim 2 implies that neither of these two sets contains \(0\) or \(2002\), so in total, these two sets have at most \(2001\) elements.
- Therefore, \(A\) and \(B\) must have some common element, meaning:
\[
10^a \equiv -10^b - 1 \pmod{2003}
\]
\[
10^a + 10^b + 1 \equiv 0 \pmod{2003}
\]
- This number has a sum of digits precisely equal to \(3\).
\(\blacksquare\)
The final answer is \( \boxed{ 3 } \).
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $A_1A_2A_3A_4A_5A_6A_7A_8$ be convex 8-gon (no three diagonals concruent).
The intersection of arbitrary two diagonals will be called "button".Consider the convex quadrilaterals formed by four vertices of $A_1A_2A_3A_4A_5A_6A_7A_8$ and such convex quadrilaterals will be called "sub quadrilaterals".Find the smallest $n$ satisfying:
We can color n "button" such that for all $i,k \in\{1,2,3,4,5,6,7,8\},i\neq k,s(i,k)$ are the same where $s(i,k)$ denote the number of the "sub quadrilaterals" has $A_i,A_k$ be the vertices and the intersection of two its diagonals is "button".
|
1. **Understanding the Problem:**
We need to find the smallest number \( n \) such that we can color \( n \) "buttons" (intersections of diagonals) in a convex 8-gon \( A_1A_2A_3A_4A_5A_6A_7A_8 \) so that for any pair of vertices \( A_i \) and \( A_k \), the number of sub-quadrilaterals containing \( A_i \) and \( A_k \) and having a "button" at the intersection of their diagonals is the same.
2. **Counting Pairs and Buttons:**
Each "button" is formed by the intersection of the diagonals of a sub-quadrilateral. A sub-quadrilateral is formed by choosing 4 vertices out of 8, which can be done in \( \binom{8}{4} \) ways. Each sub-quadrilateral has exactly one "button".
3. **Pairs of Vertices:**
There are \( \binom{8}{2} = 28 \) pairs of vertices in an 8-gon. Each pair of vertices \( (A_i, A_k) \) must be part of the same number of sub-quadrilaterals.
4. **Equating the Number of Buttons:**
Let \( s(i, k) \) denote the number of sub-quadrilaterals that have \( A_i \) and \( A_k \) as vertices and whose diagonals intersect at a "button". We need \( s(i, k) \) to be the same for all pairs \( (i, k) \).
5. **Total Number of Buttons:**
Each sub-quadrilateral has 6 pairs of vertices (since \( \binom{4}{2} = 6 \)). If we color \( n \) buttons, then:
\[
6n = \sum_{i < j} s(i, j)
\]
Since \( s(i, j) \) is the same for all pairs, let \( s(i, j) = s \). Then:
\[
6n = 28s
\]
Solving for \( n \):
\[
n = \frac{28s}{6} = \frac{14s}{3}
\]
For \( n \) to be an integer, \( s \) must be a multiple of 3. The smallest such \( s \) is 3, giving:
\[
n = \frac{14 \times 3}{3} = 14
\]
6. **Verification:**
We need to verify that \( n = 14 \) buttons can be colored such that every pair of vertices is part of exactly 3 sub-quadrilaterals with a "button". Consider the following sets of vertices for the buttons:
\[
\{1, 2, 3, 4\}, \{1, 2, 5, 6\}, \{1, 2, 7, 8\}, \{3, 4, 5, 6\}, \{3, 4, 7, 8\}, \{5, 6, 7, 8\},
\]
\[
\{1, 3, 5, 7\}, \{1, 3, 6, 8\}, \{1, 4, 5, 8\}, \{1, 4, 6, 7\}, \{2, 3, 5, 8\}, \{2, 3, 6, 7\}, \{2, 4, 5, 7\}, \{2, 4, 6, 8\}
\]
Each pair of vertices appears in exactly 3 of these sets, satisfying the condition.
The final answer is \( \boxed{14} \)
|
14
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $S$ be a set of 2006 numbers. We call a subset $T$ of $S$ [i]naughty[/i] if for any two arbitrary numbers $u$, $v$ (not neccesary distinct) in $T$, $u+v$ is [i]not[/i] in $T$. Prove that
1) If $S=\{1,2,\ldots,2006\}$ every naughty subset of $S$ has at most 1003 elements;
2) If $S$ is a set of 2006 arbitrary positive integers, there exists a naughty subset of $S$ which has 669 elements.
|
### Part 1: Prove that if \( S = \{1, 2, \ldots, 2006\} \), every naughty subset of \( S \) has at most 1003 elements.
1. **Definition and Initial Observation**:
- A subset \( T \) of \( S \) is called *naughty* if for any two arbitrary numbers \( u, v \in T \), \( u + v \notin T \).
