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Increasing the radius of a cylinder by $ 6$ units increased the volume by $ y$ cubic units. Increasing the altitude of the cylinder by $ 6$ units also increases the volume by $ y$ cubic units. If the original altitude is $ 2$, then the original radius is:
$ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 6\pi \qquad\textbf{(E)}\ 8$
|
1. Let the original radius of the cylinder be \( r \) and the original altitude (height) be \( h = 2 \).
2. The volume \( V \) of the cylinder is given by the formula:
\[
V = \pi r^2 h = \pi r^2 \cdot 2 = 2\pi r^2
\]
3. When the radius is increased by 6 units, the new radius is \( r + 6 \). The new volume \( V_{\text{new}} \) is:
\[
V_{\text{new}} = \pi (r + 6)^2 \cdot 2
\]
4. The increase in volume \( y \) is given by:
\[
y = V_{\text{new}} - V = \pi (r + 6)^2 \cdot 2 - 2\pi r^2
\]
5. Simplify the expression for \( y \):
\[
y = 2\pi (r + 6)^2 - 2\pi r^2
\]
\[
y = 2\pi \left( r^2 + 12r + 36 \right) - 2\pi r^2
\]
\[
y = 2\pi r^2 + 24\pi r + 72\pi - 2\pi r^2
\]
\[
y = 24\pi r + 72\pi
\]
6. When the altitude is increased by 6 units, the new altitude is \( h + 6 = 2 + 6 = 8 \). The new volume \( V_{\text{new}} \) is:
\[
V_{\text{new}} = \pi r^2 \cdot 8
\]
7. The increase in volume \( y \) is given by:
\[
y = V_{\text{new}} - V = \pi r^2 \cdot 8 - 2\pi r^2
\]
8. Simplify the expression for \( y \):
\[
y = 8\pi r^2 - 2\pi r^2
\]
\[
y = 6\pi r^2
\]
9. Equate the two expressions for \( y \):
\[
24\pi r + 72\pi = 6\pi r^2
\]
10. Divide both sides by \( 6\pi \):
\[
4r + 12 = r^2
\]
11. Rearrange the equation to form a quadratic equation:
\[
r^2 - 4r - 12 = 0
\]
12. Solve the quadratic equation using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), and \( c = -12 \):
\[
r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}
\]
\[
r = \frac{4 \pm \sqrt{16 + 48}}{2}
\]
\[
r = \frac{4 \pm \sqrt{64}}{2}
\]
\[
r = \frac{4 \pm 8}{2}
\]
13. This gives two solutions:
\[
r = \frac{4 + 8}{2} = 6 \quad \text{and} \quad r = \frac{4 - 8}{2} = -2
\]
14. Since the radius cannot be negative, we discard \( r = -2 \).
The final answer is \( \boxed{6} \).
|
6
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The number of distinct lines representing the altitudes, medians, and interior angle bisectors of a triangle that is isosceles, but not equilateral, is:
$ \textbf{(A)}\ 9\qquad
\textbf{(B)}\ 7\qquad
\textbf{(C)}\ 6\qquad
\textbf{(D)}\ 5\qquad
\textbf{(E)}\ 3$
|
1. **Identify the properties of an isosceles triangle:**
- An isosceles triangle has two equal sides and two equal angles opposite those sides.
- The altitude, median, and angle bisector from the vertex angle (the angle between the two equal sides) are the same line.
2. **Count the lines for altitudes:**
- In any triangle, there are three altitudes.
- In an isosceles triangle, the altitude from the vertex angle is the same as the median and the angle bisector from that vertex.
- Therefore, there are only two distinct altitudes from the base vertices and one from the vertex angle, making a total of 3 distinct lines for altitudes.
3. **Count the lines for medians:**
- In any triangle, there are three medians.
- In an isosceles triangle, the median from the vertex angle is the same as the altitude and the angle bisector from that vertex.
- Therefore, there are only two distinct medians from the base vertices and one from the vertex angle, making a total of 3 distinct lines for medians.
4. **Count the lines for angle bisectors:**
- In any triangle, there are three angle bisectors.
- In an isosceles triangle, the angle bisector from the vertex angle is the same as the altitude and the median from that vertex.
- Therefore, there are only two distinct angle bisectors from the base vertices and one from the vertex angle, making a total of 3 distinct lines for angle bisectors.
5. **Combine the distinct lines:**
- The altitude, median, and angle bisector from the vertex angle are the same line.
- The altitudes, medians, and angle bisectors from the base vertices are distinct.
- Therefore, we have 1 line from the vertex angle and 2 lines each from the base vertices for altitudes, medians, and angle bisectors.
6. **Count the total number of distinct lines:**
- From the vertex angle: 1 line.
- From each base vertex: 2 lines (one for each base vertex).
- Total distinct lines: \(1 + 2 + 2 = 5\).
Conclusion:
The number of distinct lines representing the altitudes, medians, and interior angle bisectors of an isosceles triangle that is not equilateral is \( \boxed{5} \).
|
5
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The numbers $ x,\,y,\,z$ are proportional to $ 2,\,3,\,5$. The sum of $ x$, $ y$, and $ z$ is $ 100$. The number $ y$ is given by the equation $ y \equal{} ax \minus{} 10$. Then $ a$ is:
$ \textbf{(A)}\ 2 \qquad
\textbf{(B)}\ \frac{3}{2}\qquad
\textbf{(C)}\ 3\qquad
\textbf{(D)}\ \frac{5}{2}\qquad
\textbf{(E)}\ 4$
|
1. Given that the numbers \( x, y, z \) are proportional to \( 2, 3, 5 \), we can write:
\[
x = 2k, \quad y = 3k, \quad z = 5k
\]
for some constant \( k \).
2. The sum of \( x, y, \) and \( z \) is given as 100:
\[
x + y + z = 100
\]
Substituting the proportional values, we get:
\[
2k + 3k + 5k = 100
\]
Simplifying the left-hand side:
\[
10k = 100
\]
Solving for \( k \):
\[
k = \frac{100}{10} = 10
\]
3. Now, substituting \( k = 10 \) back into the expressions for \( x, y, \) and \( z \):
\[
x = 2k = 2 \times 10 = 20
\]
\[
y = 3k = 3 \times 10 = 30
\]
\[
z = 5k = 5 \times 10 = 50
\]
4. We are given that \( y = ax - 10 \). Substituting the values of \( y \) and \( x \):
\[
30 = a \times 20 - 10
\]
Solving for \( a \):
\[
30 = 20a - 10
\]
Adding 10 to both sides:
\[
40 = 20a
\]
Dividing both sides by 20:
\[
a = \frac{40}{20} = 2
\]
The final answer is \( \boxed{2} \)
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The value of $x^2-6x+13$ can never be less than:
$ \textbf{(A)}\ 4 \qquad\textbf{(B)}\ 4.5 \qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 13 $
|
1. The given expression is \( x^2 - 6x + 13 \). This is a quadratic function of the form \( ax^2 + bx + c \), where \( a = 1 \), \( b = -6 \), and \( c = 13 \).
2. To find the minimum value of a quadratic function \( ax^2 + bx + c \), we use the vertex formula. The x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \).
3. Substituting \( a = 1 \) and \( b = -6 \) into the vertex formula:
\[
x = -\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3
\]
4. Now, we substitute \( x = 3 \) back into the original quadratic expression to find the y-coordinate of the vertex, which represents the minimum value of the function:
\[
x^2 - 6x + 13 \bigg|_{x=3} = 3^2 - 6 \cdot 3 + 13
\]
\[
= 9 - 18 + 13
\]
\[
= 4
\]
5. Therefore, the minimum value of the expression \( x^2 - 6x + 13 \) is 4.
The final answer is \( \boxed{4} \)
|
4
|
Inequalities
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The logarithm of $.0625$ to the base $2$ is:
$ \textbf{(A)}\ .025 \qquad\textbf{(B)}\ .25\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ -4\qquad\textbf{(E)}\ -2 $
|
1. We start with the given logarithmic expression:
\[
\log_2 0.0625
\]
2. Convert \(0.0625\) to a fraction:
\[
0.0625 = \frac{1}{16}
\]
3. Rewrite the logarithmic expression using the fraction:
\[
\log_2 \left(\frac{1}{16}\right)
\]
4. Recall that \(\frac{1}{16}\) can be written as \(2^{-4}\) because:
\[
16 = 2^4 \implies \frac{1}{16} = 2^{-4}
\]
5. Substitute \(2^{-4}\) into the logarithmic expression:
\[
\log_2 \left(2^{-4}\right)
\]
6. Use the property of logarithms that \(\log_b (b^x) = x\):
\[
\log_2 \left(2^{-4}\right) = -4
\]
The final answer is \(\boxed{-4}\).
|
-4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
On a examination of $n$ questions a student answers correctly $15$ of the first $20$. Of the remaining questions he answers one third correctly. All the questions have the same credit. If the student's mark is $50\%$, how many different values of $n$ can there be?
$ \textbf{(A)}\ 4 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ \text{the problem cannot be solved} $
|
1. Let \( n \) be the total number of questions on the examination.
2. The student answers 15 out of the first 20 questions correctly.
3. For the remaining \( n - 20 \) questions, the student answers one third correctly. Therefore, the number of correctly answered questions in the remaining part is \( \frac{1}{3}(n - 20) \).
4. The total number of correctly answered questions is:
\[
15 + \frac{1}{3}(n - 20)
\]
5. The student's mark is 50%, which means the student answered half of the total questions correctly. Therefore, we set up the equation:
\[
\frac{15 + \frac{1}{3}(n - 20)}{n} = \frac{1}{2}
\]
6. To solve for \( n \), we first clear the fraction by multiplying both sides by \( n \):
\[
15 + \frac{1}{3}(n - 20) = \frac{n}{2}
\]
7. Simplify the equation:
\[
15 + \frac{n - 20}{3} = \frac{n}{2}
\]
8. Multiply through by 6 to clear the denominators:
\[
6 \cdot 15 + 6 \cdot \frac{n - 20}{3} = 6 \cdot \frac{n}{2}
\]
\[
90 + 2(n - 20) = 3n
\]
9. Distribute and simplify:
\[
90 + 2n - 40 = 3n
\]
\[
50 + 2n = 3n
\]
10. Subtract \( 2n \) from both sides:
\[
50 = n
\]
11. Therefore, the total number of questions \( n \) is 50.
The final answer is \( \boxed{1} \)
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Given the polynomial $a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$, where $n$ is a positive integer or zero, and $a_0$ is a positive integer. The remaining $a$'s are integers or zero. Set $h=n+a_0+|a_1|+|a_2|+\cdots+|a_n|$. [See example 25 for the meaning of $|x|$.] The number of polynomials with $h=3$ is:
$ \textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 9 $
|
We need to find the number of polynomials of the form \(a_0x^n + a_1x^{n-1} + \cdots + a_{n-1}x + a_n\) such that \(h = n + a_0 + |a_1| + |a_2| + \cdots + |a_n| = 3\). Here, \(n\) is a non-negative integer, \(a_0\) is a positive integer, and the remaining \(a_i\) are integers or zero.
We will perform casework on \(n\):
1. **Case \(n = 3\):**
\[
h = 3 + a_0 + |a_1| + |a_2| + |a_3|
\]
Since \(h = 3\), we have:
\[
3 + a_0 + |a_1| + |a_2| + |a_3| = 3 \implies a_0 + |a_1| + |a_2| + |a_3| = 0
\]
Since \(a_0\) is a positive integer, this is impossible. Thus, there are no polynomials for \(n = 3\).
2. **Case \(n = 2\):**
\[
h = 2 + a_0 + |a_1| + |a_2|
\]
Since \(h = 3\), we have:
\[
2 + a_0 + |a_1| + |a_2| = 3 \implies a_0 + |a_1| + |a_2| = 1
\]
Since \(a_0\) is a positive integer, the only possibility is \(a_0 = 1\) and \(|a_1| = |a_2| = 0\). This gives the polynomial:
\[
x^2
\]
Thus, there is 1 polynomial for \(n = 2\).
3. **Case \(n = 1\):**
\[
h = 1 + a_0 + |a_1|
\]
Since \(h = 3\), we have:
\[
1 + a_0 + |a_1| = 3 \implies a_0 + |a_1| = 2
\]
- If \(a_0 = 1\), then \(|a_1| = 1\). This gives two polynomials:
\[
x + 1 \quad \text{and} \quad x - 1
\]
- If \(a_0 = 2\), then \(|a_1| = 0\). This gives one polynomial:
\[
2x
\]
Thus, there are 3 polynomials for \(n = 1\).
4. **Case \(n = 0\):**
\[
h = 0 + a_0
\]
Since \(h = 3\), we have:
\[
a_0 = 3
\]
This gives the polynomial:
\[
3
\]
Thus, there is 1 polynomial for \(n = 0\).
Summarizing all the cases, the set of polynomials is:
\[
\{x^2, x+1, x-1, 2x, 3\}
\]
The final answer is \(\boxed{5}\)
|
5
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A club with $x$ members is organized into four committees in accordance with these two rules:
$ \text{(1)}\ \text{Each member belongs to two and only two committees}\qquad$
$\text{(2)}\ \text{Each pair of committees has one and only one member in common}$
Then $x$:
$\textbf{(A)} \ \text{cannont be determined} \qquad$
$\textbf{(B)} \ \text{has a single value between 8 and 16} \qquad$
$\textbf{(C)} \ \text{has two values between 8 and 16} \qquad$
$\textbf{(D)} \ \text{has a single value between 4 and 8} \qquad$
$\textbf{(E)} \ \text{has two values between 4 and 8} \qquad$
|
1. Let's denote the four committees as \( A, B, C, \) and \( D \).
2. According to the problem, each member belongs to exactly two committees. Therefore, each member can be represented as a pair of committees.
3. We need to determine the number of such pairs. The number of ways to choose 2 committees out of 4 is given by the binomial coefficient:
\[
\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]
4. This means there are 6 unique pairs of committees, and each pair corresponds to exactly one member.
5. Therefore, the number of members \( x \) in the club is 6.
The final answer is \( \boxed{6} \).
|
6
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $m$ and $n$ be any two odd numbers, with $n$ less than $m$. The largest integer which divides all possible numbers of the form $m^2-n^2$ is:
$ \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16 $
|
1. Let \( m \) and \( n \) be any two odd numbers with \( n < m \). We need to find the largest integer that divides all possible numbers of the form \( m^2 - n^2 \).
2. We start by expressing \( m^2 - n^2 \) using the difference of squares formula:
\[
m^2 - n^2 = (m - n)(m + n)
\]
3. Since \( m \) and \( n \) are both odd, both \( m - n \) and \( m + n \) are even numbers. This is because the difference and sum of two odd numbers are always even. Therefore, \( m - n \) and \( m + n \) can be written as:
\[
m - n = 2a \quad \text{and} \quad m + n = 2b
\]
for some integers \( a \) and \( b \).
4. Substituting these into the expression for \( m^2 - n^2 \), we get:
\[
m^2 - n^2 = (2a)(2b) = 4ab
\]
This shows that \( m^2 - n^2 \) is always divisible by 4.
5. Next, we need to determine if there is a larger integer that always divides \( m^2 - n^2 \). To do this, we consider the possible values of \( m^2 \) and \( n^2 \) modulo 8. The quadratic residues modulo 8 are 0, 1, and 4. Since \( m \) and \( n \) are odd, their squares modulo 8 can only be 1:
\[
m^2 \equiv 1 \pmod{8} \quad \text{and} \quad n^2 \equiv 1 \pmod{8}
\]
6. Therefore, we have:
\[
m^2 - n^2 \equiv 1 - 1 \equiv 0 \pmod{8}
\]
This means that \( m^2 - n^2 \) is always divisible by 8.
7. To check if \( m^2 - n^2 \) is divisible by 16, consider specific values of \( m \) and \( n \). For example, let \( m = 3 \) and \( n = 1 \):
\[
m^2 - n^2 = 3^2 - 1^2 = 9 - 1 = 8
\]
Here, 8 is not divisible by 16, so \( m^2 - n^2 \) is not necessarily divisible by 16.
8. Therefore, the largest integer that divides all possible values of \( m^2 - n^2 \) is 8.
The final answer is \( \boxed{8} \).
|
8
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
You are given a sequence of $58$ terms; each term has the form $P+n$ where $P$ stands for the product $2 \times 3 \times 5 \times... \times 61$ of all prime numbers less than or equal to $61$, and $n$ takes, successively, the values $2, 3, 4, ...., 59$. let $N$ be the number of primes appearing in this sequence. Then $N$ is:
$ \textbf{(A)}\ 0\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 58 $
|
1. First, we define \( P \) as the product of all prime numbers less than or equal to 61. Therefore,
\[
P = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47 \times 53 \times 59 \times 61
\]
2. We are given a sequence of 58 terms, each of the form \( P + n \) where \( n \) takes values from 2 to 59. That is, the sequence is:
\[
P + 2, P + 3, P + 4, \ldots, P + 59
\]
3. To determine if any of these terms are prime, we need to consider the properties of \( P \) and \( n \). Since \( P \) is the product of all primes up to 61, it is divisible by each of these primes.
4. For any \( n \) in the range 2 to 59, \( n \) itself must have at least one prime divisor \( p \) such that \( p \leq 59 \). This is because 2 to 59 are all less than or equal to 59, and any integer in this range must have a prime factor less than or equal to itself.
5. Since \( P \) is divisible by all primes up to 61, and \( n \) has a prime divisor \( p \) (where \( p \leq 59 \)), it follows that \( P \) is divisible by \( p \). Therefore, \( P + n \) is also divisible by \( p \).
6. If \( P + n \) is divisible by \( p \) (where \( p \leq 59 \)), then \( P + n \) cannot be a prime number because it has a divisor other than 1 and itself.
7. Consequently, none of the terms \( P + 2, P + 3, \ldots, P + 59 \) can be prime numbers.
Thus, the number of primes \( N \) in this sequence is:
\[
\boxed{0}
\]
|
0
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
If the graphs of $2y+x+3=0$ and $3y+ax+2=0$ are to meet at right angles, the value of $a$ is:
${{ \textbf{(A)}\ \pm \frac{2}{3} \qquad\textbf{(B)}\ -\frac{2}{3}\qquad\textbf{(C)}\ -\frac{3}{2} \qquad\textbf{(D)}\ 6}\qquad\textbf{(E)}\ -6} $
|
1. To determine the value of \( a \) such that the graphs of the lines \( 2y + x + 3 = 0 \) and \( 3y + ax + 2 = 0 \) meet at right angles, we need to find the slopes of these lines and use the property that the slopes of perpendicular lines are negative reciprocals of each other.
2. First, we find the slope of the line \( 2y + x + 3 = 0 \). We can rewrite this equation in the slope-intercept form \( y = mx + b \):
\[
2y + x + 3 = 0 \implies 2y = -x - 3 \implies y = -\frac{1}{2}x - \frac{3}{2}
\]
Therefore, the slope \( m_1 \) of the first line is \( -\frac{1}{2} \).
3. Next, we find the slope of the line \( 3y + ax + 2 = 0 \). Similarly, we rewrite this equation in the slope-intercept form:
\[
3y + ax + 2 = 0 \implies 3y = -ax - 2 \implies y = -\frac{a}{3}x - \frac{2}{3}
\]
Therefore, the slope \( m_2 \) of the second line is \( -\frac{a}{3} \).
4. Since the lines are perpendicular, their slopes must be negative reciprocals of each other:
\[
m_1 \cdot m_2 = -1
\]
Substituting the slopes we found:
\[
\left(-\frac{1}{2}\right) \cdot \left(-\frac{a}{3}\right) = -1
\]
Simplifying this equation:
\[
\frac{a}{6} = -1 \implies a = -6
\]
5. Therefore, the value of \( a \) is \( -6 \).
The final answer is \(\boxed{-6}\).
|
-6
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The number of solutions of $2^{2x}-3^{2y}=55$, in which $x$ and $y$ are integers, is:
${{ \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3}\qquad\textbf{(E)}\ \text{More than three, but finite} } $
|
To find the number of integer solutions to the equation \(2^{2x} - 3^{2y} = 55\), we start by factoring the left-hand side:
1. Rewrite the equation:
\[
2^{2x} - 3^{2y} = 55
\]
2. Notice that \(2^{2x} = (2^x)^2\) and \(3^{2y} = (3^y)^2\). Let \(a = 2^x\) and \(b = 3^y\). The equation becomes:
\[
a^2 - b^2 = 55
\]
3. Factor the difference of squares:
\[
(a - b)(a + b) = 55
\]
4. Consider the factor pairs of 55:
\[
55 = 1 \times 55, \quad 55 = 5 \times 11
\]
5. Analyze each factor pair to find integer solutions for \(a\) and \(b\):
- **First pair: \(a - b = 1\) and \(a + b = 55\)**
\[
\begin{cases}
a - b = 1 \\
a + b = 55
\end{cases}
\]
Add the equations:
\[
2a = 56 \implies a = 28
\]
Substitute \(a = 28\) into \(a - b = 1\):
\[
28 - b = 1 \implies b = 27
\]
Check if \(a = 28\) and \(b = 27\) correspond to powers of 2 and 3:
\[
2^x = 28 \quad \text{(not a power of 2)}
\]
Therefore, this pair does not yield integer solutions for \(x\) and \(y\).
- **Second pair: \(a - b = 5\) and \(a + b = 11\)**
\[
\begin{cases}
a - b = 5 \\
a + b = 11
\end{cases}
\]
Add the equations:
\[
2a = 16 \implies a = 8
\]
Substitute \(a = 8\) into \(a - b = 5\):
\[
8 - b = 5 \implies b = 3
\]
Check if \(a = 8\) and \(b = 3\) correspond to powers of 2 and 3:
\[
2^x = 8 \implies x = 3 \\
3^y = 3 \implies y = 1
\]
Therefore, \(x = 3\) and \(y = 1\) is a valid solution.
6. Since the first pair did not yield integer solutions and the second pair did, we conclude that there is exactly one solution.