- Consider the set \( S = \{1, 2, \ldots, 2006\} \).
2. **Upper Bound on Naughty Subset**:
- Suppose \( T \) is a naughty subset of \( S \) with more than 1003 elements, i.e., \( |T| > 1003 \).
- Since \( T \) is naughty, for any \( u, v \in T \), \( u + v \notin T \).
3. **Sum of Elements**:
- The maximum element in \( S \) is 2006.
- If \( T \) contains more than 1003 elements, then there must be at least one pair \( (u, v) \) in \( T \) such that \( u + v \leq 2006 \).
4. **Contradiction**:
- If \( |T| > 1003 \), then by the pigeonhole principle, there must be at least one pair \( (u, v) \) such that \( u + v \leq 2006 \).
- This contradicts the definition of a naughty subset, where \( u + v \notin T \).
5. **Conclusion**:
- Therefore, the maximum number of elements in a naughty subset \( T \) of \( S \) is at most 1003.
\[
\boxed{1003}
\]
### Part 2: Prove that if \( S \) is a set of 2006 arbitrary positive integers, there exists a naughty subset of \( S \) which has 669 elements.
1. **Definition and Initial Observation**:
- Consider \( S \) as a set of 2006 arbitrary positive integers.
- We need to show that there exists a naughty subset of \( S \) with at least 669 elements.
2. **Construction of Naughty Subset**:
- We can use a greedy algorithm to construct a naughty subset \( T \).
- Start with an empty set \( T \).
- Iterate through the elements of \( S \) and add an element \( x \) to \( T \) if for all \( y \in T \), \( x + y \notin T \).
3. **Size of Naughty Subset**:
- By carefully selecting elements, we can ensure that \( T \) remains naughty.
- Since we are adding elements to \( T \) only if they do not violate the naughty condition, we can continue this process until \( T \) has at least 669 elements.
4. **Conclusion**:
- By the construction method, we can always find a naughty subset \( T \) of \( S \) with at least 669 elements.
\[
\boxed{669}
\]
|
669
|
Combinatorics
|
proof
|
Yes
|
Yes
|
aops_forum
| false
|
Given a regular 2007-gon. Find the minimal number $k$ such that: Among every $k$ vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.
|
1. **Define the vertices of the polygon**: Let the vertices of the regular 2007-gon be labeled as \( A_1, A_2, \ldots, A_{2007} \).
2. **Objective**: We need to find the minimal number \( k \) such that among any \( k \) vertices of the polygon, there always exist 4 vertices forming a convex quadrilateral with 3 sides being sides of the polygon.
3. **Pigeonhole Principle Application**: To ensure that any set of \( k \) vertices contains at least 4 consecutive vertices, we can use the pigeonhole principle.
4. **Stronger Form of Pigeonhole Principle**: According to the stronger form of the pigeonhole principle, if we divide the 2007 vertices into groups of 3, we get:
\[
\left\lceil \frac{2007}{3} \right\rceil = 669
\]
This means that if we select more than \( 3 \times 669 = 2007 \) vertices, we are guaranteed to have at least one group of 4 consecutive vertices.
5. **Inequality Setup**: To ensure that any set of \( k \) vertices contains at least 4 consecutive vertices, we need:
\[
4k > 3 \times 2007
\]
Simplifying this inequality:
\[
4k > 6021 \implies k > \frac{6021}{4} = 1505.25
\]
Since \( k \) must be an integer, we round up to the next whole number:
\[
k \ge 1506
\]
6. **Verification**: To show that \( k = 1506 \) works, consider any set of 1506 vertices. By the pigeonhole principle, these vertices must include at least one set of 4 consecutive vertices, ensuring that there exists a convex quadrilateral with 3 sides being sides of the polygon.
Conclusion:
\[
\boxed{1506}
\]
|
1506
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Determine the number of solutions of the simultaneous equations $ x^2 \plus{} y^3 \equal{} 29$ and $ \log_3 x \cdot \log_2 y \equal{} 1.$
|
1. **Rewrite the given equations:**
The given equations are:
\[
x^2 + y^3 = 29
\]
and
\[
\log_3 x \cdot \log_2 y = 1.
\]
2. **Express one variable in terms of the other using the logarithmic equation:**
From the second equation, we have:
\[
\log_3 x \cdot \log_2 y = 1.
\]
Let \( \log_3 x = a \). Then \( x = 3^a \).
Also, let \( \log_2 y = b \). Then \( y = 2^b \).
The equation becomes:
\[
a \cdot b = 1.
\]
Thus, \( b = \frac{1}{a} \).
3. **Substitute \( x \) and \( y \) in terms of \( a \) into the first equation:**
Substitute \( x = 3^a \) and \( y = 2^{1/a} \) into the first equation:
\[
(3^a)^2 + (2^{1/a})^3 = 29.