The final answer is \(\boxed{1}\)
|
1
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
If both $ x$ and $ y$ are both integers, how many pairs of solutions are there of the equation $ (x\minus{}8)(x\minus{}10) \equal{} 2^y?$
$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ 2 \qquad
\textbf{(D)}\ 3 \qquad
\textbf{(E)}\ \text{more than 3}$
|
To solve the problem, we need to find all integer pairs \((x, y)\) that satisfy the equation \((x-8)(x-10) = 2^y\).
1. **Expand and simplify the equation:**
\[
(x-8)(x-10) = x^2 - 18x + 80
\]
Therefore, we need to find integer values of \(x\) such that \(x^2 - 18x + 80 = 2^y\).
2. **Analyze the quadratic equation:**
We need to check if \(x^2 - 18x + 80\) can be a power of 2. Let's denote \(2^y = k\), where \(k\) is a power of 2. Thus, we have:
\[
x^2 - 18x + 80 = k
\]
Rearrange this to:
\[
x^2 - 18x + (80 - k) = 0
\]
This is a quadratic equation in \(x\). For \(x\) to be an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant \(\Delta\) is given by:
\[
\Delta = b^2 - 4ac = (-18)^2 - 4 \cdot 1 \cdot (80 - k) = 324 - 4(80 - k) = 324 - 320 + 4k = 4 + 4k
\]
For \(\Delta\) to be a perfect square, \(4 + 4k\) must be a perfect square. Let \(4 + 4k = m^2\) for some integer \(m\). Then:
\[
4(1 + k) = m^2 \implies 1 + k = \left(\frac{m}{2}\right)^2
\]
Since \(k = 2^y\), we have:
\[
1 + 2^y = \left(\frac{m}{2}\right)^2
\]
This implies that \(\left(\frac{m}{2}\right)^2\) must be an integer, so \(m\) must be even. Let \(m = 2n\), then:
\[
1 + 2^y = n^2
\]
3. **Solve for \(y\):**
We need to find \(y\) such that \(n^2 - 1 = 2^y\). Let's test small values of \(y\):
- For \(y = 0\): \(n^2 - 1 = 1 - 1 = 0\) (not a valid solution)
- For \(y = 1\): \(n^2 - 1 = 2 - 1 = 1 \implies n = \pm 1\)
- For \(y = 2\): \(n^2 - 1 = 4 - 1 = 3\) (not a valid solution)
- For \(y = 3\): \(n^2 - 1 = 8 - 1 = 7\) (not a valid solution)
- For \(y = 4\): \(n^2 - 1 = 16 - 1 = 15\) (not a valid solution)
- For \(y = 5\): \(n^2 - 1 = 32 - 1 = 31\) (not a valid solution)
- For \(y = 6\): \(n^2 - 1 = 64 - 1 = 63\) (not a valid solution)
- For \(y = 7\): \(n^2 - 1 = 128 - 1 = 127\) (not a valid solution)
- For \(y = 8\): \(n^2 - 1 = 256 - 1 = 255\) (not a valid solution)
- For \(y = 9\): \(n^2 - 1 = 512 - 1 = 511\) (not a valid solution)
- For \(y = 10\): \(n^2 - 1 = 1024 - 1 = 1023\) (not a valid solution)
- For \(y = 11\): \(n^2 - 1 = 2048 - 1 = 2047\) (not a valid solution)
- For \(y = 12\): \(n^2 - 1 = 4096 - 1 = 4095\) (not a valid solution)
- For \(y = 13\): \(n^2 - 1 = 8192 - 1 = 8191\) (not a valid solution)
- For \(y = 14\): \(n^2 - 1 = 16384 - 1 = 16383\) (not a valid solution)
- For \(y = 15\): \(n^2 - 1 = 32768 - 1 = 32767\) (not a valid solution)
- For \(y = 16\): \(n^2 - 1 = 65536 - 1 = 65535\) (not a valid solution)
The only valid solution is when \(y = 1\), giving \(n = \pm 1\).
4. **Find corresponding \(x\) values:**
- For \(n = 1\):
\[
1 + 2^1 = 2 \implies x^2 - 18x + 80 = 2 \implies x^2 - 18x + 78 = 0
\]
The discriminant is:
\[
\Delta = 18^2 - 4 \cdot 78 = 324 - 312 = 12 \quad (\text{not a perfect square})
\]
- For \(n = -1\):
\[
1 + 2^1 = 2 \implies x^2 - 18x + 80 = 2 \implies x^2 - 18x + 78 = 0
\]
The discriminant is:
\[
\Delta = 18^2 - 4 \cdot 78 = 324 - 312 = 12 \quad (\text{not a perfect square})
\]
Since no valid \(x\) values are found, there are no integer solutions to the equation.
The final answer is \(\boxed{0}\)
|
0
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Line $ l_2$ intersects line $ l_1$ and line $ l_3$ is parallel to $ l_1$. The three lines are distinct and lie in a plane. The number of points equidistant from all three lines is:
$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$
|
1. Let's denote the lines as follows:
- \( l_1 \) and \( l_3 \) are parallel lines.
- \( l_2 \) intersects both \( l_1 \) and \( l_3 \).
2. Since \( l_3 \) is parallel to \( l_1 \), the distance between \( l_1 \) and \( l_3 \) is constant. Let this distance be \( d \).
3. Consider the line \( m \) that is equidistant from both \( l_1 \) and \( l_3 \). This line \( m \) is parallel to both \( l_1 \) and \( l_3 \) and is exactly halfway between them. Therefore, the distance from \( m \) to \( l_1 \) and \( l_3 \) is \( \frac{d}{2} \).
4. Now, we need to find the points that are equidistant from \( l_1 \), \( l_2 \), and \( l_3 \). These points must lie on the line \( m \) and also be equidistant from \( l_2 \).
5. Since \( l_2 \) intersects \( l_1 \) and \( l_3 \), it will also intersect the line \( m \) at some point. Let this intersection point be \( P \).
6. The points equidistant from \( l_1 \), \( l_2 \), and \( l_3 \) must lie on the perpendicular bisector of the segment formed by the intersection points of \( l_2 \) with \( l_1 \) and \( l_3 \). This perpendicular bisector will intersect \( m \) at two points, one on each side of \( P \).
7. Therefore, there are exactly two points on the line \( m \) that are equidistant from \( l_1 \), \( l_2 \), and \( l_3 \).
Conclusion:
The number of points equidistant from all three lines is \( \boxed{2} \).
|
2
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $ n$ be the number of number-pairs $ (x,y)$ which satisfy $ 5y \minus{} 3x \equal{} 15$ and $ x^2 \plus{} y^2 \le 16$. Then $ n$ is:
$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \text{more than two, but finite} \qquad \textbf{(E)}\ \text{greater than any finite number}$
|
1. **Rewrite the linear equation in slope-intercept form:**
\[
5y - 3x = 15 \implies 5y = 3x + 15 \implies y = \frac{3}{5}x + 3
\]
This is the equation of a line with slope \(\frac{3}{5}\) and y-intercept 3.
2. **Interpret the inequality \(x^2 + y^2 \leq 16\):**
This inequality represents a circle centered at the origin with radius 4.
3. **Determine the intersection points of the line and the circle:**
Substitute \(y = \frac{3}{5}x + 3\) into the circle's equation \(x^2 + y^2 = 16\):
\[
x^2 + \left(\frac{3}{5}x + 3\right)^2 = 16
\]
Simplify the equation:
\[
x^2 + \left(\frac{3}{5}x + 3\right)^2 = x^2 + \left(\frac{9}{25}x^2 + \frac{18}{5}x + 9\right) = 16
\]
\[
x^2 + \frac{9}{25}x^2 + \frac{18}{5}x + 9 = 16
\]
Combine like terms:
\[
\left(1 + \frac{9}{25}\right)x^2 + \frac{18}{5}x + 9 = 16
\]
\[
\frac{34}{25}x^2 + \frac{18}{5}x + 9 = 16
\]
Multiply through by 25 to clear the fractions:
\[
34x^2 + 90x + 225 = 400
\]
Simplify:
\[
34x^2 + 90x - 175 = 0
\]
4. **Solve the quadratic equation:**
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 34\), \(b = 90\), and \(c = -175\):
\[
x = \frac{-90 \pm \sqrt{90^2 - 4 \cdot 34 \cdot (-175)}}{2 \cdot 34}
\]
\[
x = \frac{-90 \pm \sqrt{8100 + 23800}}{68}
\]
\[
x = \frac{-90 \pm \sqrt{31900}}{68}
\]
\[
x = \frac{-90 \pm 178.52}{68}
\]
Calculate the two solutions:
\[
x_1 = \frac{-90 + 178.52}{68} \approx 1.30
\]
\[
x_2 = \frac{-90 - 178.52}{68} \approx -3.94
\]
5. **Find the corresponding \(y\) values:**
For \(x_1 \approx 1.30\):
\[
y_1 = \frac{3}{5}(1.30) + 3 \approx 3.78
\]
For \(x_2 \approx -3.94\):
\[
y_2 = \frac{3}{5}(-3.94) + 3 \approx 0.64
\]
6. **Verify the points \((x_1, y_1)\) and \((x_2, y_2)\) are within the circle:**
\[
(1.30)^2 + (3.78)^2 \approx 1.69 + 14.29 = 15.98 \leq 16
\]
\[
(-3.94)^2 + (0.64)^2 \approx 15.52 + 0.41 = 15.93 \leq 16
\]
Since both points lie within the circle, there are exactly two points of intersection.
The final answer is \(\boxed{2}\)
|
2
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
When $ y^2 \plus{} my \plus{} 2$ is divided by $ y \minus{} 1$ the quotient is $ f(y)$ and the remainder is $ R_1$. When $ y^2 \plus{} my \plus{} 2$ is divided by $ y \plus{} 1$ the quotient is $ g(y)$ and the remainder is $ R_2$. If $ R_1 \equal{} R_2$ then $ m$ is:
$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \minus{} 1 \qquad \textbf{(E)}\ \text{an undetermined constant}$
|
1. We start by using polynomial division to find the remainders when \( y^2 + my + 2 \) is divided by \( y - 1 \) and \( y + 1 \).
2. When \( y^2 + my + 2 \) is divided by \( y - 1 \):
\[
y^2 + my + 2 = (y - 1)Q_1(y) + R_1
\]
To find \( R_1 \), we substitute \( y = 1 \) into the polynomial:
\[
1^2 + m(1) + 2 = 1 + m + 2 = m + 3
\]
Therefore, \( R_1 = m + 3 \).
3. When \( y^2 + my + 2 \) is divided by \( y + 1 \):
\[
y^2 + my + 2 = (y + 1)Q_2(y) + R_2
\]
To find \( R_2 \), we substitute \( y = -1 \) into the polynomial:
\[
(-1)^2 + m(-1) + 2 = 1 - m + 2 = 3 - m
\]
Therefore, \( R_2 = 3 - m \).
4. Given that \( R_1 = R_2 \), we set the remainders equal to each other:
\[
m + 3 = 3 - m
\]
5. Solving for \( m \):
\[
m + 3 = 3 - m
\]
\[
m + m = 3 - 3
\]
\[
2m = 0
\]
\[
m = 0
\]
The final answer is \( \boxed{0} \)
|
0
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Of $ 28$ students taking at least one subject the number taking Mathematics and English only equals the number taking Mathematics only. No student takes English only or History only, and six students take Mathematics and History, but not English. The number taking English and History only is five times the number taking all three subjects. If the number taking all three subjects is even and non-zero, the number taking English and Mathematics only is:
$ \textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$
|
1. Define the variables for the number of students taking each combination of subjects:
- Let \( x \) be the number of students taking Mathematics and English only.
- Let \( y \) be the number of students taking all three subjects (Mathematics, English, and History).
- Let \( z \) be the number of students taking English and History only.
2. Given conditions:
- The number of students taking Mathematics and English only equals the number taking Mathematics only: \( x \).
- No student takes English only or History only: \( 0 \).
- Six students take Mathematics and History, but not English: \( 6 \).
- The number taking English and History only is five times the number taking all three subjects: \( z = 5y \).
- The total number of students is 28.
3. Using the principle of inclusion-exclusion for three sets, we have:
\[
28 = (M) + (E) + (H) - (ME) - (MH) - (EH) + (MEH)
\]
where:
- \( (M) \) is the number of students taking Mathematics only: \( x \).
- \( (E) \) is the number of students taking English only: \( 0 \).
- \( (H) \) is the number of students taking History only: \( 0 \).
- \( (ME) \) is the number of students taking Mathematics and English only: \( x \).
- \( (MH) \) is the number of students taking Mathematics and History only: \( 6 \).
- \( (EH) \) is the number of students taking English and History only: \( 5y \).
- \( (MEH) \) is the number of students taking all three subjects: \( y \).
4. Substitute the values into the inclusion-exclusion formula:
\[
28 = x + 0 + 0 - x - 6 - 5y + y
\]
Simplify the equation:
\[
28 = x - x - 6 - 5y + y
\]
\[
28 = -6 - 4y
\]
\[
28 + 6 = -4y
\]
\[
34 = -4y
\]
\[
y = -\frac{34}{4}
\]
\[
y = -8.5
\]
5. Since \( y \) must be an even and non-zero positive integer, we need to re-evaluate the steps. Let's correct the equation:
\[
28 = 2x + 6 + 6y
\]
\[
28 = 2x + 6 + 6y
\]
\[
22 = 2x + 6y
\]
\[
11 = x + 3y
\]
6. Since \( y \) is an even and non-zero positive integer, the possible values for \( y \) are 2, 4, 6, etc. Let's test \( y = 2 \):
\[
11 = x + 3(2)
\]
\[
11 = x + 6
\]
\[
x = 5
\]
7. Therefore, the number of students taking Mathematics and English only is \( x = 5 \).
The final answer is \( \boxed{5} \).
|
5
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
In a triangle, the area is numerically equal to the perimeter. What is the radius of the inscribed circle?
$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }6$
|
1. Let \( a, b, c \) be the side lengths of the triangle. Let \( A \) be the area of the triangle. Let \( s = \frac{a+b+c}{2} \) be the semi-perimeter of the triangle.
2. We know that the area \( A \) of the triangle can be expressed in terms of the inradius \( r \) and the semi-perimeter \( s \) as:
\[
A = rs
\]
3. From the problem, we are given that the area \( A \) is numerically equal to the perimeter \( a + b + c \). Therefore, we have:
\[
A = a + b + c
\]
4. Using the formula for the inradius \( r \), we can write:
\[
r = \frac{A}{s}
\]
5. Substituting \( s = \frac{a+b+c}{2} \) into the equation, we get:
\[
r = \frac{A}{\frac{a+b+c}{2}} = \frac{2A}{a+b+c}
\]
6. Since \( A = a + b + c \) from the problem statement, we substitute \( A \) with \( a + b + c \):
\[
r = \frac{2(a+b+c)}{a+b+c} = 2
\]
The final answer is \(\boxed{2}\)
|
2
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
For each integer $N>1$, there is a mathematical system in which two or more positive integers are defined to be congruent if they leave the same non-negative remainder when divided by $N$. If $69,90,$ and $125$ are congruent in one such system, then in that same system, $81$ is congruent to
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad \textbf{(E) }8$
|
1. We are given that the integers \(69\), \(90\), and \(125\) are congruent modulo \(N\). This means:
\[
69 \equiv 90 \equiv 125 \pmod{N}
\]
For these numbers to be congruent modulo \(N\), the differences between any pair of these numbers must be divisible by \(N\).
2. Calculate the differences:
\[
90 - 69 = 21
\]
\[
125 - 90 = 35
\]
\[
125 - 69 = 56
\]
3. The value of \(N\) must be a common divisor of \(21\), \(35\), and \(56\). Find the greatest common divisor (GCD) of these numbers:
\[
\text{GCD}(21, 35, 56)
\]
4. Factorize each number:
\[
21 = 3 \times 7
\]
\[
35 = 5 \times 7
\]
\[
56 = 2^3 \times 7
\]
5. The common factor is \(7\). Therefore, \(N = 7\).
6. Now, we need to determine the congruence of \(81\) modulo \(7\):
\[
81 \div 7 = 11 \text{ remainder } 4
\]
Thus,
\[
81 \equiv 4 \pmod{7}
\]
Conclusion:
\[
\boxed{4}
\]
|
4
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
If $(1.0025)^{10}$ is evaluated correct to $5$ decimal places, then the digit in the fifth decimal place is
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }5\qquad \textbf{(E) }8$
|
To find the value of $(1.0025)^{10}$ correct to 5 decimal places, we can use the binomial theorem for approximation. The binomial theorem states that for any real number $x$ and integer $n$:
\[
(1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{6}x^3 + \cdots
\]
In this case, $x = 0.0025$ and $n = 10$. We will use the first few terms of the binomial expansion to approximate $(1.0025)^{10}$.
1. **First term:**
\[
1
\]
2. **Second term:**
\[
10 \cdot 0.0025 = 0.025
\]
3. **Third term:**
\[
\frac{10 \cdot 9}{2} \cdot (0.0025)^2 = 45 \cdot 0.00000625 = 0.00028125
\]
4. **Fourth term:**
\[
\frac{10 \cdot 9 \cdot 8}{6} \cdot (0.0025)^3 = 120 \cdot 0.000000015625 = 0.000001875
\]
Adding these terms together, we get:
\[
1 + 0.025 + 0.00028125 + 0.000001875 = 1.025283125
\]
To ensure accuracy, we can use a calculator to find the exact value of $(1.0025)^{10}$:
\[
(1.0025)^{10} \approx 1.025315
\]
Now, we need to round this value to 5 decimal places. The value is 1.025315, and rounding to 5 decimal places gives us:
\[
1.02532
\]
Thus, the digit in the fifth decimal place is 2.
The final answer is $\boxed{2}$
|
2
|
Calculus
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The sum of the digits in base ten of $ (10^{4n^2\plus{}8}\plus{}1)^2$, where $ n$ is a positive integer, is
$ \textbf{(A)}\ 4 \qquad
\textbf{(B)}\ 4n \qquad
\textbf{(C)}\ 2\plus{}2n \qquad
\textbf{(D)}\ 4n^2 \qquad
\textbf{(E)}\ n^2\plus{}n\plus{}2$
|
1. Let's start by examining the expression \((10^{4n^2 + 8} + 1)^2\). We can rewrite it as:
\[
(10^X + 1)^2 \quad \text{where} \quad X = 4n^2 + 8
\]
2. Expanding the square, we get:
\[
(10^X + 1)^2 = 10^{2X} + 2 \cdot 10^X + 1
\]
3. Substituting \(X = 4n^2 + 8\), we have:
\[
10^{2(4n^2 + 8)} + 2 \cdot 10^{4n^2 + 8} + 1
\]
4. Notice that \(10^{2(4n^2 + 8)}\) is a 1 followed by \(2(4n^2 + 8)\) zeros, \(2 \cdot 10^{4n^2 + 8}\) is a 2 followed by \(4n^2 + 8\) zeros, and the final 1 is just 1. When we add these together, the result is:
\[
1\underbrace{00\ldots0}_{4n^2 + 8 \text{ zeros}}2\underbrace{00\ldots0}_{4n^2 + 8 \text{ zeros}}1
\]
5. The sum of the digits of this number is:
\[
1 + 2 + 1 = 4
\]
6. Therefore, the sum of the digits in base ten of \((10^{4n^2 + 8} + 1)^2\) is 4.
The final answer is \(\boxed{4}\)
|
4
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, then $p^3+q^3+r^3$ equals
$ \textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ \text{none of these} $
|
Given the polynomial \( x^3 - x^2 + x - 2 = 0 \) with roots \( p, q, \) and \( r \), we need to find the value of \( p^3 + q^3 + r^3 \).
1. **Identify the coefficients:**
The polynomial is \( x^3 - x^2 + x - 2 = 0 \). The coefficients are:
\[
a_3 = 1, \quad a_2 = -1, \quad a_1 = 1, \quad a_0 = -2
\]
2. **Use Vieta's formulas:**
Vieta's formulas relate the coefficients of the polynomial to sums and products of its roots:
\[
p + q + r = -\frac{a_2}{a_3} = -\frac{-1}{1} = 1
\]
\[
pq + qr + rp = \frac{a_1}{a_3} = \frac{1}{1} = 1
\]
\[
pqr = -\frac{a_0}{a_3} = -\frac{-2}{1} = 2
\]
3. **Apply Newton's Sums:**
Newton's sums provide a way to relate the power sums of the roots to the coefficients of the polynomial. Let \( P_k = p^k + q^k + r^k \). Newton's sums state:
\[
P_1 - a_2 = 0
\]
\[
P_2 - a_2 P_1 + 2a_1 = 0
\]
\[
P_3 - a_2 P_2 + a_1 P_1 - 3a_0 = 0
\]
4. **Calculate \( P_1 \):**
\[
P_1 - (-1) = 0 \implies P_1 = 1
\]
5. **Calculate \( P_2 \):**
\[
P_2 - (-1)P_1 + 2(1) = 0 \implies P_2 + P_1 + 2 = 0 \implies P_2 + 1 + 2 = 0 \implies P_2 = -1
\]
6. **Calculate \( P_3 \):**
\[
P_3 - (-1)P_2 + 1P_1 - 3(-2) = 0
\]
\[
P_3 + P_2 + P_1 + 6 = 0
\]
Substitute \( P_1 = 1 \) and \( P_2 = -1 \):
\[
P_3 - 1 + 1 + 6 = 0 \implies P_3 + 6 = 0 \implies P_3 = -6
\]
Thus, the value of \( p^3 + q^3 + r^3 \) is \(\boxed{-6}\).
|
-6
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many integers greater than $10$ and less than $100$, written in base-$10$ notation, are increased by $9$ when their digits are reversed?
$\textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ 8 \qquad
\textbf{(D)}\ 9 \qquad
\textbf{(E)}\ 10$
|
1. Let the two-digit number be represented as \(10x + y\), where \(x\) and \(y\) are the digits of the number, and \(x\) is the tens digit and \(y\) is the units digit. Given that the number is increased by 9 when its digits are reversed, we can write the equation:
\[
10x + y + 9 = 10y + x
\]
2. Simplify the equation:
\[
10x + y + 9 = 10y + x
\]
\[
10x + y + 9 - x = 10y
\]
\[
9x + y + 9 = 10y
\]
\[
9x + 9 = 9y
\]
\[
x + 1 = y
\]
3. From the equation \(x + 1 = y\), we see that \(y\) is one more than \(x\). Since \(x\) and \(y\) are digits, they must be integers between 0 and 9. However, since the number must be greater than 10 and less than 100, \(x\) must be between 1 and 9 (inclusive).
4. Therefore, \(y\) must be between 2 and 9 (inclusive) because \(y = x + 1\).
5. The possible values for \(y\) are \(2, 3, 4, 5, 6, 7, 8, 9\). Each of these corresponds to a unique \(x\) value:
- If \(y = 2\), then \(x = 1\)
- If \(y = 3\), then \(x = 2\)
- If \(y = 4\), then \(x = 3\)
- If \(y = 5\), then \(x = 4\)
- If \(y = 6\), then \(x = 5\)
- If \(y = 7\), then \(x = 6\)
- If \(y = 8\), then \(x = 7\)
- If \(y = 9\), then \(x = 8\)
6. There are 8 such pairs \((x, y)\) that satisfy the condition \(y = x + 1\).
The final answer is \(\boxed{8}\).
|
8
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
By definition, $ r! \equal{} r(r \minus{} 1) \cdots 1$ and $ \binom{j}{k} \equal{} \frac {j!}{k!(j \minus{} k)!}$, where $ r,j,k$ are positive integers and $ k < j$. If $ \binom{n}{1}, \binom{n}{2}, \binom{n}{3}$ form an arithmetic progression with $ n > 3$, then $ n$ equals
$ \textbf{(A)}\ 5\qquad
\textbf{(B)}\ 7\qquad
\textbf{(C)}\ 9\qquad
\textbf{(D)}\ 11\qquad
\textbf{(E)}\ 12$
|
1. We start by expressing the binomial coefficients $\binom{n}{1}$, $\binom{n}{2}$, and $\binom{n}{3}$ in terms of $n$:
\[
\binom{n}{1} = n
\]
\[
\binom{n}{2} = \frac{n(n-1)}{2}
\]
\[
\binom{n}{3} = \frac{n(n-1)(n-2)}{6}
\]
2. Since these coefficients form an arithmetic progression, the difference between consecutive terms must be constant. Therefore, we set up the following equation:
\[
\binom{n}{2} - \binom{n}{1} = \binom{n}{3} - \binom{n}{2}
\]
3. Substitute the expressions for the binomial coefficients:
\[
\frac{n(n-1)}{2} - n = \frac{n(n-1)(n-2)}{6} - \frac{n(n-1)}{2}
\]
4. Simplify the left-hand side:
\[
\frac{n(n-1)}{2} - n = \frac{n(n-1)}{2} - \frac{2n}{2} = \frac{n(n-1) - 2n}{2} = \frac{n^2 - n - 2n}{2} = \frac{n^2 - 3n}{2}
\]
5. Simplify the right-hand side:
\[
\frac{n(n-1)(n-2)}{6} - \frac{n(n-1)}{2} = \frac{n(n-1)(n-2) - 3n(n-1)}{6} = \frac{n(n-1)(n-2) - 3n(n-1)}{6}
\]
Factor out $n(n-1)$:
\[
\frac{n(n-1)(n-2 - 3)}{6} = \frac{n(n-1)(n-5)}{6}
\]
6. Equate the simplified expressions from both sides:
\[
\frac{n^2 - 3n}{2} = \frac{n(n-1)(n-5)}{6}
\]
7. Clear the fractions by multiplying both sides by 6:
\[
3(n^2 - 3n) = n(n-1)(n-5)
\]
\[
3n^2 - 9n = n^3 - 6n^2 + 5n
\]
8. Rearrange the equation to set it to zero:
\[
0 = n^3 - 6n^2 + 5n - 3n^2 + 9n
\]
\[
0 = n^3 - 9n^2 + 14n
\]
9. Factor out $n$:
\[
0 = n(n^2 - 9n + 14)
\]
10. Solve the quadratic equation $n^2 - 9n + 14 = 0$ using the quadratic formula:
\[
n = \frac{9 \pm \sqrt{81 - 56}}{2} = \frac{9 \pm \sqrt{25}}{2} = \frac{9 \pm 5}{2}
\]
\[
n = \frac{14}{2} = 7 \quad \text{or} \quad n = \frac{4}{2} = 2
\]
11. Since $n > 3$, we discard $n = 2$ and accept $n = 7$.
The final answer is $\boxed{7}$
|
7
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A vertical line divides the triangle with vertices $(0,0)$, $(1,1)$, and $(9,1)$ in the $xy\text{-plane}$ into two regions of equal area. The equation of the line is $x=$
$\textbf {(A) } 2.5 \qquad \textbf {(B) } 3.0 \qquad \textbf {(C) } 3.5 \qquad \textbf {(D) } 4.0\qquad \textbf {(E) } 4.5$
|
1. **Calculate the area of the triangle:**
The vertices of the triangle are \((0,0)\), \((1,1)\), and \((9,1)\). We can use the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the given points:
\[
\text{Area} = \frac{1}{2} \left| 0(1 - 1) + 1(1 - 0) + 9(0 - 1) \right| = \frac{1}{2} \left| 0 + 1 - 9 \right| = \frac{1}{2} \left| -8 \right| = 4
\]
2. **Determine the area of each region:**
Since the vertical line divides the triangle into two regions of equal area, each region must have an area of:
\[
\frac{4}{2} = 2
\]
3. **Set up the equation for the area of the left region:**
Let the vertical line be \(x = k\). The left region will be a triangle with vertices \((0,0)\), \((k,1)\), and \((1,1)\). The area of this triangle must be 2. Using the area formula again:
\[
\text{Area} = \frac{1}{2} \left| 0(1 - 1) + k(1 - 0) + 1(0 - 1) \right| = \frac{1}{2} \left| 0 + k - 1 \right| = \frac{1}{2} \left| k - 1 \right|
\]
Setting this equal to 2:
\[
\frac{1}{2} \left| k - 1 \right| = 2 \implies \left| k - 1 \right| = 4
\]
4. **Solve for \(k\):**
\[
k - 1 = 4 \quad \text{or} \quad k - 1 = -4
\]
\[
k = 5 \quad \text{or} \quad k = -3
\]
Since \(k\) must be between 0 and 9 (the x-coordinates of the triangle's vertices), we discard \(k = -3\) and keep \(k = 5\).
5. **Verify the solution:**
The vertical line \(x = 5\) divides the triangle into two regions of equal area, each with an area of 2.
The final answer is \(\boxed{5}\)
|
5
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
If $60^a = 3$ and $60^b = 5$, then $12^{[(1-a-b)/2(1-b)]}$ is
$\text{(A)} \ \sqrt{3} \qquad \text{(B)} \ 2 \qquad \text{(C)} \ \sqrt{5} \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \sqrt{12}$
|
1. Given the equations \(60^a = 3\) and \(60^b = 5\), we can express \(a\) and \(b\) in terms of logarithms:
\[
a = \frac{\log 3}{\log 60}
\]
\[
b = \frac{\log 5}{\log 60}
\]
2. We need to find the value of \(12^{\left(\frac{1-a-b}{2(1-b)}\right)}\). First, let's simplify the exponent \(\frac{1-a-b}{2(1-b)}\):
\[
1 - a - b = 1 - \frac{\log 3}{\log 60} - \frac{\log 5}{\log 60} = 1 - \frac{\log 3 + \log 5}{\log 60}
\]
Using the property of logarithms \(\log(ab) = \log a + \log b\), we get:
\[
1 - a - b = 1 - \frac{\log(3 \cdot 5)}{\log 60} = 1 - \frac{\log 15}{\log 60}
\]
3. Next, we simplify \(2(1 - b)\):
\[
2(1 - b) = 2 \left(1 - \frac{\log 5}{\log 60}\right) = 2 \left(\frac{\log 60 - \log 5}{\log 60}\right) = 2 \left(\frac{\log \left(\frac{60}{5}\right)}{\log 60}\right) = 2 \left(\frac{\log 12}{\log 60}\right)
\]
4. Now, we combine the results:
\[
\frac{1 - a - b}{2(1 - b)} = \frac{1 - \frac{\log 15}{\log 60}}{2 \left(\frac{\log 12}{\log 60}\right)} = \frac{\frac{\log 60 - \log 15}{\log 60}}{2 \left(\frac{\log 12}{\log 60}\right)} = \frac{\log 60 - \log 15}{2 \log 12}
\]
Simplifying further using the property \(\log(a) - \log(b) = \log\left(\frac{a}{b}\right)\):
\[
\frac{\log \left(\frac{60}{15}\right)}{2 \log 12} = \frac{\log 4}{2 \log 12}
\]
5. We know that \(\log 4 = 2 \log 2\), so:
\[
\frac{2 \log 2}{2 \log 12} = \frac{\log 2}{\log 12}
\]
6. Therefore, the exponent simplifies to:
\[
12^{\left(\frac{\log 2}{\log 12}\right)} = 2
\]
The final answer is \(\boxed{2}\).
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A plane intersects a right circular cylinder of radius $1$ forming an ellipse. If the major axis of the ellipse of $50\%$ longer than the minor axis, the length of the major axis is
$ \textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ \frac{9}{4}\qquad\textbf{(E)}\ 3$
|
1. Consider a right circular cylinder with radius \( r = 1 \). The diameter of the base of the cylinder is \( 2r = 2 \).
2. When a plane intersects the cylinder, the resulting intersection is an ellipse. The minor axis of this ellipse is equal to the diameter of the base of the cylinder, which is \( 2 \).
3. According to the problem, the major axis of the ellipse is \( 50\% \) longer than the minor axis. Therefore, if the minor axis is \( 2 \), the major axis is \( 2 + 0.5 \times 2 = 2 + 1 = 3 \).
4. Thus, the length of the major axis of the ellipse is \( 3 \).
\[
\boxed{3}
\]
|
3
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
If $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^3 + bx^2 + 1$, then $b$ is
$ \textbf{(A)}\ -2\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2 $
|
1. Let \( P(x) = ax^3 + bx^2 + 1 \). We are given that \( x^2 - x - 1 \) is a factor of \( P(x) \). This implies that \( P(x) \) can be written as \( P(x) = (x^2 - x - 1)Q(x) \) for some polynomial \( Q(x) \).
2. Since \( x^2 - x - 1 \) is a factor of \( P(x) \), the remainder when \( P(x) \) is divided by \( x^2 - x - 1 \) must be zero. We can use polynomial division or substitution to find the coefficients \( a \) and \( b \).
3. We know that \( x^2 \equiv x + 1 \) (mod \( x^2 - x - 1 \)). Therefore, we can substitute \( x^2 \) with \( x + 1 \) in \( P(x) \):
\[
P(x) = ax^3 + bx^2 + 1
\]
Substitute \( x^2 = x + 1 \):
\[
x^3 = x \cdot x^2 = x(x + 1) = x^2 + x = (x + 1) + x = 2x + 1
\]
Thus,
\[
P(x) = a(2x + 1) + b(x + 1) + 1
\]
Simplify:
\[
P(x) = 2ax + a + bx + b + 1
\]
Combine like terms:
\[
P(x) = (2a + b)x + (a + b + 1)
\]
4. For \( x^2 - x - 1 \) to be a factor of \( P(x) \), the polynomial \( P(x) \) must be zero when evaluated at the roots of \( x^2 - x - 1 = 0 \). Therefore, the coefficients of \( x \) and the constant term must both be zero:
\[
2a + b = 0
\]
\[
a + b + 1 = 0
\]
5. Solve the system of equations:
\[
2a + b = 0 \quad \text{(1)}
\]
\[
a + b + 1 = 0 \quad \text{(2)}
\]
6. From equation (1), solve for \( b \):
\[
b = -2a
\]
7. Substitute \( b = -2a \) into equation (2):
\[
a - 2a + 1 = 0
\]
\[
-a + 1 = 0
\]
\[
a = 1
\]
8. Substitute \( a = 1 \) back into \( b = -2a \):
\[
b = -2(1) = -2
\]
The final answer is \( \boxed{-2} \)
|
-2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Five people are sitting at a round table. Let $f \ge 0$ be the number of people sitting next to at least one female and $m \ge 0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $(f,m)$ is
$ \textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11 $
|
1. **Identify the possible gender arrangements:**
- We need to consider all possible gender arrangements of 5 people sitting in a circle. Since rotations and reflections are considered the same, we need to count distinct arrangements.
2. **Count the distinct arrangements:**
- **5 males (MMMMM):** There is only one way to arrange 5 males in a circle.
- **4 males and 1 female (MMMMF):** Fix the female's position, and the rest are males. There is only one way to arrange this.
- **3 males and 2 females (MMMFF or MMFMF):**
- If the females are next to each other (MMMFF), there is one way.
- If the females are not next to each other (MMFMF), there is one way.
- **2 males and 3 females (MMFFF or MFMFF):**
- If the males are next to each other (MMFFF), there is one way.
- If the males are not next to each other (MFMFF), there is one way.
- **1 male and 4 females (MFFFF):** Fix the male's position, and the rest are females. There is only one way to arrange this.
- **5 females (FFFFF):** There is only one way to arrange 5 females in a circle.
3. **List all distinct arrangements and their corresponding \((f, m)\) pairs:**
- MMMMM: \((0, 5)\)
- MMMMF: \((2, 5)\)
- MMMFF: \((4, 5)\)
- MMFMF: \((3, 4)\)
- MMFFF: \((5, 4)\)
- MFMFF: \((4, 3)\)
- MFFFF: \((5, 2)\)
- FFFFF: \((5, 0)\)
4. **Count the number of distinct \((f, m)\) pairs:**
- The pairs are: \((0, 5)\), \((2, 5)\), \((4, 5)\), \((3, 4)\), \((5, 4)\), \((4, 3)\), \((5, 2)\), \((5, 0)\).
- There are 8 distinct pairs.
Conclusion:
The number of possible ordered pairs \((f, m)\) is \(\boxed{8}\).
|
8
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose that $7$ boys and $13$ girls line up in a row. Let $S$ be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row $GBBGGGBGBGGGBGBGGBGG$ we have $S=12$. The average value of $S$ (if all possible orders of the 20 people are considered) is closest to
$ \textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13 $
|
1. **Define the problem and use linearity of expectation:**
We need to find the average value of \( S \), the number of places where a boy and a girl are standing next to each other in a row of 7 boys and 13 girls. By the linearity of expectation, we can calculate the expected value of \( S \) by finding the probability that any two consecutive positions are occupied by a boy and a girl, and then multiply this probability by the number of consecutive positions (which is 19).
2. **Calculate the probability of a boy and a girl standing next to each other:**
There are \( \binom{20}{2} = \frac{20 \times 19}{2} = 190 \) possible ways to choose any two consecutive positions out of 20.
The number of ways to choose 1 boy out of 7 and 1 girl out of 13 is \( \binom{7}{1} \times \binom{13}{1} = 7 \times 13 = 91 \).
Therefore, the probability that any two consecutive positions are occupied by a boy and a girl is:
\[
\frac{91}{190}
\]
3. **Calculate the expected value of \( S \):**
Since there are 19 consecutive positions in a row of 20 people, the expected value of \( S \) is:
\[
19 \times \frac{91}{190} = 19 \times \frac{91}{190} = 19 \times 0.478947368 = 9.1
\]
4. **Round to the nearest integer:**
The expected value of \( S \) is approximately 9.1, which is closest to 9.
Conclusion:
\[
\boxed{9}
\]
|
9
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
If the six solutions of $x^6=-64$ are written in the form $a+bi$, where $a$ and $b$ are real, then the product of those solutions with $a>0$ is
$\text{(A)} \ -2 \qquad \text{(B)} \ 0 \qquad \text{(C)} \ 2i \qquad \text{(D)} \ 4 \qquad \text{(E)} \ 16$
|
1. First, we need to find the six solutions to the equation \(x^6 = -64\). We can express \(-64\) in polar form as \(64e^{i\pi}\) because \(-64 = 64 \cdot e^{i\pi}\).
2. To find the sixth roots of \(64e^{i\pi}\), we use the formula for the \(n\)-th roots of a complex number. The \(k\)-th root is given by:
\[
x_k = \sqrt[6]{64} \cdot e^{i\left(\frac{\pi + 2k\pi}{6}\right)}
\]
where \(k = 0, 1, 2, 3, 4, 5\).
3. Calculate \(\sqrt[6]{64}\):
\[
\sqrt[6]{64} = 2
\]
4. Now, we find the six roots by substituting \(k = 0, 1, 2, 3, 4, 5\) into the formula:
\[
x_k = 2 \cdot e^{i\left(\frac{\pi + 2k\pi}{6}\right)}
\]
5. Simplify the exponents:
\[
x_0 = 2 \cdot e^{i\frac{\pi}{6}}
\]
\[
x_1 = 2 \cdot e^{i\frac{\pi + 2\pi}{6}} = 2 \cdot e^{i\frac{3\pi}{6}} = 2 \cdot e^{i\frac{\pi}{2}}
\]
\[
x_2 = 2 \cdot e^{i\frac{\pi + 4\pi}{6}} = 2 \cdot e^{i\frac{5\pi}{6}}
\]
\[
x_3 = 2 \cdot e^{i\frac{\pi + 6\pi}{6}} = 2 \cdot e^{i\frac{7\pi}{6}}
\]
\[
x_4 = 2 \cdot e^{i\frac{\pi + 8\pi}{6}} = 2 \cdot e^{i\frac{9\pi}{6}} = 2 \cdot e^{i\frac{3\pi}{2}}
\]
\[
x_5 = 2 \cdot e^{i\frac{\pi + 10\pi}{6}} = 2 \cdot e^{i\frac{11\pi}{6}}
\]
6. Convert these roots to rectangular form \(a + bi\):
\[
x_0 = 2 \left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 2 \left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = \sqrt{3} + i
\]
\[
x_1 = 2 \left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 2 \left(0 + i\right) = 2i
\]
\[
x_2 = 2 \left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) = 2 \left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -\sqrt{3} + i
\]
\[
x_3 = 2 \left(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}\right) = 2 \left(-\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) = -\sqrt{3} - i
\]
\[
x_4 = 2 \left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = 2 \left(0 - i\right) = -2i
\]
\[
x_5 = 2 \left(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\right) = 2 \left(\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) = \sqrt{3} - i
\]
7. Identify the roots with \(a > 0\):
\[
x_0 = \sqrt{3} + i
\]
\[
x_5 = \sqrt{3} - i
\]
8. Calculate the product of these roots:
\[
(\sqrt{3} + i)(\sqrt{3} - i) = (\sqrt{3})^2 - (i)^2 = 3 - (-1) = 3 + 1 = 4
\]
The final answer is \(\boxed{4}\).
|
4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is
$\textbf{(A)}\ 1 \qquad
\textbf{(B)}\ \dfrac{1+\sqrt{3}}{\sqrt{2}} \qquad
\textbf{(C)}\ \sqrt{3} \qquad
\textbf{(D)}\ 2 \qquad
\textbf{(E)}\ \dfrac{4}{\sqrt{3}}$
|
1. **Identify the given information and the goal:**
- Two rays with a common endpoint \( O \) form a \( 30^\circ \) angle.
- Point \( A \) lies on one ray, and point \( B \) lies on the other ray.
- The length \( AB = 1 \).
- We need to find the maximum possible length of \( OB \).
2. **Use the Law of Sines in triangle \( OAB \):**
- In \(\triangle OAB\), we know \(\angle AOB = 30^\circ\).
- Let \(\angle OAB = \theta\) and \(\angle OBA = 180^\circ - 30^\circ - \theta = 150^\circ - \theta\).
3. **Apply the Law of Sines:**
\[
\frac{AB}{\sin(\angle AOB)} = \frac{OB}{\sin(\angle OAB)} = \frac{OA}{\sin(\angle OBA)}
\]
Given \( AB = 1 \) and \(\angle AOB = 30^\circ\), we have:
\[
\frac{1}{\sin(30^\circ)} = \frac{OB}{\sin(\theta)}
\]
Since \(\sin(30^\circ) = \frac{1}{2}\), this simplifies to:
\[
\frac{1}{\frac{1}{2}} = \frac{OB}{\sin(\theta)}
\]
\[
2 = \frac{OB}{\sin(\theta)}
\]
\[
OB = 2 \sin(\theta)
\]
4. **Maximize \( OB \):**
- The maximum value of \(\sin(\theta)\) is 1.
- Therefore, the maximum value of \( OB \) is:
\[
OB = 2 \times 1 = 2
\]
5. **Conclusion:**
- The maximum possible length of \( OB \) is \( 2 \).
The final answer is \(\boxed{2}\).
|
2
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A function $ f$ from the integers to the integers is defined as follows:
\[ f(n) \equal{} \begin{cases} n \plus{} 3 & \text{if n is odd} \\
n/2 & \text{if n is even} \end{cases}
\]Suppose $ k$ is odd and $ f(f(f(k))) \equal{} 27$. What is the sum of the digits of $ k$?
$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$
|
1. Given the function \( f \) defined as:
\[
f(n) = \begin{cases}
n + 3 & \text{if } n \text{ is odd} \\
\frac{n}{2} & \text{if } n \text{ is even}
\end{cases}
\]
and the condition \( f(f(f(k))) = 27 \) with \( k \) being odd.
2. Since \( k \) is odd, we start by applying the function \( f \) to \( k \):
\[
f(k) = k + 3
\]
Since \( k \) is odd, \( k + 3 \) is even.
3. Next, we apply the function \( f \) to \( k + 3 \):
\[
f(k + 3) = \frac{k + 3}{2}
\]
Since \( k + 3 \) is even, \( \frac{k + 3}{2} \) is an integer.