\]
Simplify the equation:
\[
9^a + 8^{1/a} = 29.
\]
4. **Analyze the equation \( 9^a + 8^{1/a} = 29 \):**
We need to find the values of \( a \) that satisfy this equation. This is a transcendental equation and can be challenging to solve analytically. However, we can analyze it graphically or numerically.
5. **Graphical or numerical solution:**
By plotting the functions \( 9^a \) and \( 29 - 8^{1/a} \) on the same graph, we can find the points of intersection. Alternatively, we can use numerical methods to solve for \( a \).
6. **Determine the number of solutions:**
Upon graphing or using numerical methods, we find that there are exactly two values of \( a \) that satisfy the equation \( 9^a + 8^{1/a} = 29 \). These correspond to two pairs of \( (x, y) \) values.
The final answer is \( \boxed{2} \).
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ m \equal{} 2007^{2008}$, how many natural numbers n are there such that $ n < m$ and $ n(2n \plus{} 1)(5n \plus{} 2)$ is divisible by $ m$ (which means that $ m \mid n(2n \plus{} 1)(5n \plus{} 2)$) ?
|
1. **Define \( m \) and factorize it:**
\[
m = 2007^{2008}
\]
We can factorize \( 2007 \) as:
\[
2007 = 3^2 \times 223
\]
Therefore,
\[
m = (3^2 \times 223)^{2008} = 3^{4016} \times 223^{2008}
\]
Let \( a = 3^{4016} \) and \( b = 223^{2008} \). Thus, \( m = ab \) and \( \gcd(a, b) = 1 \).
2. **Conditions for divisibility:**
We need \( n(2n + 1)(5n + 2) \) to be divisible by \( m \). This means:
\[
m \mid n(2n + 1)(5n + 2)
\]
Since \( a \) and \( b \) are coprime, at least one of \( n \), \( 2n + 1 \), or \( 5n + 2 \) must be divisible by \( a \), and at least one of them must be divisible by \( b \).
3. **Set up congruences:**
We need to solve the following congruences:
\[
sn + t \equiv 0 \pmod{a}
\]
\[
un + v \equiv 0 \pmod{b}
\]
where \( (s, t) \) and \( (u, v) \) are one of \( (1, 0) \), \( (2, 1) \), and \( (5, 2) \).
4. **Uniqueness of solutions:**
Since \( s \) and \( u \) are relatively prime to both \( a \) and \( b \), each system of linear congruences has a unique solution modulo \( m \). There are 9 combinations for \( (s, t) \) and \( (u, v) \), leading to 9 potential solutions for \( n \).
5. **Check for overlap:**
We need to ensure that no two solutions coincide. If two solutions coincide, then \( a \) or \( b \) would divide both \( sn + t \) and \( un + v \) for different pairs \( (s, t) \) and \( (u, v) \), which is not possible since these numbers are relatively prime.
6. **Special case for \( n = 0 \):**
When both \( (s, t) = (1, 0) \) and \( (u, v) = (1, 0) \), the solution is \( n \equiv 0 \pmod{m} \). However, since \( n < m \) and \( n \) must be a natural number, \( n = 0 \) is not allowed.
7. **Conclusion:**
Since \( n = 0 \) is not allowed, we have 8 unique solutions for \( n \) in the range \( 0 < n < m \).
The final answer is \( \boxed{ 8 } \).
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\{x\}$ be a sequence of positive reals $x_1, x_2, \ldots, x_n$, defined by: $x_1 = 1, x_2 = 9, x_3=9, x_4=1$. And for $n \geq 1$ we have:
\[x_{n+4} = \sqrt[4]{x_{n} \cdot x_{n+1} \cdot x_{n+2} \cdot x_{n+3}}.\]
Show that this sequence has a finite limit. Determine this limit.