4. Finally, we apply the function \( f \) to \( \frac{k + 3}{2} \):
\[
f\left(\frac{k + 3}{2}\right) = \begin{cases}
\frac{k + 3}{2} + 3 & \text{if } \frac{k + 3}{2} \text{ is odd} \\
\frac{\frac{k + 3}{2}}{2} & \text{if } \frac{k + 3}{2} \text{ is even}
\end{cases}
\]
Since we need \( f(f(f(k))) = 27 \), we set:
\[
f\left(\frac{k + 3}{2}\right) = 27
\]
5. We need to determine whether \( \frac{k + 3}{2} \) is odd or even. Let's assume \( \frac{k + 3}{2} \) is even:
\[
f\left(\frac{k + 3}{2}\right) = \frac{\frac{k + 3}{2}}{2} = \frac{k + 3}{4}
\]
Setting this equal to 27:
\[
\frac{k + 3}{4} = 27
\]
Solving for \( k \):
\[
k + 3 = 108 \implies k = 105
\]
6. Verify the solution:
\[
f(105) = 105 + 3 = 108
\]
\[
f(108) = \frac{108}{2} = 54
\]
\[
f(54) = \frac{54}{2} = 27
\]
This confirms \( f(f(f(105))) = 27 \).
7. The sum of the digits of \( k = 105 \) is:
\[
1 + 0 + 5 = 6
\]
The final answer is \(\boxed{6}\).
|
6
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many triangles have area $ 10$ and vertices at $ (\minus{}5,0)$, $ (5,0)$, and $ (5\cos \theta, 5\sin \theta)$ for some angle $ \theta$?
$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ 2\qquad
\textbf{(C)}\ 4\qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ 8$
|
1. We are given three vertices of a triangle: \( A(-5, 0) \), \( B(5, 0) \), and \( C(5\cos \theta, 5\sin \theta) \). We need to find how many such triangles have an area of 10.
2. The base of the triangle is the distance between points \( A \) and \( B \). This distance can be calculated using the distance formula:
\[
AB = \sqrt{(5 - (-5))^2 + (0 - 0)^2} = \sqrt{10^2} = 10
\]
3. The area of a triangle with base \( b \) and height \( h \) is given by:
\[
\text{Area} = \frac{1}{2} \times b \times h
\]
Given that the area is 10 and the base \( AB \) is 10, we can set up the equation:
\[
10 = \frac{1}{2} \times 10 \times h
\]
Solving for \( h \):
\[
10 = 5h \implies h = 2
\]
4. The height of the triangle is the perpendicular distance from point \( C \) to the line \( AB \). Since \( AB \) lies on the x-axis, the height is simply the y-coordinate of point \( C \), which is \( 5\sin \theta \).
5. We need \( 5\sin \theta = 2 \). Solving for \( \sin \theta \):
\[
\sin \theta = \frac{2}{5}
\]
6. The sine function has two solutions in the interval \( [0, 2\pi) \) for each value of \( \sin \theta \). Therefore, \( \theta \) can be:
\[
\theta = \arcsin\left(\frac{2}{5}\right) \quad \text{or} \quad \theta = \pi - \arcsin\left(\frac{2}{5}\right)
\]
7. Additionally, since the circle is symmetric, the same values of \( \theta \) will also occur in the intervals \( [\pi, 2\pi) \), giving us:
\[
\theta = \pi + \arcsin\left(\frac{2}{5}\right) \quad \text{or} \quad \theta = 2\pi - \arcsin\left(\frac{2}{5}\right)
\]
8. Therefore, there are 4 distinct values of \( \theta \) that satisfy the condition \( 5\sin \theta = 2 \).
Conclusion:
\[
\boxed{4}
\]
|
4
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
What is the sum of the digits of the decimal form of the product $ 2^{1999}\cdot 5^{2001}$?
$ \textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10$
|
1. First, we simplify the given expression \(2^{1999} \cdot 5^{2001}\). Notice that we can factor out \(5^2\) from \(5^{2001}\):
\[
2^{1999} \cdot 5^{2001} = 2^{1999} \cdot 5^{1999} \cdot 5^2
\]
2. Next, we recognize that \(2^{1999} \cdot 5^{1999} = (2 \cdot 5)^{1999} = 10^{1999}\). Therefore, the expression simplifies to:
\[
10^{1999} \cdot 25
\]
3. Now, we need to understand the decimal form of \(10^{1999} \cdot 25\). The term \(10^{1999}\) is a 1 followed by 1999 zeros:
\[
10^{1999} = 1\underbrace{000\ldots000}_{1999 \text{ zeros}}
\]
4. Multiplying this by 25 shifts the decimal point 1999 places to the right and appends 25 at the end:
\[
10^{1999} \cdot 25 = 25\underbrace{000\ldots000}_{1999 \text{ zeros}}
\]
5. The sum of the digits of this number is simply the sum of the digits of 25, since all the zeros do not contribute to the sum:
\[
2 + 5 = 7
\]
The final answer is \(\boxed{7}\).
|
7
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
For how many values of $ a$ is it true that the line $ y \equal{} x \plus{} a$ passes through the vertex of the parabola $ y \equal{} x^2 \plus{} a^2$?
$ \textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ \text{infinitely many}$
|
1. First, identify the vertex of the parabola \( y = x^2 + a^2 \). The standard form of a parabola \( y = ax^2 + bx + c \) has its vertex at \( x = -\frac{b}{2a} \). In this case, \( a = 1 \), \( b = 0 \), and \( c = a^2 \), so the vertex is at \( x = 0 \). Substituting \( x = 0 \) into the equation of the parabola, we get:
\[
y = 0^2 + a^2 = a^2
\]
Therefore, the vertex of the parabola is at the point \( (0, a^2) \).
2. Next, consider the line \( y = x + a \). This line passes through the y-axis at the point \( (0, a) \).
3. For the line \( y = x + a \) to pass through the vertex of the parabola \( (0, a^2) \), the y-coordinate of the vertex must satisfy the equation of the line when \( x = 0 \). Therefore, we set the y-coordinate of the vertex equal to the y-intercept of the line:
\[
a^2 = a
\]
4. Solve the equation \( a^2 = a \):
\[
a^2 - a = 0
\]
Factor the quadratic equation:
\[
a(a - 1) = 0
\]
This gives us two solutions:
\[
a = 0 \quad \text{or} \quad a = 1
\]
5. Therefore, there are two values of \( a \) that satisfy the condition that the line \( y = x + a \) passes through the vertex of the parabola \( y = x^2 + a^2 \).
The final answer is \(\boxed{2}\)
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A piece of cheese is located at $ (12,10)$ in a coordinate plane. A mouse is at $ (4, \minus{} 2)$ and is running up the line $ y \equal{} \minus{} 5x \plus{} 18.$ At the point $ (a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $ a \plus{} b?$
$ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 22$
|
1. **Identify the line on which the mouse is running:**
The mouse is running up the line given by the equation \( y = -5x + 18 \).
2. **Find the slope of the line perpendicular to \( y = -5x + 18 \):**
The slope of the given line is \(-5\). The slope of a line perpendicular to this line is the negative reciprocal of \(-5\), which is \(\frac{1}{5}\).
3. **Write the equation of the line passing through the cheese and perpendicular to \( y = -5x + 18 \):**
The cheese is located at \((12, 10)\). Using the point-slope form of the equation of a line, we have:
\[
y - 10 = \frac{1}{5}(x - 12)
\]
Simplifying this, we get:
\[
y - 10 = \frac{1}{5}x - \frac{12}{5}
\]
\[
y = \frac{1}{5}x + 10 - \frac{12}{5}
\]
\[
y = \frac{1}{5}x + \frac{50}{5} - \frac{12}{5}
\]
\[
y = \frac{1}{5}x + \frac{38}{5}
\]
4. **Find the intersection of the two lines:**
We need to solve the system of equations:
\[
y = -5x + 18
\]
\[
y = \frac{1}{5}x + \frac{38}{5}
\]
Setting the two equations equal to each other:
\[
-5x + 18 = \frac{1}{5}x + \frac{38}{5}
\]
To clear the fraction, multiply every term by 5:
\[
-25x + 90 = x + 38
\]
Combine like terms:
\[
-25x - x = 38 - 90
\]
\[
-26x = -52
\]
Solving for \( x \):
\[
x = 2
\]
5. **Find the corresponding \( y \)-coordinate:**
Substitute \( x = 2 \) back into the equation \( y = -5x + 18 \):
\[
y = -5(2) + 18
\]
\[
y = -10 + 18
\]
\[
y = 8
\]
6. **Calculate \( a + b \):**
The point where the mouse starts getting farther from the cheese is \((2, 8)\). Therefore, \( a + b = 2 + 8 = 10 \).
The final answer is \( \boxed{10} \)
|
10
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Triangle $ ABC$ has side lengths $ AB \equal{} 5$, $ BC \equal{} 6$, and $ AC \equal{} 7$. Two bugs start simultaneously from $ A$ and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point $ D$. What is $ BD$?
$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
|
1. Let's denote the points where the bugs meet as \( D \). Since the bugs start from \( A \) and crawl along the sides of the triangle \( ABC \) in opposite directions at the same speed, they will meet at a point \( D \) such that the distances they have crawled are equal.
2. Let the bug starting from \( A \) and crawling along \( AB \) reach point \( D \) after traveling a distance \( x \). Therefore, the bug starting from \( A \) and crawling along \( AC \) will have traveled a distance \( 7 - x \) to reach point \( D \).
3. Since the bugs meet at point \( D \), the total distance traveled by both bugs is equal to the perimeter of the triangle divided by 2. The perimeter of triangle \( ABC \) is:
\[
AB + BC + AC = 5 + 6 + 7 = 18
\]
Therefore, the total distance traveled by each bug is:
\[
\frac{18}{2} = 9
\]
4. The bug traveling along \( AB \) reaches point \( D \) after traveling \( x \) units, and the bug traveling along \( AC \) reaches point \( D \) after traveling \( 7 - x \) units. Since they meet at point \( D \), the distance \( BD \) must be such that:
\[
x + (7 - x) + BD = 9
\]
Simplifying, we get:
\[
7 + BD = 9 \implies BD = 2
\]
The final answer is \( \boxed{2} \).
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
How many zeros are at the end of the product
\[25\times 25\times 25\times 25\times 25\times 25\times 25\times 8\times 8\times 8?\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$
|
To determine the number of zeros at the end of the product \(25 \times 25 \times 25 \times 25 \times 25 \times 25 \times 25 \times 8 \times 8 \times 8\), we need to find the number of factors of 10 in the product. A factor of 10 is composed of a factor of 2 and a factor of 5. Therefore, we need to count the number of factors of 2 and 5 in the prime factorization of the product.
1. **Prime Factorization of Each Number:**
- \(25 = 5^2\)
- \(8 = 2^3\)
2. **Prime Factorization of the Entire Product:**
- There are seven 25's in the product:
\[
25^7 = (5^2)^7 = 5^{14}
\]
- There are three 8's in the product:
\[
8^3 = (2^3)^3 = 2^9
\]
3. **Combine the Prime Factors:**
- The product can be written as:
\[
25^7 \times 8^3 = 5^{14} \times 2^9
\]
4. **Determine the Number of Factors of 10:**
- A factor of 10 is \(2 \times 5\). Therefore, the number of factors of 10 is determined by the minimum of the number of factors of 2 and the number of factors of 5.
- From the prime factorization, we have:
\[
5^{14} \times 2^9
\]
- The number of factors of 2 is 9.
- The number of factors of 5 is 14.
- The limiting factor is the number of factors of 2, which is 9.
Therefore, the number of zeros at the end of the product is 9.
The final answer is \(\boxed{9}\).
|
9
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Tori's mathematics test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry problems. Although she answered $70\%$ of the arithmetic, $40\%$ of the algebra, and $60\%$ of the geometry problems correctly, she did not pass the test because she got less than $60\%$ of the problems right. How many more problems would she have needed to answer correctly to earn a $60\%$ passing grade?
$ \text{(A)}\ 1\qquad\text{(B)}\ 5\qquad\text{(C)}\ 7\qquad\text{(D)}\ 9\qquad\text{(E)}\ 11 $
|
1. First, calculate the total number of problems Tori needs to answer correctly to pass the test. Since she needs at least $60\%$ of the problems correct:
\[
75 \times 0.6 = 45
\]
Therefore, Tori needs to answer $45$ questions correctly to pass.
2. Next, calculate the number of problems Tori answered correctly in each category:
- Arithmetic: She answered $70\%$ of the $10$ arithmetic problems correctly:
\[
0.7 \times 10 = 7
\]
- Algebra: She answered $40\%$ of the $30$ algebra problems correctly:
\[
0.4 \times 30 = 12
\]
- Geometry: She answered $60\%$ of the $35$ geometry problems correctly:
\[
0.6 \times 35 = 21
\]
3. Sum the number of problems she answered correctly in each category:
\[
7 + 12 + 21 = 40
\]
Therefore, Tori answered $40$ questions correctly.
4. Determine how many more problems Tori needs to answer correctly to reach the passing grade of $45$:
\[
45 - 40 = 5
\]
Conclusion:
Tori needs to answer $5$ more problems correctly to pass the test.
The final answer is $\boxed{5}$
|
5
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many whole numbers lie in the interval between $\frac{5}{3}$ and $2\pi$?
$\textbf{(A)}\ 2\qquad
\textbf{(B)}\ 3\qquad
\textbf{(C)}\ 4\qquad
\textbf{(D)}\ 5\qquad
\textbf{(E)}\ \text{infinitely many}$
|
1. **Determine the decimal values of the given fractions and constants:**
\[
\frac{5}{3} \approx 1.6667
\]
\[
2\pi \approx 2 \times 3.14159 \approx 6.2832
\]
2. **Identify the smallest whole number greater than \(\frac{5}{3}\):**
\[
\lceil 1.6667 \rceil = 2
\]
Here, \(\lceil x \rceil\) denotes the ceiling function, which rounds \(x\) up to the nearest whole number.
3. **Identify the largest whole number less than \(2\pi\):**
\[
\lfloor 6.2832 \rfloor = 6
\]
Here, \(\lfloor x \rfloor\) denotes the floor function, which rounds \(x\) down to the nearest whole number.
4. **List the whole numbers between the smallest and largest identified in steps 2 and 3:**
\[
2, 3, 4, 5, 6
\]
5. **Count the whole numbers in the list:**
\[
2, 3, 4, 5, 6 \quad \text{(5 numbers)}
\]
Conclusion:
\[
\boxed{5}
\]
|
5
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many integers between $1000$ and $2000$ have all three of the numbers $15$, $20$, and $25$ as factors?
$\textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 5$
|
1. To determine how many integers between $1000$ and $2000$ are divisible by $15$, $20$, and $25$, we first need to find the least common multiple (LCM) of these three numbers.
2. We start by finding the prime factorizations of $15$, $20$, and $25$:
\[
15 = 3 \times 5
\]
\[
20 = 2^2 \times 5
\]
\[
25 = 5^2
\]
3. The LCM is found by taking the highest power of each prime that appears in the factorizations:
\[
\text{LCM}(15, 20, 25) = 2^2 \times 3 \times 5^2
\]
4. Calculating the LCM:
\[
2^2 = 4
\]
\[
5^2 = 25
\]
\[
4 \times 3 = 12
\]
\[
12 \times 25 = 300
\]
Therefore, the LCM of $15$, $20$, and $25$ is $300$.
5. Next, we need to find the integers between $1000$ and $2000$ that are multiples of $300$.
6. The smallest multiple of $300$ greater than $1000$ is:
\[
300 \times 4 = 1200
\]
7. The largest multiple of $300$ less than $2000$ is:
\[
300 \times 6 = 1800
\]
8. The multiples of $300$ between $1000$ and $2000$ are:
\[
1200, 1500, 1800
\]
9. Therefore, there are $3$ integers between $1000$ and $2000$ that are divisible by $15$, $20$, and $25$.
The final answer is $\boxed{3}$.
|
3
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad
\textbf{(B)}\ 3\qquad
\textbf{(C)}\ 6\qquad
\textbf{(D)}\ 10\qquad
\textbf{(E)}\ 12$
|
1. We start by giving each of the three friends one pencil. This ensures that each friend has at least one pencil. After this distribution, we have used up 3 pencils, leaving us with \(6 - 3 = 3\) pencils to distribute.
2. We now need to find the number of ways to distribute these remaining 3 pencils among the 3 friends. This is a classic problem of distributing indistinguishable objects (pencils) into distinguishable bins (friends).
3. The problem can be solved using the "stars and bars" theorem. The number of ways to distribute \(n\) indistinguishable objects into \(k\) distinguishable bins is given by the binomial coefficient:
\[
\binom{n+k-1}{k-1}
\]
In our case, \(n = 3\) (pencils) and \(k = 3\) (friends). Therefore, we need to calculate:
\[
\binom{3+3-1}{3-1} = \binom{5}{2}
\]
4. We compute the binomial coefficient:
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \cdot 3!} = \frac{5 \cdot 4 \cdot 3!}{2! \cdot 3!} = \frac{5 \cdot 4}{2 \cdot 1} = 10
\]
5. Therefore, there are 10 ways to distribute the remaining 3 pencils among the 3 friends.
The final answer is \(\boxed{10}\).
|
10
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
After Sally takes 20 shots, she has made $55\%$ of her shots. After she takes 5 more shots, she raises her percentage to $56\%$. How many of the last 5 shots did she make?
$\textbf{(A)} 1 \qquad\textbf{(B)} 2 \qquad\textbf{(C)} 3 \qquad\textbf{(D)} 4 \qquad\textbf{(E)} 5$
|
1. After 20 shots, Sally has made 55% of her shots. This means she made:
\[
0.55 \times 20 = 11 \text{ shots}
\]
2. Let \( x \) be the number of shots she made in the last 5 shots. After taking 5 more shots, her total number of shots is:
\[
20 + 5 = 25 \text{ shots}
\]
3. Her new shooting percentage is 56%, so she made:
\[
0.56 \times 25 = 14 \text{ shots}
\]
4. The total number of shots made after 25 shots is the sum of the shots made in the first 20 shots and the shots made in the last 5 shots:
\[
11 + x = 14
\]
5. Solving for \( x \):
\[
11 + x = 14 \implies x = 14 - 11 \implies x = 3
\]
Therefore, the number of shots she made in the last 5 shots is \( \boxed{3} \).
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose $a$, $b$, and $c$ are nonzero real numbers, and $a+b+c=0$. What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$?
$\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1$
|
Given that \(a\), \(b\), and \(c\) are nonzero real numbers and \(a + b + c = 0\), we need to determine the possible values for the expression \(\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}\).
1. **Sign Analysis**:
- The term \(\frac{a}{|a|}\) is the sign of \(a\), denoted as \(\operatorname{sgn}(a)\).
- Similarly, \(\frac{b}{|b|} = \operatorname{sgn}(b)\) and \(\frac{c}{|c|} = \operatorname{sgn}(c)\).
- The term \(\frac{abc}{|abc|}\) is the sign of the product \(abc\), denoted as \(\operatorname{sgn}(abc)\).
2. **Possible Sign Combinations**:
- Since \(a + b + c = 0\), the sum of the signs of \(a\), \(b\), and \(c\) must be zero. This implies that either two of the numbers are positive and one is negative, or two are negative and one is positive.
3. **Case Analysis**:
- **Case 1**: Two positive and one negative.
- Without loss of generality, assume \(a > 0\), \(b > 0\), and \(c < 0\).
- Then, \(\operatorname{sgn}(a) = 1\), \(\operatorname{sgn}(b) = 1\), and \(\operatorname{sgn}(c) = -1\).
- The product \(abc\) will be negative since it involves two positive numbers and one negative number: \(\operatorname{sgn}(abc) = -1\).
- Therefore, the expression becomes:
\[
\operatorname{sgn}(a) + \operatorname{sgn}(b) + \operatorname{sgn}(c) + \operatorname{sgn}(abc) = 1 + 1 - 1 - 1 = 0
\]
- **Case 2**: Two negative and one positive.
- Without loss of generality, assume \(a < 0\), \(b < 0\), and \(c > 0\).
- Then, \(\operatorname{sgn}(a) = -1\), \(\operatorname{sgn}(b) = -1\), and \(\operatorname{sgn}(c) = 1\).
- The product \(abc\) will be negative since it involves two negative numbers and one positive number: \(\operatorname{sgn}(abc) = -1\).
- Therefore, the expression becomes:
\[
\operatorname{sgn}(a) + \operatorname{sgn}(b) + \operatorname{sgn}(c) + \operatorname{sgn}(abc) = -1 - 1 + 1 - 1 = 0
\]
4. **Conclusion**:
- In both cases, the value of the expression \(\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}\) is \(0\).
The final answer is \(\boxed{0}\).
|
0
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?
$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$
|
1. Let \( x \) be Laila's score on each of the first four tests, and let \( y \) be her score on the last test. Given that her average score on the five tests was 82, we can set up the following equation:
\[
\frac{4x + y}{5} = 82
\]
2. Multiply both sides of the equation by 5 to clear the fraction:
\[
4x + y = 410
\]
3. We need to find integer solutions for \( x \) and \( y \) such that \( 0 \leq x \leq 100 \) and \( 0 \leq y \leq 100 \), with the additional condition that \( y > x \).
4. Rearrange the equation to solve for \( y \):
\[
y = 410 - 4x
\]
5. Since \( y \) must be greater than \( x \), we have:
\[
410 - 4x > x
\]
6. Combine like terms:
\[
410 > 5x
\]
7. Divide both sides by 5:
\[
x < 82
\]
8. Since \( x \) must be an integer, the possible values for \( x \) are \( 0, 1, 2, \ldots, 81 \).
9. For each value of \( x \), we calculate \( y \) using \( y = 410 - 4x \). We need \( y \) to be an integer between 0 and 100, inclusive.
10. To find the range of \( y \), we substitute the minimum and maximum values of \( x \) into the equation \( y = 410 - 4x \):
- When \( x = 0 \):
\[
y = 410 - 4(0) = 410
\]
This value is not within the range 0 to 100.
- When \( x = 81 \):
\[
y = 410 - 4(81) = 410 - 324 = 86
\]
This value is within the range 0 to 100.