|
To show that the sequence $\{x_n\}$ has a finite limit and to determine this limit, we will proceed as follows:
1. **Define the sequence and its properties:**
Given the sequence $\{x_n\}$ with initial values:
\[
x_1 = 1, \quad x_2 = 9, \quad x_3 = 9, \quad x_4 = 1
\]
and the recurrence relation for $n \geq 1$:
\[
x_{n+4} = \sqrt[4]{x_n \cdot x_{n+1} \cdot x_{n+2} \cdot x_{n+3}}
\]
2. **Transform the sequence using logarithms:**
Define a new sequence $\{y_n\}$ such that:
\[
y_n = \log_3 x_n
\]
This transforms the recurrence relation into:
\[
y_{n+4} = \frac{y_n + y_{n+1} + y_{n+2} + y_{n+3}}{4}
\]
with initial values:
\[
y_1 = \log_3 1 = 0, \quad y_2 = \log_3 9 = 2, \quad y_3 = \log_3 9 = 2, \quad y_4 = \log_3 1 = 0
\]
3. **Analyze the transformed sequence:**
The recurrence relation for $\{y_n\}$ is a linear homogeneous recurrence relation. We can write it in matrix form:
\[
\begin{pmatrix}
y_{n+4} \\
y_{n+3} \\
y_{n+2} \\
y_{n+1}
\end{pmatrix}
=
\begin{pmatrix}
\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
y_n \\
y_{n+1} \\
y_{n+2} \\
y_{n+3}
\end{pmatrix}
\]
4. **Find the eigenvalues of the matrix:**
The characteristic polynomial of the matrix is:
\[
\det\left(
\begin{pmatrix}
\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{pmatrix}
- \lambda I
\right) = 0
\]
Solving this, we find the eigenvalues to be $\lambda = 1, -i, i, -1$.
5. **Analyze the long-term behavior:**
The eigenvalue $\lambda = 1$ indicates that the sequence $\{y_n\}$ will converge to a constant value as $n \to \infty$. The other eigenvalues ($-i, i, -1$) have magnitudes less than or equal to 1, which means their contributions will diminish over time.
6. **Determine the limit:**
Since $\{y_n\}$ converges to a constant value, let $L$ be this limit. Then:
\[
L = \frac{L + L + L + L}{4} \implies L = L
\]
This confirms that $\{y_n\}$ converges to a constant value. Given the initial values and the symmetry, the limit $L$ must be the average of the initial values:
\[
L = \frac{0 + 2 + 2 + 0}{4} = 1
\]
7. **Transform back to the original sequence:**
Since $y_n = \log_3 x_n$ and $y_n \to 1$, we have:
\[
\log_3 x_n \to 1 \implies x_n \to 3^1 = 3
\]
The final answer is $\boxed{3}$
|
3
|
Other
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
We call a rectangle of size $2 \times 3$ (or $3 \times 2$) without one cell in corner a $P$-rectangle. We call a rectangle of size $2 \times 3$ (or $3 \times 2$) without two cells in opposite (under center of rectangle) corners a $S$-rectangle. Using some squares of size $2 \times 2$, some $P$-rectangles and some $S$-rectangles, one form one rectangle of size $1993 \times 2000$ (figures don’t overlap each other). Let $s$ denote the sum of numbers of squares and $S$-rectangles used in such tiling. Find the maximal value of $s$.
|
1. **Understanding the Problem:**
We need to tile a $1993 \times 2000$ rectangle using $2 \times 2$ squares, $P$-rectangles, and $S$-rectangles. We need to find the maximum value of $s$, where $s$ is the sum of the number of $2 \times 2$ squares and $S$-rectangles used.
2. **Area Calculation:**
The total area of the $1993 \times 2000$ rectangle is:
\[
1993 \times 2000 = 3986000
\]
3. **Area of Each Tile:**
- A $2 \times 2$ square has an area of $4$.
- A $P$-rectangle (a $2 \times 3$ or $3 \times 2$ rectangle missing one corner cell) has an area of $5$.
- An $S$-rectangle (a $2 \times 3$ or $3 \times 2$ rectangle missing two opposite corner cells) has an area of $4$.
4. **Equation for Total Area:**
Let $a$ be the number of $2 \times 2$ squares, $b$ be the number of $P$-rectangles, and $c$ be the number of $S$-rectangles. The total area equation is:
\[
4a + 5b + 4c = 3986000
\]
5. **Expression for $s$:**
We need to maximize $s = a + c$. Rearranging the total area equation, we get:
\[
4(a + c) + 5b = 3986000
\]
Let $s = a + c$. Then:
\[
4s + 5b = 3986000
\]
6. **Minimizing $b$:**
To maximize $s$, we need to minimize $b$. Since $b$ must be a non-negative integer, we solve for $b$:
\[
b = \frac{3986000 - 4s}{5}
\]
For $b$ to be an integer, $3986000 - 4s$ must be divisible by $5$. Therefore:
\[
3986000 \equiv 4s \pmod{5}
\]
Simplifying $3986000 \pmod{5}$:
\[
3986000 \div 5 = 797200 \quad \text{(remainder 0)}
\]
Thus:
\[
0 \equiv 4s \pmod{5} \implies 4s \equiv 0 \pmod{5} \implies s \equiv 0 \pmod{5}
\]
Therefore, $s$ must be a multiple of $5$.
7. **Maximizing $s$:**
The maximum value of $s$ occurs when $b$ is minimized. The smallest possible value for $b$ is $0$ (if possible). Solving for $s$ when $b = 0$:
\[
4s = 3986000 \implies s = \frac{3986000}{4} = 996500
\]
8. **Verification:**
We need to check if $b = 0$ is possible. If $b = 0$, then:
\[
4s = 3986000 \implies s = 996500
\]
This means we can tile the entire $1993 \times 2000$ rectangle using only $2 \times 2$ squares and $S$-rectangles, which is feasible.
Conclusion:
\[
\boxed{996500}
\]
|
996500
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
There are $ 25$ towns in a country. Find the smallest $ k$ for which one can set up two-direction flight routes connecting these towns so that the following conditions are satisfied:
1) from each town there are exactly $ k$ direct routes to $ k$ other towns;
2) if two towns are not connected by a direct route, then there is a town which has direct routes to these two towns.
|
1. **Define the problem in terms of graph theory:**
- We have 25 towns, which we can represent as vertices in a graph.
- We need to find the smallest \( k \) such that each vertex has exactly \( k \) edges (degree \( k \)).
- Additionally, for any two vertices that are not directly connected, there must be a vertex that is directly connected to both.
2. **Set up the problem:**
- Let \( A \) be an arbitrary town.
- Let \( R(A) \) be the set of towns directly connected to \( A \), so \( |R(A)| = k \).
- Let \( N(A) \) be the set of towns not directly connected to \( A \), so \( |N(A)| = 24 - k \).
3. **Analyze the connections:**
- For every town in \( N(A) \), there must be a direct route to at least one town in \( R(A) \) to satisfy the condition that there is a common town connected to both.
- The total number of possible connections from towns in \( R(A) \) to towns in \( N(A) \) is \( k(k-1) \) because each of the \( k \) towns in \( R(A) \) can connect to \( k-1 \) other towns (excluding \( A \)).
4. **Set up the inequality:**
- The number of connections from \( R(A) \) to \( N(A) \) must be at least \( 24 - k \) to ensure that every town in \( N(A) \) is connected to at least one town in \( R(A) \).
- Therefore, we have the inequality:
\[
k(k-1) \geq 24 - k
\]
5. **Solve the inequality:**
\[
k(k-1) \geq 24 - k
\]
\[
k^2 - k \geq 24 - k
\]
\[
k^2 \geq 24
\]
\[
k \geq \sqrt{24}
\]
\[
k \geq 5
\]
6. **Check the smallest integer \( k \):**
- Since \( k \) must be an integer, we check \( k = 5 \) and \( k = 6 \).
7. **Verify \( k = 5 \):**
- If \( k = 5 \), then \( k(k-1) = 5 \times 4 = 20 \).
- We need \( 20 \geq 24 - 5 \), which simplifies to \( 20 \geq 19 \), which is true.
- However, we need to ensure that the graph structure satisfies the condition for all pairs of towns.
8. **Verify \( k = 6 \):**
- If \( k = 6 \), then \( k(k-1) = 6 \times 5 = 30 \).
- We need \( 30 \geq 24 - 6 \), which simplifies to \( 30 \geq 18 \), which is true.
- Petersen's graph is a known graph that satisfies these conditions with \( k = 6 \).
Therefore, the smallest \( k \) that satisfies the conditions is \( k = 6 \).
The final answer is \( \boxed{6} \).
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A collection of $2000$ congruent circles is given on the plane such that no
two circles are tangent and each circle meets at least two other circles.
Let $N$ be the number of points that belong to at least two of the circles.
Find the smallest possible value of $N$.
|
1. Let \( f(n) \) denote the smallest number of points that belong to at least two of the \( n \) circles. We aim to show that \( f(n) \geq 2(n - 2) + 1 \).
2. **Base Case: \( n = 3 \)**
- For \( n = 3 \), consider three circles arranged such that each circle intersects the other two circles at distinct points. This configuration results in exactly 3 points of intersection. Therefore, \( f(3) = 3 \).
- This satisfies the inequality \( f(3) \geq 2(3 - 2) + 1 = 3 \).
3. **Inductive Hypothesis:**
- Assume that for some \( k \geq 3 \), the inequality \( f(k) \geq 2(k - 2) + 1 \) holds true.
4. **Inductive Step:**
- We need to show that \( f(k + 1) \geq 2((k + 1) - 2) + 1 \).
- Consider adding one more circle to the existing \( k \) circles. This new circle must intersect at least two of the existing circles at distinct points.
- By the inductive hypothesis, the \( k \) circles have at least \( f(k) \geq 2(k - 2) + 1 \) points of intersection.
- The new circle intersects at least two of the existing circles, adding at least 2 new points of intersection.
- Therefore, the total number of points of intersection is at least \( f(k) + 2 \).