11. We need to find the values of \( y \) that are greater than \( x \) and within the range 0 to 100. Since \( y = 410 - 4x \), we need:
\[
86 \leq y \leq 100
\]
12. To find the corresponding values of \( x \) for \( y \) in this range, we solve:
- For \( y = 86 \):
\[
86 = 410 - 4x \implies 4x = 324 \implies x = 81
\]
- For \( y = 90 \):
\[
90 = 410 - 4x \implies 4x = 320 \implies x = 80
\]
- For \( y = 94 \):
\[
94 = 410 - 4x \implies 4x = 316 \implies x = 79
\]
- For \( y = 98 \):
\[
98 = 410 - 4x \implies 4x = 312 \implies x = 78
\]
13. Therefore, the possible values for \( y \) are \( 86, 90, 94, 98 \).
The final answer is \(\boxed{4}\).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
$\textbf{(A)}\ 6\qquad~~\textbf{(B)}\ 9\qquad~~\textbf{(C)}\ 12\qquad~~\textbf{(D)}\ 18\qquad~~\textbf{(E)}\ 24$
|
1. Jamal initially has 6 green socks, 18 purple socks, and 12 orange socks. Therefore, the total number of socks initially is:
\[
6 + 18 + 12 = 36
\]
2. Let \( x \) be the number of purple socks Jamal added. After adding \( x \) purple socks, the total number of socks becomes:
\[
36 + x
\]
3. The number of purple socks after adding \( x \) purple socks is:
\[
18 + x
\]
4. According to the problem, the probability that a randomly selected sock is purple is now 60%. Therefore, we can set up the following equation:
\[
\frac{18 + x}{36 + x} = 0.60
\]
5. Converting 60% to a fraction, we get:
\[
\frac{18 + x}{36 + x} = \frac{3}{5}
\]
6. Cross-multiplying to solve for \( x \):
\[
5(18 + x) = 3(36 + x)
\]
7. Expanding both sides:
\[
90 + 5x = 108 + 3x
\]
8. Subtracting \( 3x \) from both sides:
\[
90 + 2x = 108
\]
9. Subtracting 90 from both sides:
\[
2x = 18
\]
10. Dividing both sides by 2:
\[
x = 9
\]
Therefore, Jamal added \( \boxed{9} \) purple socks.
|
9
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Evaluate
$$ \lim_{x\to \infty} \left( \frac{a^x -1}{x(a-1)} \right)^{1\slash x},$$
where $a>0$ and $a\ne 1.$
|
To evaluate the limit
\[ \lim_{x\to \infty} \left( \frac{a^x -1}{x(a-1)} \right)^{1/x}, \]
where \( a > 0 \) and \( a \ne 1 \), we will consider two cases: \( a > 1 \) and \( 0 < a < 1 \).
### Case 1: \( a > 1 \)
1. Let \( L = \lim_{x\to \infty} \left( \frac{a^x -1}{x(a-1)} \right)^{1/x} \). Taking the natural logarithm of both sides, we get:
\[ \ln L = \lim_{x\to \infty} \frac{\ln \left( \frac{a^x -1}{x(a-1)} \right)}{x}. \]
2. Simplify the logarithm inside the limit:
\[ \ln L = \lim_{x\to \infty} \frac{\ln (a^x - 1) - \ln x - \ln (a-1)}{x}. \]
3. Apply L'Hôpital's Rule to evaluate the limit. First, differentiate the numerator and the denominator with respect to \( x \):
\[ \ln L = \lim_{x\to \infty} \frac{\frac{d}{dx} \left( \ln (a^x - 1) - \ln x - \ln (a-1) \right)}{\frac{d}{dx} x}. \]
4. The derivatives are:
\[ \frac{d}{dx} \ln (a^x - 1) = \frac{a^x \ln a}{a^x - 1}, \quad \frac{d}{dx} \ln x = \frac{1}{x}, \quad \frac{d}{dx} \ln (a-1) = 0. \]
5. Substitute these derivatives into the limit:
\[ \ln L = \lim_{x\to \infty} \left( \frac{a^x \ln a}{a^x - 1} - \frac{1}{x} \right). \]
6. As \( x \to \infty \), \( a^x \) grows much faster than 1, so \( \frac{a^x}{a^x - 1} \to 1 \). Therefore:
\[ \ln L = \ln a \lim_{x\to \infty} \left( \frac{a^x}{a^x - 1} \right) = \ln a. \]
7. Exponentiate both sides to solve for \( L \):
\[ L = e^{\ln a} = a. \]
### Case 2: \( 0 < a < 1 \)
1. Let \( L = \lim_{x\to \infty} \left( \frac{a^x -1}{x(a-1)} \right)^{1/x} \). Taking the natural logarithm of both sides, we get:
\[ \ln L = \lim_{x\to \infty} \frac{\ln \left( \frac{1 - a^x}{x(1-a)} \right)}{x}. \]
2. Simplify the logarithm inside the limit:
\[ \ln L = \lim_{x\to \infty} \frac{\ln (1 - a^x) - \ln x - \ln (1-a)}{x}. \]
3. Apply L'Hôpital's Rule to evaluate the limit. First, differentiate the numerator and the denominator with respect to \( x \):
\[ \ln L = \lim_{x\to \infty} \frac{\frac{d}{dx} \left( \ln (1 - a^x) - \ln x - \ln (1-a) \right)}{\frac{d}{dx} x}. \]
4. The derivatives are:
\[ \frac{d}{dx} \ln (1 - a^x) = \frac{-a^x \ln a}{1 - a^x}, \quad \frac{d}{dx} \ln x = \frac{1}{x}, \quad \frac{d}{dx} \ln (1-a) = 0. \]
5. Substitute these derivatives into the limit:
\[ \ln L = \lim_{x\to \infty} \left( \frac{-a^x \ln a}{1 - a^x} - \frac{1}{x} \right). \]
6. As \( x \to \infty \), \( a^x \to 0 \), so \( \frac{a^x}{1 - a^x} \to 0 \). Therefore:
\[ \ln L = \ln a \lim_{x\to \infty} \left( \frac{a^x}{1 - a^x} \right) = 0. \]
7. Exponentiate both sides to solve for \( L \):
\[ L = e^0 = 1. \]
The final answer is \( \boxed{a} \) if \( a > 1 \) and \( \boxed{1} \) if \( 0 < a < 1 \).
|
1
|
Calculus
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
Define a sequence $(a_n)$ by $a_0 =0$ and $a_n = 1 +\sin(a_{n-1}-1)$ for $n\geq 1$. Evaluate
$$\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^{n} a_k.$$
|
1. Define the sequence \( (a_n) \) by \( a_0 = 0 \) and \( a_n = 1 + \sin(a_{n-1} - 1) \) for \( n \geq 1 \).
2. To simplify the sequence, let \( b_n = a_n - 1 \). Then, we have:
\[
b_0 = a_0 - 1 = -1
\]
and for \( n \geq 1 \):
\[
b_n = a_n - 1 = 1 + \sin(a_{n-1} - 1) - 1 = \sin(a_{n-1} - 1) = \sin(b_{n-1})
\]
3. We need to evaluate:
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} a_k
\]
Using the substitution \( b_n = a_n - 1 \), we rewrite the sum:
\[
\frac{1}{n} \sum_{k=1}^{n} a_k = \frac{1}{n} \sum_{k=1}^{n} (b_k + 1) = \frac{1}{n} \sum_{k=1}^{n} b_k + \frac{1}{n} \sum_{k=1}^{n} 1 = \frac{1}{n} \sum_{k=1}^{n} b_k + 1
\]
4. We need to find:
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} b_k
\]
5. Note that \( b_n = \sin(b_{n-1}) \). Since \( |\sin(x)| < |x| \) for all \( x \neq 0 \), the sequence \( (b_n) \) is decreasing in absolute value and converges to 0. Therefore, \( \lim_{n \to \infty} b_n = 0 \).
6. By the Stolz-Cesaro theorem, if \( \lim_{n \to \infty} b_n = 0 \), then:
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} b_k = 0
\]
7. Therefore:
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} a_k = \lim_{n \to \infty} \left( \frac{1}{n} \sum_{k=1}^{n} b_k + 1 \right) = 0 + 1 = 1
\]
The final answer is \(\boxed{1}\).
|
1
|
Calculus
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
Let a convex polygon $P$ be contained in a square of side one. Show that the sum of the sides of $P$ is less than or equal to $4$.
|
1. **Understanding the Problem:**
We need to show that the perimeter of a convex polygon \( P \) contained within a square of side length 1 is less than or equal to 4.
2. **Bounding the Perimeter:**
Consider the square with side length 1. The perimeter of this square is \( 4 \times 1 = 4 \).
3. **Convex Polygon Property:**
A convex polygon \( P \) contained within this square must have its vertices either on or inside the boundary of the square. The perimeter of \( P \) is the sum of the lengths of its sides.
4. **Perimeter Comparison:**
Since \( P \) is convex and contained within the square, the perimeter of \( P \) cannot exceed the perimeter of the square. This is because the shortest path that encloses the maximum area within the square is the square itself.
5. **Formal Argument:**
- Let \( P \) be a convex polygon with vertices \( A_1, A_2, \ldots, A_n \).
- Each side of \( P \) is a line segment connecting two vertices.
- Since \( P \) is contained within the square, the total length of these line segments (the perimeter of \( P \)) must be less than or equal to the perimeter of the square.
6. **Using the Triangle Inequality:**
- For any convex polygon, the sum of the lengths of its sides is minimized when the polygon is a straight line (degenerate case) and maximized when it is a shape that closely follows the boundary of the containing shape.
- In this case, the containing shape is a square with a perimeter of 4.
7. **Conclusion:**
Therefore, the perimeter of the convex polygon \( P \) must be less than or equal to the perimeter of the square, which is 4.
\[
\boxed{4}
\]
|
4
|
Geometry
|
proof
|
Yes
|
Yes
|
aops_forum
| false
|
Consider polynomial functions $ax^2 -bx +c$ with integer coefficients which have two distinct zeros in the open interval $(0,1).$ Exhibit with proof the least positive integer value of $a$ for which such a polynomial exists.
|
1. **Identify the conditions for the polynomial to have two distinct zeros in the interval \((0,1)\):**
Let the polynomial be \( f(x) = ax^2 - bx + c \) with integer coefficients \(a\), \(b\), and \(c\). Suppose the polynomial has two distinct zeros \(0 < x_1 < x_2 < 1\). The vertex of the parabola, given by \( x_m = \frac{x_1 + x_2}{2} \), must lie in the interval \((0,1)\). The vertex formula for a quadratic function \(ax^2 - bx + c\) is \( x_m = \frac{b}{2a} \).
Therefore, we have:
\[
0 < \frac{b}{2a} < 1 \implies 0 < b < 2a \implies 1 \leq b \leq 2a - 1
\]
2. **Express \(b\) in terms of \(a\):**
Let \( b = a + k \) where \( k \in \mathbb{Z} \) and \( |k| \leq a - 1 \).
3. **Determine the conditions for \(c\):**
- For \(f(0) > 0\):
\[
c > 0 \implies c \geq 1
\]
- For \(f(1) > 0\):
\[
a - b + c > 0 \implies c \geq b - a + 1 \implies c \geq k + 1
\]
4. **Ensure the discriminant is positive:**
The discriminant \(\Delta\) of the quadratic equation \(ax^2 - bx + c = 0\) must be positive for the polynomial to have two distinct real roots:
\[
\Delta = b^2 - 4ac > 0
\]
5. **Combine the conditions:**
- From \(b = a + k\) and \(c \geq \max(1, k + 1)\):
\[
\Delta = (a + k)^2 - 4ac
\]
- Using \(c \geq 1\):
\[
\Delta \leq (a + k)^2 - 4a \cdot 1 = (a + k)^2 - 4a
\]
- Using \(c \geq k + 1\):
\[
\Delta \leq (a + k)^2 - 4a(k + 1) = (a + k)^2 - 4ak - 4a = (a - k)^2 - 4a
\]
Therefore:
\[
\Delta \leq \min((a + k)^2 - 4a, (a - k)^2 - 4a)
\]
6. **Evaluate for small values of \(a\):**
- For \(a = 1, 2, 3, 4\):
\[
\Delta \leq a^2 - 4a \leq 0 \quad \text{(no positive discriminant)}
\]
- For \(a = 5\):
\[
\Delta \leq (5 - |k|)^2 - 20
\]
The only possibility for \(\Delta > 0\) is \(k = 0\):
\[
b = a + k = 5, \quad c = 1
\]
7. **Verify the polynomial:**
The polynomial \(f(x) = 5x^2 - 5x + 1\) has the discriminant:
\[
\Delta = 5^2 - 4 \cdot 5 \cdot 1 = 25 - 20 = 5 > 0
\]
The roots are:
\[
x_{1,2} = \frac{5 \pm \sqrt{5}}{10}
\]
Both roots lie in the interval \((0,1)\).
Conclusion:
The minimum possible positive integer value of \(a\) is \(5\).
The final answer is \(\boxed{5}\)
|
5
|
Algebra
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
Call a set of positive integers "conspiratorial" if no three of them are pairwise relatively prime. What is the largest number of elements in any "conspiratorial" subset of the integers $1$ to $16$?
|
1. **Identify the problem constraints**: We need to find the largest subset of the integers from 1 to 16 such that no three elements in the subset are pairwise relatively prime.
2. **Consider the set of prime numbers within the range**: The prime numbers between 1 and 16 are \( \{2, 3, 5, 7, 11, 13\} \). Any three of these numbers are pairwise relatively prime.
3. **Calculate the maximum size of the conspiratorial subset**: If a subset contains more than \(16 - 6 + 2 = 12\) elements, it must contain at least three of the prime numbers, which would be pairwise relatively prime. Therefore, the maximum size of a conspiratorial subset is \(11\).
4. **Construct a candidate subset**: Consider the set \( S = \{2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16\} \). We need to verify that no three elements in \( S \) are pairwise relatively prime.
5. **Verify the candidate subset**:
- Every element in \( S \) is divisible by 2, 3, or 5.
- Since 5 is not in \( S \), we only need to consider divisibility by 2 and 3.
- The elements of \( S \) are:
- Divisible by 2: \( \{2, 4, 6, 8, 10, 12, 14, 16\} \)
- Divisible by 3: \( \{3, 6, 9, 12, 15\} \)
- Any three elements in \( S \) will share a common factor of either 2 or 3, ensuring they are not pairwise relatively prime.
6. **Conclusion**: Since \( S \) is a valid conspiratorial subset and has 11 elements, and we have shown that no larger conspiratorial subset can exist, we conclude that the largest conspiratorial subset of the integers from 1 to 16 has 11 elements.
The final answer is \( \boxed{ 11 } \).
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the largest $a$ for which there exists a polynomial
$$P(x) =a x^4 +bx^3 +cx^2 +dx +e$$
with real coefficients which satisfies $0\leq P(x) \leq 1$ for $-1 \leq x \leq 1.$
|
1. We start by considering the polynomial \( P(x) = ax^4 + bx^3 + cx^2 + dx + e \) with real coefficients that satisfies \( 0 \leq P(x) \leq 1 \) for \( -1 \leq x \leq 1 \).
2. We decompose \( P(x) \) into its even and odd parts:
\[
P(x) = P_1(x) + P_2(x)
\]
where \( P_1(x) = ax^4 + cx^2 + e \) is an even function and \( P_2(x) = bx^3 + dx \) is an odd function.
3. For any \( x_1 \in [-1, 1] \), we have:
\[
0 \leq P(x_1) = P_1(x_1) + P_2(x_1)
\]
and
\[
0 \leq P(-x_1) = P_1(x_1) - P_2(x_1)
\]
4. This implies:
\[
P_1(x_1) \geq \max\{-P_2(x_1), P_2(x_1)\} = |P_2(x_1)| \geq 0
\]
The equality occurs when \( P(x_1) = P(-x_1) = 0 \) and \( P_2(x_1) = 0 \).
5. Similarly, we have:
\[
1 \geq P(x_1) = P_1(x_1) + P_2(x_1)
\]
and
\[
1 \geq P(-x_1) = P_1(x_1) - P_2(x_1)
\]
6. This implies:
\[
P_1(x_1) \leq \min\{1 - P_2(x_1), 1 + P_2(x_1)\} = 1 - |P_2(x_1)| \leq 1
\]
The equality occurs when \( P(x_1) = P(-x_1) = 1 \) and \( P_2(x_1) = 0 \).
7. Therefore, \( P_1(x) \in [0, 1] \) for all \( x \in [-1, 1] \). Since \( P_2(x) = 0 \) for all \( x \) where \( P(x) = 0 \) or \( P(x) = 1 \), we can simplify by taking \( P_2(x) = 0 \) for all \( x \in [-1, 1] \). This implies \( b = d = 0 \) and \( P(x) = P_1(x) \).
8. With the notation \( y = x^2 \in [0, 1] \), we obtain:
\[
P(x) = ax^4 + cx^2 + e = ay^2 + cy + e = Q(y) \in [0, 1] \text{ for all } y \in [0, 1]
\]
9. We now search for the possible values of \( a > 0 \). The quadratic function \( f(y) = ay^2 + cy + e \) attains its minimum at \( y_m = -\frac{c}{2a} \). It is strictly decreasing on \( (-\infty, y_m] \) and strictly increasing on \( [y_m, +\infty) \).
10. **Case 1**: \( y_m \leq 0 \Rightarrow c \geq 0 \). \( Q(y) \) is strictly increasing on \( [0, 1] \). Therefore:
\[
Q(0) = e \geq 0
\]
and
\[
Q(1) = a + c + e \leq 1 \Rightarrow a \leq 1
\]
11. **Case 2**: \( y_m \in (0, 1) \). We have:
\[
Q(y) = a(y - y_m)^2 + e - \frac{c^2}{4a}
\]
The minimum value is:
\[
\min_{y \in [0, 1]} Q(y) = Q(y_m) = e - \frac{c^2}{4a} \geq 0
\]
The maximum value is:
\[
1 \geq \max_{y \in [0, 1]} Q(y) = \max\{Q(0), Q(1)\} \geq \frac{a}{4} + e - \frac{c^2}{4a} \geq \frac{a}{4} \Rightarrow a \leq 4
\]
The equality occurs for \( y_m = \frac{1}{2} \) and \( e - \frac{c^2}{4a} = 0 \). This implies \( a = 4 \), \( c = -4 \), and \( e = 1 \). In this case:
\[
Q(y) = 4y^2 - 4y + 1 \Rightarrow P(x) = 4x^4 - 4x^2 + 1
\]
12. **Case 3**: \( y_m \geq 1 \Rightarrow c \leq -2a \). \( Q(y) \) is strictly decreasing on \( [0, 1] \). Therefore:
\[
Q(0) = e \leq 1
\]
and
\[
Q(1) = a + c + e \geq 0 \Rightarrow a \leq 1
\]
13. Hence, the maximum value of \( a \) for which there exists a polynomial with real coefficients \( P(x) = ax^4 + bx^3 + cx^2 + dx + e \) such that \( P(x) \in [0, 1] \) for all \( x \in [-1, 1] \) is \( a = 4 \). The polynomial \( P(x) = 4x^4 - 4x^2 + 1 \) satisfies the conditions.
The final answer is \( \boxed{4} \).
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Two distinct squares of the $8\times8$ chessboard $C$ are said to be adjacent if they have a vertex or side in common.
Also, $g$ is called a $C$-gap if for every numbering of the squares of $C$ with all the integers $1, 2, \ldots, 64$ there exist twoadjacent squares whose numbers differ by at least $g$. Determine the largest $C$-gap $g$.
|
To determine the largest $C$-gap $g$ for an $8 \times 8$ chessboard, we need to show that for any numbering of the squares with integers $1, 2, \ldots, 64$, there exist two adjacent squares whose numbers differ by at least $g$. We will show that the largest such $g$ is $9$.
1. **Upper Bound Argument:**
- Consider the maximum difference between two adjacent squares. If we can find a path of adjacent squares from $1$ to $64$ with at most $6$ squares in between, then the difference between the numbers on these squares must be at least $9$.
- To see why, note that if there are $6$ squares between $1$ and $64$, then the sequence of numbers could be something like $1, 2, 3, \ldots, 7, 64$. The difference between $1$ and $64$ is $63$, and if we distribute this difference among $7$ steps (including $1$ and $64$), the average difference per step is $\frac{63}{7} = 9$.
- Therefore, there must be at least one step where the difference is at least $9$.
2. **Lower Bound Argument:**
- We need to show that it is possible to label the squares such that the difference between any two adjacent squares is at most $9$.
- One way to label the squares is to number them in order by row. For example, the first row is numbered $1, 2, 3, \ldots, 8$, the second row is numbered $9, 10, 11, \ldots, 16$, and so on.
- In this labeling, the difference between any two horizontally adjacent squares is $1$, and the difference between any two vertically adjacent squares is $8$.
- The maximum difference between any two adjacent squares in this labeling is $8$, which is less than $9$.
3. **Conclusion:**
- From the upper bound argument, we know that there must be at least one pair of adjacent squares with a difference of at least $9$.
- From the lower bound argument, we know that it is possible to label the squares such that the difference between any two adjacent squares is at most $8$.
- Therefore, the largest $C$-gap $g$ is $9$.
The final answer is $\boxed{9}$.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $B(n)$ be the number of ones in the base two expression for the positive integer $n.$ Determine whether
$$\exp \left( \sum_{n=1}^{\infty} \frac{ B(n)}{n(n+1)} \right)$$
is a rational number.
|
1. We start by considering the function \( B(n) \), which counts the number of ones in the binary representation of the positive integer \( n \).
2. We need to evaluate the expression:
\[
\exp \left( \sum_{n=1}^{\infty} \frac{B(n)}{n(n+1)} \right)
\]
3. First, we analyze the contribution of each digit in the binary representation of \( n \) to \( B(n) \). The last digit (least significant bit) contributes to \( B(n) \) if \( n \) is odd. Therefore, the sum of contributions from the last digit is:
\[
\sum_{\substack{n=1 \\ n \text{ odd}}}^{\infty} \frac{1}{n(n+1)}
\]
4. We can rewrite the sum for odd \( n \) as:
\[
\frac{1}{1 \cdot 2} + \frac{1}{3 \cdot 4} + \frac{1}{5 \cdot 6} + \cdots
\]
5. This series can be simplified using partial fractions:
\[
\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}
\]
6. Therefore, the series becomes:
\[
\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{5} - \frac{1}{6} \right) + \cdots
\]
7. This is a telescoping series, and it sums to:
\[
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \log(2)
\]
(This result follows from the Taylor series expansion of \(\log(1+x)\) evaluated at \(x=1\)).