5. **Calculation:**
\[
f(k + 1) \geq f(k) + 2 \geq 2(k - 2) + 1 + 2 = 2(k - 2) + 3 = 2(k - 1) - 1 + 3 = 2(k - 1) + 1
\]
- This completes the inductive step.
6. **Conclusion:**
- By induction, we have shown that \( f(n) \geq 2(n - 2) + 1 \) for all \( n \geq 3 \).
7. **Application to \( n = 2000 \):**
- For \( n = 2000 \), we have:
\[
f(2000) \geq 2(2000 - 2) + 1 = 2 \cdot 1998 + 1 = 3996 + 1 = 3997
\]
The final answer is \( \boxed{3997} \).
|
3997
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $A$ be the set of all permutations $a = (a_1, a_2, \ldots, a_{2003})$ of the 2003 first positive integers such that each permutation satisfies the condition: there is no proper subset $S$ of the set $\{1, 2, \ldots, 2003\}$ such that $\{a_k | k \in S\} = S.$
For each $a = (a_1, a_2, \ldots, a_{2003}) \in A$, let $d(a) = \sum^{2003}_{k=1} \left(a_k - k \right)^2.$
[b]I.[/b] Find the least value of $d(a)$. Denote this least value by $d_0$.
[b]II.[/b] Find all permutations $a \in A$ such that $d(a) = d_0$.
|
To solve the problem, we need to find the least value of \( d(a) \) for permutations \( a \) in the set \( A \), and then identify all permutations that achieve this least value.
### Part I: Finding the Least Value of \( d(a) \)
1. **Understanding the Condition**:
The condition states that there is no proper subset \( S \) of \( \{1, 2, \ldots, 2003\} \) such that \( \{a_k \mid k \in S\} = S \). This implies that the permutation \( a \) must be such that no subset of indices maps exactly to itself.
2. **Rewriting the Problem**:
We need to minimize \( d(a) = \sum_{k=1}^{2003} (a_k - k)^2 \). This is equivalent to minimizing the sum of squared differences between the permutation \( a \) and the identity permutation \( (1, 2, \ldots, 2003) \).
3. **Using Symmetry and Properties of Permutations**:
Consider the sum of squared differences for a permutation \( a \). The minimum value of \( d(a) \) occurs when the differences \( a_k - k \) are minimized. The identity permutation \( (1, 2, \ldots, 2003) \) would give \( d(a) = 0 \), but it does not satisfy the condition since the entire set \( \{1, 2, \ldots, 2003\} \) maps to itself.
4. **Analyzing the Permutations**:
To satisfy the condition, we need to consider permutations where no proper subset maps to itself. One such permutation is the cyclic permutation where each element is shifted by one position. For example, \( a = (2, 3, \ldots, 2003, 1) \).
5. **Calculating \( d(a) \) for the Cyclic Permutation**:
For the cyclic permutation \( a = (2, 3, \ldots, 2003, 1) \):
\[
d(a) = \sum_{k=1}^{2003} (a_k - k)^2 = (2-1)^2 + (3-2)^2 + \cdots + (2003-2002)^2 + (1-2003)^2
\]
\[
= 1^2 + 1^2 + \cdots + 1^2 + (1-2003)^2
\]
\[
= 2002 \cdot 1^2 + (-2002)^2 = 2002 + 2002^2 = 2002 + 4008004 = 4010006
\]
6. **Verifying the Minimum**:
We need to verify if this is indeed the minimum value. Given the constraints and the nature of the problem, the cyclic permutation provides a structure that avoids any proper subset mapping to itself and minimizes the squared differences.
### Part II: Finding All Permutations \( a \in A \) Such That \( d(a) = d_0 \)
1. **Identifying Permutations**:
The cyclic permutation \( a = (2, 3, \ldots, 2003, 1) \) achieves the minimum value \( d_0 = 4010006 \). We need to check if there are other permutations that achieve this value.
2. **Considering Other Cyclic Shifts**:
Any cyclic shift of the permutation \( (2, 3, \ldots, 2003, 1) \) will also achieve the same value. For example, \( (3, 4, \ldots, 2003, 1, 2) \), \( (4, 5, \ldots, 2003, 1, 2, 3) \), etc.
3. **General Form**:
The general form of such permutations can be written as \( a = (k+1, k+2, \ldots, 2003, 1, 2, \ldots, k) \) for \( k = 1, 2, \ldots, 2002 \).
Conclusion:
\[
\boxed{4010006}
\]
|
4010006
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let us consider a set $S = \{ a_1 < a_2 < \ldots < a_{2004}\}$, satisfying the following properties: $f(a_i) < 2003$ and $f(a_i) = f(a_j) \quad \forall i, j$ from $\{1, 2,\ldots , 2004\}$, where $f(a_i)$ denotes number of elements which are relatively prime with $a_i$. Find the least positive integer $k$ for which in every $k$-subset of $S$, having the above mentioned properties there are two distinct elements with greatest common divisor greater than 1.
|
1. **Define the Set and Function:**
Let \( S = \{ a_1 < a_2 < \ldots < a_{2004} \} \) be a set of 2004 elements. Each element \( a_i \) satisfies \( f(a_i) < 2003 \) and \( f(a_i) = f(a_j) \) for all \( i, j \) in \( \{1, 2, \ldots, 2004\} \), where \( f(a_i) \) denotes the number of elements that are relatively prime to \( a_i \).
2. **Graph Representation:**
Construct a graph \( G_{2004} \) with 2004 vertices, where each vertex \( u_i \) corresponds to an element \( a_i \) from the set \( S \). Connect vertices \( u_i \) and \( u_j \) if \( \gcd(a_i, a_j) > 1 \).
3. **Degree of Vertices:**
Since \( f(a_i) \) is the number of elements relatively prime to \( a_i \), the number of elements not relatively prime to \( a_i \) is \( 2004 - f(a_i) \). Given \( f(a_i) < 2003 \), we have \( 2004 - f(a_i) \geq 1 \). Thus, each vertex in \( G_{2004} \) has a degree \( t \geq 1 \).
4. **Finding the Least \( k \):**
We need to find the smallest \( k \) such that in every \( k \)-subset of \( S \), there are two distinct elements with a greatest common divisor greater than 1. This translates to finding the smallest \( k \) such that every subgraph \( G_k \) of \( G_{2004} \) has at least one vertex with a non-zero degree.
5. **Analyzing the Degree:**
If \( t = 2 \), then the average degree of the graph is 2. For a subgraph \( G_{1003} \), if all vertices had degree 0, they would only be connected to vertices in \( G_{1001} \) (the remaining vertices in \( G_{2004} \)). This would imply \( 1003t \) edges connecting \( G_{1003} \) to \( G_{1001} \), but since each vertex in \( G_{1001} \) has degree \( t \), there can be at most \( 1001t \) such edges, leading to a contradiction.
6. **Conclusion:**
Therefore, in every subgraph \( G_{1003} \) of \( G_{2004} \), there must be at least one vertex with a degree not less than 1. Hence, the least positive integer \( k \) for which every \( k \)-subset of \( S \) contains two distinct elements with a greatest common divisor greater than 1 is \( k = 1003 \).
The final answer is \( \boxed{1003} \)
|
1003
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $p\in \mathbb P,p>3$. Calcute:
a)$S=\sum_{k=1}^{\frac{p-1}{2}} \left[\frac{2k^2}{p}\right]-2 \cdot \left[\frac{k^2}{p}\right]$ if $ p\equiv 1 \mod 4$
b) $T=\sum_{k=1}^{\frac{p-1}{2}} \left[\frac{k^2}{p}\right]$ if $p\equiv 1 \mod 8$
|
### Part (a)
Given \( p \in \mathbb{P} \) and \( p > 3 \), we need to calculate:
\[ S = \sum_{k=1}^{\frac{p-1}{2}} \left\lfloor \frac{2k^2}{p} \right\rfloor - 2 \cdot \left\lfloor \frac{k^2}{p} \right\rfloor \]
if \( p \equiv 1 \pmod{4} \).
1. **Identify Quadratic Residues:**
Let \( r_i \) for \( 1 \le i \le \frac{p-1}{2} \) be the quadratic residues modulo \( p \). These are the distinct values of \( k^2 \mod p \) for \( k = 1, 2, \ldots, \frac{p-1}{2} \).
2. **Rewrite the Sum:**
The sum can be rewritten in terms of the quadratic residues:
\[
S = \sum_{i=1}^{\frac{p-1}{2}} \left\lfloor \frac{2r_i}{p} \right\rfloor - 2 \cdot \left\lfloor \frac{r_i}{p} \right\rfloor
\]
3. **Analyze the Terms:**
Each term \( \left\lfloor \frac{r_i}{p} \right\rfloor \) is 0 if \( r_i < p \) (which is always true since \( r_i \) are residues modulo \( p \)). Therefore, \( 2 \cdot \left\lfloor \frac{r_i}{p} \right\rfloor = 0 \).
4. **Simplify the Sum:**
The sum simplifies to:
\[
S = \sum_{i=1}^{\frac{p-1}{2}} \left\lfloor \frac{2r_i}{p} \right\rfloor
\]
5. **Determine the Value of Each Term:**
Each term \( \left\lfloor \frac{2r_i}{p} \right\rfloor \) is 0 or 1. It is 1 if \( 2r_i \ge p \) and 0 otherwise. This happens if \( r_i \ge \frac{p}{2} \).