8. Next, we consider the contribution of the next-to-last digit. This digit contributes to \( B(n) \) if \( n \equiv 2 \) or \( 3 \pmod{4} \). The sum of contributions from the next-to-last digit is:
\[
\sum_{\substack{n=1 \\ n \equiv 2,3 \pmod{4}}}^{\infty} \frac{1}{n(n+1)}
\]
9. This series can be written as:
\[
\frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{6 \cdot 7} + \frac{1}{7 \cdot 8} + \cdots
\]
10. Using partial fractions again, we get:
\[
\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \cdots = \frac{1}{2} \log(2)
\]
11. By repeating this argument for each digit in the binary representation, we find that the \( n \)-th last digit contributes:
\[
\frac{1}{2^{n-1}} \log(2)
\]
12. Summing these contributions, we get:
\[
\sum_{n=1}^{\infty} \frac{1}{2^{n-1}} \log(2) = \log(2) \sum_{n=1}^{\infty} \frac{1}{2^{n-1}}
\]
13. The geometric series \(\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}\) sums to:
\[
\sum_{n=1}^{\infty} \frac{1}{2^{n-1}} = 2
\]
14. Therefore, the entire sum is:
\[
\log(2) \cdot 2 = 2 \log(2)
\]
15. Finally, we need to evaluate the exponential of this sum:
\[
\exp \left( 2 \log(2) \right) = \exp \left( \log(4) \right) = 4
\]
16. Since 4 is a rational number, we conclude that the given expression is a rational number.
The final answer is \( \boxed{ 4 } \)
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $C$ be the unit circle $x^{2}+y^{2}=1 .$ A point $p$ is chosen randomly on the circumference $C$ and another point $q$ is chosen randomly from the interior of $C$ (these points are chosen independently and uniformly over their domains). Let $R$ be the rectangle with sides parallel to the $x$ and $y$-axes with diagonal $p q .$ What is the probability that no point of $R$ lies outside of $C ?$
|
1. **Define the points \( p \) and \( q \):**
- Let \( p = (\cos \theta, \sin \theta) \) where \( \theta \) is uniformly distributed over \([0, 2\pi)\).
- Let \( q = (r \cos \phi, r \sin \phi) \) where \( r \) is uniformly distributed over \([0, 1]\) and \( \phi \) is uniformly distributed over \([0, 2\pi)\).
2. **Determine the coordinates of the vertices of the rectangle \( R \):**
- The rectangle \( R \) has sides parallel to the \( x \)- and \( y \)-axes with diagonal \( pq \).
- The vertices of \( R \) are:
\[
(\cos \theta, \sin \theta), (r \cos \phi, \sin \theta), (\cos \theta, r \sin \phi), (r \cos \phi, r \sin \phi)
\]
3. **Condition for \( R \) to lie inside \( C \):**
- For \( R \) to lie entirely within the unit circle, all vertices must satisfy the equation \( x^2 + y^2 \leq 1 \).
- The vertices are:
\[
(\cos \theta, \sin \theta), (r \cos \phi, \sin \theta), (\cos \theta, r \sin \phi), (r \cos \phi, r \sin \phi)
\]
- We need to check the conditions:
\[
(r \cos \phi)^2 + (\sin \theta)^2 \leq 1
\]
\[
(\cos \theta)^2 + (r \sin \phi)^2 \leq 1
\]
\[
(r \cos \phi)^2 + (r \sin \phi)^2 \leq 1
\]
4. **Simplify the conditions:**
- The third condition simplifies to:
\[
r^2 (\cos^2 \phi + \sin^2 \phi) \leq 1 \implies r^2 \leq 1 \implies r \leq 1
\]
This condition is always satisfied since \( r \) is uniformly distributed over \([0, 1]\).
5. **Evaluate the remaining conditions:**
- The first condition:
\[
r^2 \cos^2 \phi + \sin^2 \theta \leq 1
\]
Since \( \sin^2 \theta \leq 1 \), this condition is always satisfied for \( r \leq 1 \).
- The second condition:
\[
\cos^2 \theta + r^2 \sin^2 \phi \leq 1
\]
Since \( \cos^2 \theta \leq 1 \), this condition is always satisfied for \( r \leq 1 \).
6. **Conclusion:**
- Since all conditions are satisfied for \( r \leq 1 \), the probability that no point of \( R \) lies outside of \( C \) is 1.
The final answer is \(\boxed{1}\).
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $G$ be a finite set of real $n \times n$ matrices $\left\{M_{i}\right\}, 1 \leq i \leq r,$ which form a group under matrix multiplication. Suppose that $\textstyle\sum_{i=1}^{r} \operatorname{tr}\left(M_{i}\right)=0,$ where $\operatorname{tr}(A)$ denotes the trace of the matrix $A .$ Prove that $\textstyle\sum_{i=1}^{r} M_{i}$ is the $n \times n$ zero matrix.
|
1. **Define the sum of the elements of \( G \):**
Let \( S = \sum_{i=1}^{r} M_i \). We need to show that \( S \) is the \( n \times n \) zero matrix.
2. **Injectivity and surjectivity of the map \( f(M) = M_i M \):**
Since \( G \) is a finite group under matrix multiplication, for any \( M_i \in G \), the map \( f(M) = M_i M \) is injective. In a finite set, an injective map is also surjective. Therefore, for any \( M_i \in G \), there exists a permutation of the elements of \( G \) such that:
\[
\sum_{i=1}^{r} M_i = M_k \sum_{i=1}^{r} M_i \quad \text{for all } k.
\]
3. **Summing the equalities:**
Summing the above equalities for all \( k \) from 1 to \( r \), we get:
\[
\sum_{k=1}^{r} \sum_{i=1}^{r} M_i = \sum_{k=1}^{r} M_k \sum_{i=1}^{r} M_i.
\]
This simplifies to:
\[
rS = S^2.
\]
4. **Trace of \( S \):**
Given that \( \sum_{i=1}^{r} \operatorname{tr}(M_i) = 0 \), we have:
\[
\operatorname{tr}(S) = \operatorname{tr}\left(\sum_{i=1}^{r} M_i\right) = \sum_{i=1}^{r} \operatorname{tr}(M_i) = 0.
\]
5. **Trace of \( S^2 \):**
Since \( S^2 = rS \), taking the trace on both sides gives:
\[
\operatorname{tr}(S^2) = \operatorname{tr}(rS) = r \operatorname{tr}(S) = r \cdot 0 = 0.
\]
6. **Generalizing to higher powers:**
Similarly, for any positive integer \( k \), we have:
\[
S^k = r^{k-1}S.
\]
Taking the trace, we get:
\[
\operatorname{tr}(S^k) = \operatorname{tr}(r^{k-1}S) = r^{k-1} \operatorname{tr}(S) = r^{k-1} \cdot 0 = 0.
\]
7. **Eigenvalues of \( S \):**
Since the trace of any power of \( S \) is zero, all eigenvalues of \( S \) must be zero. This implies that \( S \) is a nilpotent matrix.
8. **Nilpotency and the zero matrix:**
Since \( S \) is nilpotent and \( S^n = r^{n-1}S = 0 \), it follows that \( S \) must be the zero matrix. Therefore:
\[
S = 0.
\]
The final answer is \( \boxed{0} \).
|
0
|
Algebra
|
proof
|
Yes
|
Yes
|
aops_forum
| false
|
Label the vertices of a trapezoid $T$ inscribed in the unit circle as $A,B,C,D$ counterclockwise with $AB\parallel CD$. Let $s_1,s_2,$ and $d$ denote the lengths of $AB$, $CD$, and $OE$, where $E$ is the intersection of the diagonals of $T$, and $O$ is the center of the circle. Determine the least upper bound of $\frac{s_1-s_2}d$ over all $T$ for which $d\ne0$, and describe all cases, if any, in which equality is attained.
|
1. **Case 1: \( s_1 = s_2 \)**
If \( s_1 = s_2 \), then the trapezoid becomes an isosceles trapezoid with \( AB = CD \). In this case, the diagonals intersect at the center of the circle, making \( d = 0 \). This contradicts the given condition \( d \neq 0 \). Therefore, this case is not possible.
2. **Case 2: \( s_1 < s_2 \)**
If \( s_1 < s_2 \), then \( \frac{s_1 - s_2}{d} < 0 \). Since we are looking for the least upper bound, negative values are not relevant for our maximum.
3. **Case 3: \( s_1 > s_2 \)**
Let \( s_1 = 2L = 2\sin\left(\frac{\angle AOB}{2}\right) = 2\sin\alpha \) and \( s_2 = 2l = 2\sin\left(\frac{\angle COD}{2}\right) = 2\sin\beta \). Given \( s_1 > s_2 \), we have \( 0 < \beta < \alpha \leq \frac{\pi}{2} \).
Let \( M \) and \( N \) be the midpoints of segments \( AB \) and \( CD \), respectively. The points \( O, M, N \) are collinear. We have:
\[
OM = \sqrt{1 - L^2} = \cos\alpha \quad \text{and} \quad ON = \sqrt{1 - l^2} = \cos\beta
\]
The height of the trapezoid is \( h = MN \).
**Case 3.1: \( O \in [MN] \)**
In this case, \( h = OM + ON = \cos\alpha + \cos\beta \). The length \( ME \) is given by:
\[
ME = (\cos\alpha + \cos\beta) \cdot \frac{\sin\alpha}{\sin\alpha + \sin\beta}
\]
Therefore, \( d = OE = ME - OM \):
\[
d = (\cos\alpha + \cos\beta) \cdot \frac{\sin\alpha}{\sin\alpha + \sin\beta} - \cos\alpha = \frac{\sin(\alpha - \beta)}{\sin\alpha + \sin\beta}
\]
Now, we calculate \( \frac{s_1 - s_2}{d} \):
\[
\frac{s_1 - s_2}{d} = \frac{2(\sin\alpha - \sin\beta) \cdot (\sin\alpha + \sin\beta)}{\sin(\alpha - \beta)}
\]
Using the product-to-sum identities:
\[
\sin\alpha - \sin\beta = 2\cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right)
\]
\[
\sin\alpha + \sin\beta = 2\sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right)
\]
Substituting these into the expression:
\[
\frac{s_1 - s_2}{d} = \frac{2 \cdot 2\cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) \cdot 2\sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right)}{2\sin\left(\frac{\alpha - \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right)}
\]
Simplifying:
\[
\frac{s_1 - s_2}{d} = 2\sin(\alpha + \beta)
\]
The maximum value of \( 2\sin(\alpha + \beta) \) for \( 0 < \beta < \alpha \leq \frac{\pi}{2} \) is \( 2 \), which occurs when \( \alpha + \beta = \frac{\pi}{2} \).
**Case 3.2: \( M \in [ON] \)**
In this case, \( h = ON - OM = \cos\beta - \cos\alpha \). The length \( ME \) is given by:
\[
ME = (\cos\beta - \cos\alpha) \cdot \frac{\sin\alpha}{\sin\alpha + \sin\beta}
\]
Therefore, \( d = OE = ME + OM \):
\[
d = (\cos\beta - \cos\alpha) \cdot \frac{\sin\alpha}{\sin\alpha + \sin\beta} + \cos\alpha = \frac{\sin(\alpha + \beta)}{\sin\alpha + \sin\beta}
\]
Now, we calculate \( \frac{s_1 - s_2}{d} \):
\[
\frac{s_1 - s_2}{d} = \frac{2(\sin\alpha - \sin\beta) \cdot (\sin\alpha + \sin\beta)}{\sin(\alpha + \beta)}
\]
Using the product-to-sum identities:
\[
\sin\alpha - \sin\beta = 2\cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right)
\]
\[
\sin\alpha + \sin\beta = 2\sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right)
\]
Substituting these into the expression:
\[
\frac{s_1 - s_2}{d} = \frac{2 \cdot 2\cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) \cdot 2\sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right)}{2\sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha + \beta}{2}\right)}
\]
Simplifying:
\[
\frac{s_1 - s_2}{d} = 2\sin(\alpha - \beta)
\]
The supremum of \( 2\sin(\alpha - \beta) \) for \( 0 < \beta < \alpha \leq \frac{\pi}{2} \) is \( 2 \), but it cannot be attained since \( 0 < \alpha - \beta < \frac{\pi}{2} \).
4. **Final Conclusion:**
The least upper bound of \( \frac{s_1 - s_2}{d} \) is \( 2 \). The equality \( \frac{s_1 - s_2}{d} = 2 \) occurs when \( \alpha + \beta = \frac{\pi}{2} \).
The final answer is \( \boxed{ 2 } \)
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Consider a paper punch that can be centered at any point
of the plane and that, when operated, removes from the
plane precisely those points whose distance from the
center is irrational. How many punches are needed to
remove every point?
|
1. **One punch is not enough.**
- If we center a punch at any point, it will remove all points whose distance from the center is irrational. However, there will always be points at rational distances from the center that remain. Therefore, one punch cannot remove every point.
2. **Two punches are not enough.**
- Suppose we center the first punch at \((0,0)\) and the second punch at \((a,0)\) for some \(a > 0\).
- Consider the point \((u,v)\) where \(u = \frac{a}{2}\) and let \(p\) be a rational number such that \(p > \frac{a}{2}\). Let \(v = \sqrt{p^2 - u^2}\).
- The distance from \((u,v)\) to \((0,0)\) is:
\[
\sqrt{u^2 + v^2} = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\sqrt{p^2 - \left(\frac{a}{2}\right)^2}\right)^2} = \sqrt{\frac{a^2}{4} + p^2 - \frac{a^2}{4}} = p
\]
which is rational.
- The distance from \((u,v)\) to \((a,0)\) is:
\[
\sqrt{(u - a)^2 + v^2} = \sqrt{\left(\frac{a}{2} - a\right)^2 + \left(\sqrt{p^2 - \left(\frac{a}{2}\right)^2}\right)^2} = \sqrt{\left(-\frac{a}{2}\right)^2 + p^2 - \left(\frac{a}{2}\right)^2} = p
\]
which is also rational.
- Therefore, the point \((u,v)\) cannot be removed by either punch, proving that two punches are not enough.
3. **Three punches are enough.**
- Consider punches centered at \((0,0)\), \((1,0)\), and \((\pi,0)\).
- Suppose there is a point \((u,v)\) such that both \(u^2 + v^2\) and \((u-1)^2 + v^2\) are rational.
- Then:
\[
u^2 + v^2 - (u-1)^2 - v^2 = 2u - 1
\]
is rational, implying \(u\) is rational.
- Now consider the distance from \((u,v)\) to \((\pi,0)\):
\[
(u-\pi)^2 + v^2
\]
If \(u\) is rational, then \((u-\pi)^2 + v^2\) cannot be rational because \(\pi\) is transcendental (not algebraic).
- Therefore, \(u^2 + v^2\) and \((u-\pi)^2 + v^2\) cannot both be rational, meaning the point \((u,v)\) will be removed by at least one of the three punches.
Thus, three punches are sufficient to remove every point from the plane.
The final answer is \(\boxed{3}\).
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
If $X$ is a finite set, let $X$ denote the number of elements in $X$. Call an ordered pair $(S,T)$ of subsets of $ \{ 1, 2, \cdots, n \} $ $ \emph {admissible} $ if $ s > |T| $ for each $ s \in S $, and $ t > |S| $ for each $ t \in T $. How many admissible ordered pairs of subsets $ \{ 1, 2, \cdots, 10 \} $ are there? Prove your answer.
|
To solve the problem, we need to count the number of admissible ordered pairs \((S, T)\) of subsets of \(\{1, 2, \cdots, 10\}\) such that \(s > |T|\) for each \(s \in S\) and \(t > |S|\) for each \(t \in T\).
1. **Understanding the conditions**:
- For each element \(s \in S\), \(s > |T|\).
- For each element \(t \in T\), \(t > |S|\).
2. **Analyzing the constraints**:
- If \(S\) is non-empty, let \(m = |S|\). Then, for each \(t \in T\), \(t > m\). This means \(T\) can only contain elements greater than \(m\).
- Similarly, if \(T\) is non-empty, let \(n = |T|\). Then, for each \(s \in S\), \(s > n\). This means \(S\) can only contain elements greater than \(n\).
3. **Considering the possible values for \(S\) and \(T\)**:
- If \(S\) and \(T\) are both non-empty, the smallest element in \(S\) must be greater than the size of \(T\), and the smallest element in \(T\) must be greater than the size of \(S\). This creates a contradiction unless both sets are empty.
4. **Counting the admissible pairs**:
- The only way to satisfy both conditions without contradiction is if both \(S\) and \(T\) are empty. Therefore, the only admissible pair is \((\emptyset, \emptyset)\).
5. **Conclusion**:
- There is exactly one admissible ordered pair of subsets of \(\{1, 2, \cdots, 10\}\).
The final answer is \(\boxed{1}\).
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A deck of $2n$ cards numbered from $1$ to $2n$ is shuffled and n cards are dealt to $A$ and $B$. $A$ and $B$ alternately discard a card face up, starting with $A$. The game when the sum of the discards is first divisible by $2n + 1$, and the last person to discard wins. What is the probability that $A$ wins if neither player makes a mistake?
|
1. **Initial Setup:**
- We have a deck of $2n$ cards numbered from $1$ to $2n$.
- The deck is shuffled, and $n$ cards are dealt to player $A$ and $n$ cards to player $B$.
- Players $A$ and $B$ alternately discard a card face up, starting with player $A$.
- The game ends when the sum of the discarded cards is first divisible by $2n + 1$, and the last person to discard wins.
2. **Observation:**
- Player $A$ cannot win on his first move because the sum of a single card cannot be divisible by $2n + 1$ (since $1 \leq k \leq 2n$ for any card $k$).
3. **Lemma:**
- Player $B$ can prevent player $A$ from making the sum of the discarded cards divisible by $2n + 1$.
4. **Proof of Lemma:**
- Before player $B$ makes his move, he has one more card than player $A$.
- Suppose the sum of the discarded cards before player $B$'s move is $S$.
- Player $B$ can always choose a card such that the new sum $S + b$ (where $b$ is the card discarded by player $B$) is not divisible by $2n + 1$.
- This is because there are $2n$ possible sums modulo $2n + 1$, and player $B$ has $n$ cards to choose from. Since $n < 2n + 1$, there will always be a card that does not make the sum divisible by $2n + 1$.
5. **Continuation of the Process:**
- Player $B$ can continue this strategy until his last turn.
- On his last turn, player $B$ will discard his final card, ensuring that the sum of all discarded cards is $2n^2 + n$.
- Note that $2n^2 + n$ is divisible by $2n + 1$ because:
\[
2n^2 + n = (2n + 1) \cdot n
\]
- Therefore, the sum of the discarded cards will be divisible by $2n + 1$ on player $B$'s last turn, making player $B$ the winner.
6. **Conclusion:**
- Since player $B$ can always ensure that the sum of the discarded cards is not divisible by $2n + 1$ until his last turn, player $A$ cannot win the game.
The final answer is $\boxed{0}$.
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $N$ be the positive integer with 1998 decimal digits, all of them 1; that is,
\[N=1111\cdots 11.\]
Find the thousandth digit after the decimal point of $\sqrt N$.
|
To find the thousandth digit after the decimal point of \(\sqrt{N}\), where \(N\) is a number with 1998 digits, all of them being 1, we can proceed as follows:
1. **Express \(N\) in a more manageable form:**
\[
N = 111\ldots111 \quad \text{(1998 digits)}
\]
This can be written as:
\[
N = \frac{10^{1998} - 1}{9}
\]
2. **Approximate \(\sqrt{N}\):**
We need to find \(\sqrt{N}\). Using the expression for \(N\):
\[
\sqrt{N} = \sqrt{\frac{10^{1998} - 1}{9}}
\]
For large \(n\), \(\sqrt{10^{2n} - 1} \approx 10^n\). Therefore:
\[
\sqrt{N} \approx \sqrt{\frac{10^{1998}}{9}} = \frac{10^{999}}{\sqrt{9}} = \frac{10^{999}}{3}
\]
3. **Decimal expansion of \(\frac{10^{999}}{3}\):**
To find the thousandth digit after the decimal point of \(\frac{10^{999}}{3}\), we need to understand the decimal expansion of \(\frac{1}{3}\):
\[
\frac{1}{3} = 0.\overline{3}
\]
Therefore:
\[
\frac{10^{999}}{3} = 10^{999-1} \times \frac{10}{3} = 10^{998} \times 3.\overline{3} = 3.\overline{3} \times 10^{998}
\]
4. **Determine the thousandth digit:**
The decimal expansion of \(3.\overline{3} \times 10^{998}\) will be:
\[
3.\underbrace{333\ldots333}_{998 \text{ threes}}333\ldots
\]
The thousandth digit after the decimal point is the 1000th digit in the sequence of threes.
Therefore, the thousandth digit after the decimal point of \(\sqrt{N}\) is \(3\).
The final answer is \(\boxed{3}\)
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Right triangle $ABC$ has right angle at $C$ and $\angle BAC=\theta$; the point $D$ is chosen on $AB$ so that $|AC|=|AD|=1$; the point $E$ is chosen on $BC$ so that $\angle CDE=\theta$. The perpendicular to $BC$ at $E$ meets $AB$ at $F$. Evaluate $\lim_{\theta\to 0}|EF|$.
|
1. **Identify the given information and setup the problem:**
- Right triangle \(ABC\) with \(\angle C = 90^\circ\) and \(\angle BAC = \theta\).
- Point \(D\) on \(AB\) such that \(|AC| = |AD| = 1\).
- Point \(E\) on \(BC\) such that \(\angle CDE = \theta\).
- Perpendicular from \(E\) to \(BC\) meets \(AB\) at \(F\).
2. **Determine the coordinates of points \(A\), \(B\), and \(C\):**
- Let \(C\) be at the origin \((0,0)\).
- Let \(A\) be at \((1,0)\) since \(|AC| = 1\).
- Let \(B\) be at \((1, \tan \theta)\) since \(\angle BAC = \theta\) and \(|BC| = \tan \theta\).
3. **Find the coordinates of point \(D\):**
- Since \(|AD| = 1\) and \(D\) lies on \(AB\), \(D\) must be at \((1,0)\) because \(|AD| = |AC| = 1\).
4. **Determine the coordinates of point \(E\):**
- Point \(E\) lies on \(BC\) such that \(\angle CDE = \theta\).
- Since \(\angle CDE = \theta\) and \(\angle DCE = 90^\circ - \theta\), \(E\) must be at \((1, \tan \theta)\).
5. **Find the coordinates of point \(F\):**
- Point \(F\) is the intersection of the perpendicular from \(E\) to \(BC\) with \(AB\).
- The slope of \(BC\) is \(\tan \theta\), so the slope of the perpendicular is \(-\cot \theta\).
- The equation of the line through \(E\) with slope \(-\cot \theta\) is:
\[
y - \tan \theta = -\cot \theta (x - 1)
\]
- Simplifying, we get:
\[
y = -\cot \theta (x - 1) + \tan \theta
\]
- Since \(F\) lies on \(AB\), \(y = 0\):
\[
0 = -\cot \theta (x - 1) + \tan \theta
\]
\[
\cot \theta (x - 1) = \tan \theta
\]
\[
x - 1 = \tan^2 \theta
\]
\[
x = 1 + \tan^2 \theta
\]
- Therefore, \(F\) is at \((1 + \tan^2 \theta, 0)\).
6. **Calculate the length \(EF\):**
- The coordinates of \(E\) are \((1, \tan \theta)\) and \(F\) are \((1 + \tan^2 \theta, 0)\).
- Using the distance formula:
\[
EF = \sqrt{(1 + \tan^2 \theta - 1)^2 + (\tan \theta - 0)^2}
\]
\[
EF = \sqrt{(\tan^2 \theta)^2 + (\tan \theta)^2}
\]
\[
EF = \sqrt{\tan^4 \theta + \tan^2 \theta}
\]
\[
EF = \tan \theta \sqrt{\tan^2 \theta + 1}
\]
\[
EF = \tan \theta \sec \theta
\]
\[
EF = \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\cos \theta}
\]
\[
EF = \frac{\sin \theta}{\cos^2 \theta}
\]
7. **Evaluate the limit as \(\theta \to 0\):**
\[
\lim_{\theta \to 0} EF = \lim_{\theta \to 0} \frac{\sin \theta}{\cos^2 \theta}
\]
- Using L'Hospital's Rule:
\[
\lim_{\theta \to 0} \frac{\sin \theta}{\cos^2 \theta} = \lim_{\theta \to 0} \frac{\cos \theta}{-2 \cos \theta \sin \theta} = \lim_{\theta \to 0} \frac{1}{-2 \sin \theta} = \frac{1}{0} = \infty
\]
However, the correct approach should be:
\[
\lim_{\theta \to 0} \frac{\sin \theta}{\cos^2 \theta} = \lim_{\theta \to 0} \frac{\sin \theta}{1} = 0
\]
The final answer is \(\boxed{0}\)
|
0
|
Geometry
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola $ xy\equal{}1$ and both branches of the hyperbola $ xy\equal{}\minus{}1.$ (A set $ S$ in the plane is called [i]convex[/i] if for any two points in $ S$ the line segment connecting them is contained in $ S.$)
|
1. **Identify the problem and the constraints:**
We need to find the least possible area of a convex set in the plane that intersects both branches of the hyperbolas \(xy = 1\) and \(xy = -1\). A convex set is defined such that for any two points in the set, the line segment connecting them is also contained in the set.
2. **Determine the points of intersection:**
The convex set must contain at least one point on each of the four branches of the hyperbolas. Let these points be:
\[
(x_1, y_1), (-x_2, y_2), (-x_3, -y_3), (x_4, -y_4)
\]
where \(x_1, x_2, x_3, x_4, y_1, y_2, y_3, y_4\) are positive numbers and satisfy the hyperbola equations:
\[
x_1 y_1 = 1, \quad x_2 y_2 = 1, \quad x_3 y_3 = 1, \quad x_4 y_4 = 1
\]
3. **Calculate the area of the convex hull:**
The smallest convex set containing these points is the convex hull, which in this case is a quadrilateral. The area \(A\) of a quadrilateral with vertices \((x_1, y_1), (-x_2, y_2), (-x_3, -y_3), (x_4, -y_4)\) can be calculated using the shoelace formula:
\[
A = \frac{1}{2} \left| x_1 y_2 + (-x_2) y_3 + (-x_3) (-y_4) + x_4 y_1 - (y_1 (-x_2) + y_2 (-x_3) + y_3 x_4 + y_4 x_1) \right|
\]
4. **Simplify the area expression:**
Substitute \(y_i = \frac{1}{x_i}\) for \(i = 1, 2, 3, 4\):
\[
A = \frac{1}{2} \left| x_1 \frac{1}{x_2} + (-x_2) \frac{1}{x_3} + (-x_3) \frac{1}{x_4} + x_4 \frac{1}{x_1} - \left( \frac{1}{x_1} (-x_2) + \frac{1}{x_2} (-x_3) + \frac{1}{x_3} x_4 + \frac{1}{x_4} x_1 \right) \right|
\]
\[
= \frac{1}{2} \left| \frac{x_1}{x_2} - \frac{x_2}{x_3} - \frac{x_3}{x_4} + \frac{x_4}{x_1} - \left( -\frac{x_2}{x_1} - \frac{x_3}{x_2} + \frac{x_4}{x_3} + \frac{x_1}{x_4} \right) \right|
\]
\[
= \frac{1}{2} \left| \frac{x_1}{x_2} + \frac{x_2}{x_1} + \frac{x_2}{x_3} + \frac{x_3}{x_2} + \frac{x_3}{x_4} + \frac{x_4}{x_3} + \frac{x_4}{x_1} + \frac{x_1}{x_4} \right|
\]
5. **Apply the AM-GM inequality:**
For any positive \(u\), we have \(u + \frac{1}{u} \geq 2\). Applying this to each term:
\[
\frac{x_1}{x_2} + \frac{x_2}{x_1} \geq 2, \quad \frac{x_2}{x_3} + \frac{x_3}{x_2} \geq 2, \quad \frac{x_3}{x_4} + \frac{x_4}{x_3} \geq 2, \quad \frac{x_4}{x_1} + \frac{x_1}{x_4} \geq 2
\]
Summing these inequalities:
\[
\frac{x_1}{x_2} + \frac{x_2}{x_1} + \frac{x_2}{x_3} + \frac{x_3}{x_2} + \frac{x_3}{x_4} + \frac{x_4}{x_3} + \frac{x_4}{x_1} + \frac{x_1}{x_4} \geq 8
\]
6. **Determine the minimum area:**
Therefore, the minimum area is:
\[
A \geq \frac{1}{2} \times 8 = 4
\]
Equality holds when \(x_1 = x_2 = x_3 = x_4\), meaning the quadrilateral is a rectangle with sides parallel to the axes.
The final answer is \(\boxed{4}\).
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the least positive integer \(M\) for which there exist a positive integer \(n\) and polynomials \(P_1(x)\), \(P_2(x)\), \(\ldots\), \(P_n(x)\) with integer coefficients satisfying \[Mx=P_1(x)^3+P_2(x)^3+\cdots+P_n(x)^3.\]
[i]Proposed by Karthik Vedula[/i]
|
1. **Claim 1: \(3 \mid M\)**
Differentiate the given equation \(Mx = P_1(x)^3 + P_2(x)^3 + \cdots + P_n(x)^3\) with respect to \(x\):
\[
M = \sum_{i=1}^n 3P_i'(x)P_i(x)^2.
\]
Each term on the right-hand side is \(3\) times an integer polynomial, hence \(3 \mid M\).
2. **Claim 2: \(2 \mid M\)**
Consider the roots of the polynomial \(x^2 + x + 1 = 0\), which are \(\alpha = -\frac{1}{2} + \frac{i\sqrt{3}}{2}\) and \(\beta = -\frac{1}{2} - \frac{i\sqrt{3}}{2}\). Evaluate the given equation at \(x = \alpha\) and \(x = \beta\):
\[
M\alpha = \sum_{i=1}^n P_i(\alpha)^3 \quad \text{and} \quad M\beta = \sum_{i=1}^n P_i(\beta)^3.
\]
Adding these two equations, we get:
\[
M(\alpha + \beta) = \sum_{i=1}^n \left( P_i(\alpha)^3 + P_i(\beta)^3 \right).
\]
Since \(\alpha + \beta = -1\), we have:
\[
-M = \sum_{i=1}^n \left( P_i(\alpha)^3 + P_i(\beta)^3 \right).
\]
Reduce \(P_i(\alpha)\) to the linear integer polynomial \(r_i\alpha + s_i\) via repeated applications of the identity \(\alpha^2 = -\alpha - 1\). Since \(\alpha\) and \(\beta\) are conjugates, it follows that \(P_i(\beta) = r_i\beta + s_i\) as well. Thus,
\[
-M = \sum_{i=1}^n \left( (r_i\alpha + s_i)^3 + (r_i\beta + s_i)^3 \right).
\]
Expanding the cubes, we get:
\[
(r_i\alpha + s_i)^3 + (r_i\beta + s_i)^3 = 2r_i^3 + 2s_i^3 - 3r_is_i(r_i + s_i).
\]
Observe that \(r_is_i(r_i + s_i)\) is always even, so every term in the right-hand side summation is even, hence \(2 \mid M\).
3. **Finding the least \(M\)**
We need to find the least positive integer \(M\) such that \(M\) is divisible by both 2 and 3. The least common multiple of 2 and 3 is 6. We need to verify if \(M = 6\) works:
\[
6x = (x)^3 + (-x)^3 + (x + 1)^3 + (x - 1)^3.
\]
Simplifying the right-hand side:
\[
6x = x^3 + (-x)^3 + (x + 1)^3 + (x - 1)^3 = x^3 - x^3 + (x + 1)^3 + (x - 1)^3.
\]
\[
(x + 1)^3 = x^3 + 3x^2 + 3x + 1 \quad \text{and} \quad (x - 1)^3 = x^3 - 3x^2 + 3x - 1.
\]
Adding these:
\[
(x + 1)^3 + (x - 1)^3 = (x^3 + 3x^2 + 3x + 1) + (x^3 - 3x^2 + 3x - 1) = 2x^3 + 6x.
\]
Therefore,
\[
6x = 6x.
\]
This confirms that \(M = 6\) works.
The final answer is \(\boxed{6}\).
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Elmo has 2023 cookie jars, all initially empty. Every day, he chooses two distinct jars and places a cookie in each. Every night, Cookie Monster finds a jar with the most cookies and eats all of them. If this process continues indefinitely, what is the maximum possible number of cookies that the Cookie Monster could eat in one night?
[i]Proposed by Espen Slettnes[/i]
|
1. **Define the function and initial setup:**
Let \( f(n) = 2^n \) if \( n \neq 0 \) and \( f(0) = 0 \). Suppose at a moment the jars contain \( a_1, a_2, \ldots, a_{2023} \) cookies. Define \( d = \max a_i \) and set \( L = \sum f(a_i) \).
2. **Show that \( L \) does not increase:**
We need to show that whenever \( d \geq 2 \), the next value of \( L \) (after both Elmo and Cookie Monster play) is at most \( L \).
- **Case 1: Elmo puts cookies into two non-empty jars \( i \) and \( j \):**
- If \( a_i = a_j = d \), then Cookie Monster will eat all the cookies from one of these jars. Thus,
\[
L' = L + 2^{d+1} - 2^d - 2^d = L.
\]
- If \( a_i = d \) and \( a_j < d \), then
\[
L' = L + 2^{a_j+1} - 2^{a_j} - 2^{d+1} < L.
\]
- If \( a_i, a_j < d \), then
\[
L' = L + 2^{a_i+1} + 2^{a_j+1} - 2^d - 2^{a_i} - 2^{a_j} \leq L.
\]
- **Case 2: One jar \( i \) was empty but \( j \) was not:**
- If \( a_j = d \), then
\[
L' = L - 2^{d+1} + 2 < L.
\]
- If \( a_j < d \), then
\[
L' = L + 2^{a_j+1} - 2^{a_j} - 2^d + 2 \leq L.
\]
- **Case 3: Both jars \( i \) and \( j \) were empty:**
\[
L' = L + 2 + 2 - 2^d \leq L \quad \text{since} \quad d \geq 2.
\]
3. **Claim: \( L < 2^{12} \) every morning (before Elmo plays and after Cookie Monster's turn):**
- **Proof:**
Suppose that at some moment, for the first time, we have \( L < 2^{12} \) and \( L' \geq 2^{12} \). Since \( L' > L \), we have \( d = 1 \). Thus,
\[
L \leq 2^1 + 2^1 + \ldots + 2^1 = 2 \cdot 2023 = 4046,
\]
and thus,
\[
L' \leq L + 2 < 2^{12},
\]
which is a contradiction. Therefore, \( L < 2^{12} \).
4. **Conclusion:**
Since \( L < 2^{12} \), we have that \( a_i \leq 11 \) every morning. Therefore, \( a_i \geq 12 \) at each time, and so Cookie Monster can eat at most 12 cookies on a single night.
The final answer is \( \boxed{12} \).
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Two triangles intersect to form seven finite disjoint regions, six of which are triangles with area 1. The last region is a hexagon with area \(A\). Compute the minimum possible value of \(A\).
[i]Proposed by Karthik Vedula[/i]
|
1. **Define Variables and Setup**:
Let the desired area of the hexagon be \( S \). We are given that there are six triangular regions each with area 1, and one hexagonal region with area \( S \). The total area of the two intersecting triangles is thus \( 6 + S \).
2. **Use of Convex Quadrilateral Property**:
We use the fact that for any convex quadrilateral \( WXYZ \), if \( P \) is the intersection of the diagonals \( WY \) and \( XZ \), then:
\[
S_{PWX} \cdot S_{PYZ} = S_{PXY} \cdot S_{PZW}
\]
This property will help us relate the areas of the triangles formed by the intersection of the two main triangles.
3. **Assign Areas to Triangles**:
Let the areas of the six triangular regions be denoted as follows:
\[
S_{AKF} = a, \quad S_{FLB} = f, \quad S_{BGD} = b, \quad S_{DHC} = d, \quad S_{CIE} = c, \quad S_{EJA} = e
\]
Notice that each of these areas is given to be 1.
4. **Apply the Convex Quadrilateral Property**:
Using the property mentioned, we can write the areas of the larger triangles formed by the intersection points:
\[
S_{ALD} = \frac{(a+1)(b+1)}{f} \geq \frac{4\sqrt{ab}}{f}
\]
\[
S_{FGC} = \frac{(f+1)(d+1)}{b} \geq \frac{4\sqrt{fd}}{b}
\]
\[
S_{BHE} = \frac{(b+1)(c+1)}{d} \geq \frac{4\sqrt{bc}}{d}
\]
\[
S_{DIA} = \frac{(d+1)(e+1)}{c} \geq \frac{4\sqrt{de}}{c}
\]
\[
S_{CJF} = \frac{(c+1)(a+1)}{e} \geq \frac{4\sqrt{ca}}{e}
\]
\[
S_{EKB} = \frac{(e+1)(f+1)}{a} \geq \frac{4\sqrt{ef}}{a}
\]
5. **Sum of Areas**:
Summing these inequalities, we get:
\[
3S + 6 = S_{ALD} + S_{FGC} + S_{BHE} + S_{DIA} + S_{CJF} + S_{EKB} \geq \sum \frac{4\sqrt{ab}}{f} \geq 24
\]
This implies:
\[
3S + 6 \geq 24 \implies S \geq 6
\]
6. **Conclusion**:
Therefore, the minimum possible value of \( S \) is 6.
The final answer is \( \boxed{6} \).
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
In a volleyball tournament for the Euro-African cup, there were nine more teams from Europe than from Africa. Each pair of teams played exactly once and the Europeans teams won precisely nine times as many matches as the African teams, overall. What is the maximum number of matches that a single African team might have won?
|
1. Let \( n \) be the number of African teams. Then, the number of European teams is \( n + 9 \).
2. Each pair of teams plays exactly once. Therefore, the total number of matches played is:
\[
\binom{n}{2} + \binom{n+9}{2} + n(n+9)
\]
where \(\binom{n}{2}\) represents the matches between African teams, \(\binom{n+9}{2}\) represents the matches between European teams, and \(n(n+9)\) represents the matches between African and European teams.
3. Let \( k \) be the number of games in which a European team defeated an African team. The condition states that the Europeans won precisely nine times as many matches as the African teams. Therefore, we have:
\[
\binom{n+9}{2} + k = 9 \left( \binom{n}{2} + (n(n+9) - k) \right)
\]
4. Expanding the binomial coefficients, we get:
\[
\frac{(n+9)(n+8)}{2} + k = 9 \left( \frac{n(n-1)}{2} + n(n+9) - k \right)
\]
5. Simplifying the equation:
\[
\frac{(n+9)(n+8)}{2} + k = 9 \left( \frac{n(n-1)}{2} + n(n+9) - k \right)
\]
\[
\frac{(n+9)(n+8)}{2} + k = 9 \left( \frac{n^2 - n}{2} + n^2 + 9n - k \right)
\]
\[
\frac{(n+9)(n+8)}{2} + k = 9 \left( \frac{n^2 - n + 2n^2 + 18n - 2k}{2} \right)
\]
\[
\frac{(n+9)(n+8)}{2} + k = 9 \left( \frac{3n^2 + 17n - 2k}{2} \right)
\]
\[
\frac{(n+9)(n+8)}{2} + k = \frac{9(3n^2 + 17n - 2k)}{2}
\]
\[
(n+9)(n+8) + 2k = 9(3n^2 + 17n - 2k)
\]
\[
n^2 + 17n + 72 + 2k = 27n^2 + 153n - 18k
\]
\[
2k + 18k = 27n^2 - n^2 + 153n - 17n - 72
\]
\[
20k = 26n^2 + 136n - 72
\]
\[
k = \frac{13n^2 + 68n - 36}{10}
\]
6. Clearly, \( k \leq n(n+9) \), so:
\[
10k + 36 \leq 10n^2 + 90n + 36
\]
\[
13n^2 + 68n \leq 10n^2 + 90n + 36
\]
\[
3n^2 - 22n \leq 36
\]
\[
3n^2 - 22n - 36 \leq 0
\]
7. Solving the quadratic inequality:
\[
n = \frac{22 \pm \sqrt{484 + 432}}{6}
\]
\[
n = \frac{22 \pm \sqrt{916}}{6}
\]
\[
n = \frac{22 \pm 2\sqrt{229}}{6}
\]
\[
n = \frac{11 \pm \sqrt{229}}{3}
\]
Since \( n \) must be an integer, we test values from 1 to 8. Only \( n = 6 \) and \( n = 8 \) give integer \( k \), which are \( k = 84 \) and \( k = 134 \) respectively.
8. If \( n = 6 \) and \( k = 84 \), there are 6 African teams and 15 European teams. To maximize the number of wins of one African team, have all 6 of these wins be attributed to the same African team. Since they can have up to 5 more wins (from other African teams), the maximum possible is 11.
9. In the other case of \( n = 8 \) and \( k = 134 \), the maximum is \( 7 + 2 = 9 \), which is not as high.
Thus, the answer is 11, achieved when:
- There are 15 European teams and 6 African teams.
- One of the African teams wins against all 5 other African teams.
- That same African team beats 6 out of the 15 European teams. All other games between African and European teams result in a win for the European team.
This clearly works since:
\[
\binom{15}{2} + 84 = 189 = 9 \left( \binom{6}{2} + 6 \right)
\]
The final answer is \(\boxed{11}\).
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
There are 51 senators in a senate. The senate needs to be divided into $n$ committees so that each senator is on one committee. Each senator hates exactly three other senators. (If senator A hates senator B, then senator B does [i]not[/i] necessarily hate senator A.) Find the smallest $n$ such that it is always possible to arrange the committees so that no senator hates another senator on his or her committee.
|
1. **Restate the problem in graph theory terms:**
- We have a directed graph \( G \) with 51 vertices, where each vertex has an out-degree of exactly 3.
- We need to find the smallest number \( n \) such that we can partition the vertices into \( n \) committees (or color the vertices with \( n \) colors) such that no two vertices in the same committee (or with the same color) are connected by a directed edge.
2. **Generalize the problem:**
- Let \( G \) be a directed graph on \( n \) vertices where no vertex has an out-degree greater than \( k \).
- We need to show that the vertices of \( G \) can be colored with \( 2k+1 \) colors such that no two vertices of the same color are connected by a directed edge.
3. **Base case:**
- For \( n = 1 \), the graph has only one vertex, and it can be colored with one color. This is trivially true.
4. **Inductive step:**
- Assume the statement is true for a graph with \( n \) vertices.
- Consider a graph \( G' \) with \( n+1 \) vertices where no vertex has an out-degree greater than \( k \).
- We need to show that \( G' \) can be colored with \( 2k+1 \) colors.
5. **Find a vertex with in-degree \(\le k\):**
- In \( G' \), there must exist a vertex \( \nu \) with in-degree \(\le k\). If every vertex had an in-degree greater than \( k \), the total number of edges would exceed \( (n+1)k \), which is a contradiction since each vertex has an out-degree of at most \( k \).
6. **Partition the graph:**
- Remove the vertex \( \nu \) from \( G' \), resulting in a subgraph \( G' \setminus \nu \) with \( n \) vertices.
- By the induction hypothesis, \( G' \setminus \nu \) can be colored with \( 2k+1 \) colors such that no two adjacent vertices share the same color.
7. **Color the removed vertex:**
- The vertex \( \nu \) has at most \( 2k \) neighbors in \( G' \setminus \nu \) (since it has an out-degree of at most \( k \) and an in-degree of at most \( k \)).
- Since \( 2k+1 \) colors are used to color \( G' \setminus \nu \), there is at least one color available that is not used by any of \( \nu \)'s neighbors.
- Assign this available color to \( \nu \).
8. **Conclusion:**
- By induction, the graph \( G \) with \( n+1 \) vertices can be colored with \( 2k+1 \) colors such that no two adjacent vertices share the same color.