6. **Count the Number of Terms:**
Since \( -1 \) is a quadratic residue and \( p \equiv 1 \pmod{4} \), the number of quadratic residues greater than \( \frac{p-1}{2} \) is equal to the number of those less than \( \frac{p-1}{2} \). Therefore, there are \( \frac{\frac{p-1}{2}}{2} = \frac{p-1}{4} \) such residues.
7. **Sum the Values:**
Hence, the sum \( S \) is:
\[
S = \frac{p-1}{4}
\]
### Part (b)
Given \( p \in \mathbb{P} \) and \( p > 3 \), we need to calculate:
\[ T = \sum_{k=1}^{\frac{p-1}{2}} \left\lfloor \frac{k^2}{p} \right\rfloor \]
if \( p \equiv 1 \pmod{8} \).
1. **Identify Quadratic Residues:**
Let \( r_i \) for \( 1 \le i \le \frac{p-1}{2} \) be the quadratic residues modulo \( p \). These are the distinct values of \( k^2 \mod p \) for \( k = 1, 2, \ldots, \frac{p-1}{2} \).
2. **Rewrite the Sum:**
The sum can be rewritten in terms of the quadratic residues:
\[
T = \sum_{i=1}^{\frac{p-1}{2}} \left\lfloor \frac{r_i}{p} \right\rfloor
\]
3. **Analyze the Terms:**
Each term \( \left\lfloor \frac{r_i}{p} \right\rfloor \) is 0 if \( r_i < p \) (which is always true since \( r_i \) are residues modulo \( p \)).
4. **Simplify the Sum:**
Since all terms are 0, the sum \( T \) is:
\[
T = 0
\]
The final answer is \( \boxed{\frac{p-1}{4}} \) for part (a) and \( \boxed{0} \) for part (b).
|
0
|
Number Theory
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
In the space are given $2006$ distinct points, such that no $4$ of them are coplanar. One draws a segment between each pair of points.
A natural number $m$ is called [i]good[/i] if one can put on each of these segments a positive integer not larger than $m$, so that every triangle whose three vertices are among the given points has the property that two of this triangle's sides have equal numbers put on, while the third has a larger number put on.
Find the minimum value of a [i]good[/i] number $m$.
|
1. **Define the problem and notation:**
Let \( A(n) \) be the minimum value of a good number if we have \( n \) points in space. We need to find \( A(2006) \).
2. **Initial observation:**
In the minimal case, there is obviously a segment on which the number \( 1 \) is put. Let its endpoints be \( X \) and \( Y \). Each of the other \( 2004 \) points is either linked to \( X \) by a segment whose number is \( 1 \) or to \( Y \). Otherwise, we could take a point which is linked neither to \( X \) nor to \( Y \) by a segment with number \( 1 \), as well as \( X \) and \( Y \), so we would have a triangle which would not satisfy the condition.
3. **Partition the points:**
Let the set of points which are linked to \( X \) by a segment with number \( 1 \) be \( A \) and the set of points which are linked to \( Y \) by a segment with number \( 1 \) be \( B \). No point belongs to both \( A \) and \( B \), or we would have a triangle with three equal numbers, so \( |A| + |B| = 2004 \).
4. **Connecting points between sets:**
Take any point in \( A \) and any in \( B \). On the segment which links both points must be put number \( 1 \), or we could take these two points and \( X \) (or \( Y \), if \( X \) is one of the two points) and we would have three points which don't satisfy the condition. So, each point in \( A \) is linked by a segment with number \( 1 \) with each point in \( B \).
5. **Internal connections within sets:**
Two points in \( A \) can't be linked by a segment with number \( 1 \) or we would have a triangle with three equal numbers on its sides. If we write a number greater than \( 1 \) on each other segment, and if we take two points out of \( A \) and one out of \( B \), they satisfy the condition, because on two of the three segments would be number \( 1 \) and on the third would be a greater number. Similarly, if we take two points out of \( B \) and one out of \( A \).
6. **Recursive argument:**
Now, consider we take three points out of \( A \). None of the three sides has a \( 1 \) put on it. We need \( A(|A|) \) numbers greater than \( 1 \) for these \( |A| \) points. Similarly, we need \( A(|B|) \) numbers for the \( |B| \) points in the set \( B \), but those could be the same as in set \( A \). So, we have (since if \( |A| \geq |B| \), then \( |A| \geq 1002 \)):
\[
A(2006) \geq 1 + A(1003) \geq 2 + A(502) \geq \ldots \geq 10 + A(2) \geq 11
\]
7. **Verification:**
It is easy to see that we can really do it with just \( 11 \) numbers if we take \( |A| = |B| \) and so on, so \( 11 \) is the value we look for.
The final answer is \( \boxed{11} \)
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
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