Therefore, for the given problem with \( k = 3 \), the smallest number \( n \) such that it is always possible to arrange the committees so that no senator hates another senator on his or her committee is \( 2k+1 = 2 \cdot 3 + 1 = 7 \).
The final answer is \( \boxed{7} \).
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Mr. Fat and Ms. Taf play a game. Mr. Fat chooses a sequence of positive integers $ k_1, k_2, \ldots , k_n$. Ms. Taf must guess this sequence of integers. She is allowed to give Mr. Fat a red card and a blue card, each with an integer written on it. Mr. Fat replaces the number on the red card with $ k_1$ times the number on the red card plus the number on the blue card, and replaces the number on the blue card with the number originally on the red card. He repeats this process with number $ k_2$. (That is, he replaces the number on the red card with $ k_2$ times the number now on the red card plus the number now on the blue card, and replaces the number on the blue card with the number that was just placed on the red card.) He then repeats this process with each of the numbers $ k_3, \ldots k_n$, in this order. After has has gone through the sequence of integers, Mr. Fat then gives the cards back to Ms. Taf. How many times must Ms. Taf submit the red and blue cards in order to be able to determine the sequence of integers $ k_1, k_2, \ldots k_n$?
|
1. **Initial Setup:**
Let \( r_i \) and \( b_i \) denote the numbers on the red and blue cards at stage \( i \) (after Mr. Fat has just carried out the rewriting process with number \( k_i \)), where \( r_0 \) and \( b_0 \) are the initial numbers. Ms. Taf can choose \( r_0 > b_0 \).
2. **Transformation Process:**
The given algorithm translates to the following transformations for \( 0 \leq i < n \):
\[
r_{i+1} = k_{i+1} r_i + b_i
\]
\[
b_{i+1} = r_i
\]
3. **Induction Hypothesis:**
We will prove by induction that \( r_i > b_i \) for all \( i \).
- **Base Case:** For \( i = 0 \), we have \( r_0 > b_0 \) by choice.
- **Inductive Step:** Assume \( r_i > b_i \). Then,
\[
r_{i+1} = k_{i+1} r_i + b_i > r_i > b_i
\]
and
\[
b_{i+1} = r_i > b_i
\]
Thus, \( r_{i+1} > b_{i+1} \).
4. **Backward Induction to Determine \( r_i \) and \( b_i \):**
We will show by backward induction that Ms. Taf can determine \( r_i \) and \( b_i \) for \( 0 \leq i \leq n \).
- **Base Case:** Ms. Taf knows \( r_n \) and \( b_n \).
- **Inductive Step:** Suppose she knows \( r_i \) and \( b_i \). By the division algorithm, there is a unique integer \( k \) and \( r < b_i \) such that:
\[
r_i = k b_i + r = k r_{i-1} + r
\]
But we also have:
\[
r_i = k_i r_{i-1} + b_{i-1}
\]
with integer \( k_i \) and \( b_{i-1} < r_{i-1} \). Thus, \( k = k_i \), and Ms. Taf can find \( k_i \) by division. She can then find:
\[
r_{i-1} = b_i
\]
and
\[
b_{i-1} = r_i - k_i r_{i-1}
\]
This completes the induction step.
5. **Conclusion:**
Thus, Ms. Taf can determine all numbers \( k_1, k_2, \ldots, k_n \) from just one submission of the red and blue cards.
The final answer is \( \boxed{1} \)
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose $a_{1} < a_{2}< \cdots < a_{2024}$ is an arithmetic sequence of positive integers, and $b_{1} <b_{2} < \cdots <b_{2024}$ is a geometric sequence of positive integers. Find the maximum possible number of integers that could appear in both sequences, over all possible choices of the two sequences.
[i]Ray Li[/i]
|
1. **Define the sequences**: Let \( a_n \) be an arithmetic sequence and \( b_n \) be a geometric sequence. Specifically, we have:
\[
a_n = a_1 + (n-1)d
\]
\[
b_n = b_1 \cdot r^{n-1}
\]
where \( d \) is the common difference of the arithmetic sequence, and \( r \) is the common ratio of the geometric sequence.
2. **Set up the problem**: We need to find the maximum number of integers that can appear in both sequences. Suppose there are \( k \) common terms between the sequences. Let these common terms be \( c_1, c_2, \ldots, c_k \).
3. **Common terms in both sequences**: For each common term \( c_i \), there exist integers \( m_i \) and \( n_i \) such that:
\[
c_i = a_1 + (m_i - 1)d = b_1 \cdot r^{n_i - 1}
\]
Since \( a_1 \) and \( b_1 \) are positive integers, and \( d \) and \( r \) are positive integers, we need to find the maximum \( k \) such that the above equation holds for distinct \( m_i \) and \( n_i \).
4. **Example with specific sequences**: Consider the arithmetic sequence \( a_n = n \) and the geometric sequence \( b_n = 2^{n-1} \). The common terms are:
\[
1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024
\]
These are the first 11 powers of 2, which are common to both sequences.
5. **General case analysis**: Suppose there are more than 11 common terms. Let \( c_1, c_2, \ldots, c_{12} \) be 12 common terms. Then:
\[
c_i = a_1 + (m_i - 1)d = b_1 \cdot r^{n_i - 1}
\]
for \( i = 1, 2, \ldots, 12 \).
6. **Prime factor analysis**: Let \( r = \frac{q}{p} \) where \( q \) and \( p \) are relatively prime positive integers with \( q > p \geq 1 \). Then:
\[
b_i = b_1 \left( \frac{q}{p} \right)^{i-1}
\]
For \( b_i \) to be an integer, \( b_1 \) must be a multiple of \( p^{i-1} \). Let \( b_1 = p^{2023} x \) for some integer \( x \). Then:
\[
b_i = x p^{2024-i} q^{i-1}
\]
7. **Divisibility constraints**: For the common terms \( c_i \) to be in both sequences, the differences \( c_{i+1} - c_i \) must be divisible by \( d \). Specifically:
\[
d \mid c_{i+1} - c_i = x p^{2024-i_{i+1}} q^{i_{i+1}-1} - x p^{2024-i_i} q^{i_i-1}
\]
Simplifying, we get:
\[
d \mid x p^{2024-i_{i+1}} q^{i_i-1} (q^{i_{i+1}-i_i} - p^{i_{i+1}-i_i})
\]
8. **Prime factor bounds**: For any prime \( r \) dividing \( q \):
\[
\nu_r(d) \leq \nu_r(x q^{i_i-1})
\]
Repeating this for all 12 terms, we find that:
\[
\nu_r(b_{i_{k+1}} - b_{i_k}) - \nu_r(d) \geq \nu_r(q^{k-1})
\]
Thus:
\[
b_{i_{k+1}} - b_{i_k} \geq d q^{k-1}
\]
9. **Summing the differences**: Summing these differences for 12 terms:
\[
b_{i_{12}} - b_{i_1} \geq d \sum_{k=1}^{11} q^{k-1} = d (q^{11} - 1) / (q - 1)
\]
Since \( q \geq 2 \):
\[
b_{i_{12}} - b_{i_1} \geq d (2^{11} - 1) = 2047d
\]
But \( b_{i_{12}} - b_{i_1} \leq 2023d \), a contradiction.
10. **Conclusion**: Therefore, there cannot be more than 11 common terms. We have already shown that 11 common terms are possible.
The final answer is \(\boxed{11}\)
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
[b]p1.[/b] Kevin plants corn and cotton. Once he harvests the crops, he has $30$ pounds of corn and $x$ pounds of cotton. Corn sells for $\$5$ per pound and cotton sells for $\$10$ per pound. If Kevin sells all his corn and cotton for a total of $\$640$, then compute $x$.
[b]p2.[/b] $ABCD$ is a square where $AB =\sqrt{2016}$. Let $X$ be a point on $AB$ and $Y$ be a point on $CD$ such that $AX = CY$ . Compute the area of trapezoid $AXYD$.
[b]p3.[/b] If the integer $n$ leaves a remainder of 4 when divided by $5$, then what is the largest possible remainder when $2n$ is divided by $15$?
[b]p4.[/b] Let $d(n)$ represent the sum of the digits of the integer $n$. For example, $d(2016) = 2 + 0 + 1 + 6 = 9$. For how many positive 3-digit integers k is it true that $d(k) > d(k + 1)$?
[b]p5.[/b] Let $A, B$, and $C$ be three points on circle $O$ such that $AC$ is a diameter of $O$. Extend line $AC$ to a point $D$ such that $DB$ is tangent to $O$ at $B$, and suppose that $\angle ADB = 20^o$. Compute $\angle ACB$.
[b]p6.[/b] A group of n people, including Caroline and her best friend Catherine, stand in random order around a circle, with each order being equally likely. If the probability that Catherine is adjacent to Caroline is $1/9$ , then what is the value of $n$?
[b]p7.[/b] The polynomial $P(x) = x^4 + 4x^3 + 8x^2 + 8x + 4$ is the square of another polynomial $Q(x) = ax^2 + bx + c$ with positive coefficients $a, b$, and $c$. Compute $4a + 2b + c$.
[b]p8.[/b] A physics class has $25$ students, two of whom are Alex and Justin. Three students are chosen uniformly at random for a lab demonstration. What is the probability that at least one of Alex and Justin is chosen?
[b]p9.[/b] Natasha carries a paintbrush of length $1$ and walks around the perimeter of a regular $2016$-gon with side length $1$, painting all of the area outside the polygon that she can reach. What is the area that Natasha paints?
[b]p10.[/b] Let $S$ be the set of values which can be written as the sum of five consecutive perfect squares. What is the smallest element of $S$ which is divisible by $17$?
[b]p11.[/b] Consider $6$ points, two of which are $A$ and $B$, such that each pair of points is connected by a line segment and no three points are collinear. Compute the number of distinct paths from $A$ to $B$ such that no point is visited more than once.
[b]p12.[/b] Let $f(x) = 3x^2 + 2x+ 6$. Two distinct lines $l_1$ and $l_2$ exist that are tangent to $f(x)$ and intersect at the origin. Given that l1 is tangent to $f(x)$ at $(x_1, f(x_1))$ and that $l_2$ is tangent to $f(x)$ at $(x_2, f(x_2))$, compute $x_1x_2$.
[b]p13.[/b] One day, Connie plays a game with a fair $6$-sided die. Connie rolls the die until she rolls a $6$, at which point the game ends. If she rolls a $6$ on her first turn, Connie wins $6$ dollars. For each subsequent turn, Connie wins $1/6$ of the amount she would have won the previous turn. What is Connie’s expected earnings from the game?
[b]p14.[/b] Calculate the positive root of the polynomial $(x - 1)(x - 2)(x - 3)(x - 4) - 63$.
[b]p15.[/b] Consider an equilateral triangle 4ABC with unit side length. Let $M$ be the midpoint of side $AB$. A ball is released in a straight line from $M$ and bounces off the side $BC$ at a point $D$. Then, it bounces off the side $CA$ at a point $E$ and lands exactly at $B$. By the law of reflection, we have $\angle BDM = \angle CDE$ and $\angle CED = \angle AEB$. Calculate $MD + DE + EB$, the distance that the ball travels before reaching $B$.
PS. You should use hide for answers. Problems 16-27 have been posted [url=https://artofproblemsolving.com/community/c4h2780524p24413447]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
|
1. **Problem 1:**
Kevin plants corn and cotton. Once he harvests the crops, he has 30 pounds of corn and \( x \) pounds of cotton. Corn sells for \$5 per pound and cotton sells for \$10 per pound. If Kevin sells all his corn and cotton for a total of \$640, then compute \( x \).
Let's denote the total revenue from selling corn and cotton as \( R \). We know:
\[
R = 5 \times 30 + 10 \times x = 640
\]
Simplifying the equation:
\[
150 + 10x = 640
\]
Subtract 150 from both sides:
\[
10x = 490
\]
Divide by 10:
\[
x = 49
\]
The final answer is \( \boxed{ 8 } \)
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
You are tossing an unbiased coin. The last $ 28 $ consecutive flips have all resulted in heads. Let $ x $ be the expected number of additional tosses you must make before you get $ 60 $ consecutive heads. Find the sum of all distinct prime factors in $ x $.
|
1. **Understanding the Problem:**
We need to find the expected number of additional tosses required to get 60 consecutive heads, given that the last 28 consecutive flips have all resulted in heads. Let \( x \) be this expected number.
2. **Expected Number of Tosses:**
The problem can be approached using the concept of Markov chains. Let \( E_n \) be the expected number of additional tosses to get 60 consecutive heads, given that we already have \( n \) consecutive heads.
We need to find \( E_{28} \).
3. **Recursive Relation:**
The recursive relation for \( E_n \) can be derived as follows:
\[
E_n = 1 + \frac{1}{2}E_{n+1} + \frac{1}{2}E_0
\]
This is because:
- With probability \( \frac{1}{2} \), the next toss is heads, and we move to \( n+1 \) consecutive heads.
- With probability \( \frac{1}{2} \), the next toss is tails, and we reset to 0 consecutive heads.
4. **Base Case:**
The base case is \( E_{60} = 0 \), since if we already have 60 consecutive heads, no more tosses are needed.
5. **Solving the Recurrence:**
We solve the recurrence relation starting from \( E_{59} \) down to \( E_{28} \):
\[
E_{59} = 1 + \frac{1}{2}E_{60} + \frac{1}{2}E_0 = 1 + \frac{1}{2}(0) + \frac{1}{2}E_0 = 1 + \frac{1}{2}E_0
\]
\[
E_{58} = 1 + \frac{1}{2}E_{59} + \frac{1}{2}E_0 = 1 + \frac{1}{2}(1 + \frac{1}{2}E_0) + \frac{1}{2}E_0 = 1 + \frac{1}{2} + \frac{1}{4}E_0 + \frac{1}{2}E_0 = 1.5 + \frac{3}{4}E_0
\]
Continuing this process, we can generalize:
\[
E_n = 2^{60-n} - 1 + \left(1 - 2^{-(60-n)}\right)E_0
\]
6. **Finding \( E_0 \):**
To find \( E_0 \), we use the fact that \( E_0 \) is the expected number of tosses to get 60 consecutive heads starting from 0 heads:
\[
E_0 = 1 + \frac{1}{2}E_1 + \frac{1}{2}E_0
\]
Solving for \( E_0 \):
\[
E_0 = 2 + E_1
\]
Using the general form:
\[
E_1 = 2^{59} - 1 + \left(1 - 2^{-59}\right)E_0
\]
Substituting back:
\[
E_0 = 2 + 2^{59} - 1 + \left(1 - 2^{-59}\right)E_0
\]
\[
E_0 = 1 + 2^{59} + \left(1 - 2^{-59}\right)E_0
\]
\[
E_0 - \left(1 - 2^{-59}\right)E_0 = 1 + 2^{59}
\]
\[
2^{-59}E_0 = 1 + 2^{59}
\]
\[
E_0 = 2^{59}(1 + 2^{-59}) = 2^{59} + 1
\]
7. **Finding \( E_{28} \):**
Using the general form:
\[
E_{28} = 2^{32} - 1 + \left(1 - 2^{-32}\right)E_0
\]
\[
E_{28} = 2^{32} - 1 + \left(1 - 2^{-32}\right)(2^{59} + 1)
\]
\[
E_{28} = 2^{32} - 1 + 2^{59} + 1 - 2^{27}
\]
\[
E_{28} = 2^{59} + 2^{32} - 2^{27}
\]
8. **Prime Factors:**
The prime factors of \( 2^{59} + 2^{32} - 2^{27} \) are 2 and 3.
9. **Sum of Distinct Prime Factors:**
The sum of all distinct prime factors is \( 2 + 3 = 5 \).
The final answer is \(\boxed{5}\)
|
5
|
Other
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Square $ABCD$ has side length $5$ and arc $BD$ with center $A$. $E$ is the midpoint of $AB$ and $CE$ intersects arc $BD$ at $F$. $G$ is placed onto $BC$ such that $FG$ is perpendicular to $BC$. What is the length of $FG$?
|
1. **Determine the length of \( AE \) and \( BE \):**
Since \( E \) is the midpoint of \( \overline{AB} \), we have:
\[
AE = BE = \frac{5}{2}
\]
2. **Calculate \( CE \) using the Pythagorean Theorem:**
\[
CE^2 = BC^2 + BE^2
\]
Given \( BC = 5 \) and \( BE = \frac{5}{2} \):
\[
CE^2 = 5^2 + \left( \frac{5}{2} \right)^2 = 25 + \frac{25}{4} = \frac{100}{4} + \frac{25}{4} = \frac{125}{4}
\]
Therefore:
\[
CE = \sqrt{\frac{125}{4}} = \frac{5\sqrt{5}}{2}
\]
3. **Use the Power of a Point theorem:**
Since \( AB \) is a chord and \( E \) is a point on \( AB \):
\[
\left( \frac{5}{2} \right)^2 = \frac{5\sqrt{5}}{2} \cdot EF
\]
Solving for \( EF \):
\[
\frac{25}{4} = \frac{5\sqrt{5}}{2} \cdot EF
\]
\[
EF = \frac{25}{4} \cdot \frac{2}{5\sqrt{5}} = \frac{25 \cdot 2}{4 \cdot 5\sqrt{5}} = \frac{50}{20\sqrt{5}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}
\]
4. **Calculate \( CF \):**
Since \( CE = \frac{5\sqrt{5}}{2} \) and \( EF = \frac{\sqrt{5}}{2} \):
\[
CF = CE - EF = \frac{5\sqrt{5}}{2} - \frac{\sqrt{5}}{2} = \frac{4\sqrt{5}}{2} = 2\sqrt{5}
\]
5. **Use similarity of triangles \( \triangle CBE \) and \( \triangle CGF \):**
By AA similarity, we have:
\[
\frac{BE}{CE} = \frac{GF}{CF}
\]
Substituting the known values:
\[
\frac{\frac{5}{2}}{\frac{5\sqrt{5}}{2}} = \frac{FG}{2\sqrt{5}}
\]
Simplifying:
\[
\frac{5}{2} \cdot \frac{2}{5\sqrt{5}} = \frac{FG}{2\sqrt{5}}
\]
\[
\frac{1}{\sqrt{5}} = \frac{FG}{2\sqrt{5}}
\]
Solving for \( FG \):
\[
FG = 2\sqrt{5} \cdot \frac{1}{\sqrt{5}} = 2
\]
The final answer is \( \boxed{2} \).
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose three boba drinks and four burgers cost $28$ dollars, while two boba drinks and six burgers cost $\$ 37.70$. If you paid for one boba drink using only pennies, nickels, dimes, and quarters, determine the least number of coins you could use.
|
1. **Set up the system of linear equations:**
We are given two equations based on the cost of boba drinks (denoted as \(d\)) and burgers (denoted as \(b\)):
\[
3d + 4b = 28 \quad \text{(1)}
\]
\[
2d + 6b = 37.70 \quad \text{(2)}
\]
2. **Solve the system of equations:**
To eliminate one of the variables, we can multiply the first equation by 2 and the second equation by 3 to make the coefficients of \(d\) the same:
\[
2(3d + 4b) = 2 \cdot 28 \implies 6d + 8b = 56 \quad \text{(3)}
\]
\[
3(2d + 6b) = 3 \cdot 37.70 \implies 6d + 18b = 113.10 \quad \text{(4)}
\]
3. **Subtract equation (3) from equation (4):**
\[
(6d + 18b) - (6d + 8b) = 113.10 - 56
\]
\[
10b = 57.10
\]
\[
b = \frac{57.10}{10} = 5.71
\]
4. **Substitute \(b = 5.71\) back into one of the original equations to find \(d\):**
Using equation (1):
\[
3d + 4(5.71) = 28
\]
\[
3d + 22.84 = 28
\]
\[
3d = 28 - 22.84
\]
\[
3d = 5.16
\]
\[
d = \frac{5.16}{3} = 1.72
\]
5. **Determine the least number of coins to make $1.72:**
- Quarters: \(0.25\) dollars each
- Dimes: \(0.10\) dollars each
- Nickels: \(0.05\) dollars each
- Pennies: \(0.01\) dollars each
To minimize the number of coins:
- Use as many quarters as possible: \(1.72 \div 0.25 = 6.88\), so use 6 quarters (\(6 \times 0.25 = 1.50\))
- Remaining amount: \(1.72 - 1.50 = 0.22\)
- Use as many dimes as possible: \(0.22 \div 0.10 = 2.2\), so use 2 dimes (\(2 \times 0.10 = 0.20\))
- Remaining amount: \(0.22 - 0.20 = 0.02\)
- Use pennies: \(0.02 \div 0.01 = 2\), so use 2 pennies
Total number of coins:
\[
6 \text{ quarters} + 2 \text{ dimes} + 2 \text{ pennies} = 10 \text{ coins}
\]
The final answer is \(\boxed{10}\).
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. If the maximum and minimum values of $x^2+y^2$ are $M$ and $m$ respectively, compute the numerical value of $M-m$.
|
1. Start with the given equation:
\[
x^2 - 4x + y^2 + 3 = 0
\]
2. Rearrange the equation to complete the square for \(x\):
\[
x^2 - 4x + 4 + y^2 + 3 - 4 = 0
\]
\[
(x - 2)^2 + y^2 - 1 = 0
\]
\[
(x - 2)^2 + y^2 = 1
\]
This represents a circle with center \((2, 0)\) and radius \(1\).
3. We need to find the maximum and minimum values of \(x^2 + y^2\). This is the square of the distance from any point \((x, y)\) on the circle to the origin \((0, 0)\).
4. The distance \(d\) from the origin to the center of the circle \((2, 0)\) is:
\[
d = \sqrt{(2 - 0)^2 + (0 - 0)^2} = \sqrt{4} = 2
\]
5. The maximum distance from the origin to a point on the circle is the sum of the distance from the origin to the center and the radius of the circle:
\[
d_{\text{max}} = 2 + 1 = 3
\]
Therefore, the maximum value of \(x^2 + y^2\) is:
\[
M = d_{\text{max}}^2 = 3^2 = 9
\]
6. The minimum distance from the origin to a point on the circle is the difference between the distance from the origin to the center and the radius of the circle:
\[
d_{\text{min}} = 2 - 1 = 1
\]
Therefore, the minimum value of \(x^2 + y^2\) is:
\[
m = d_{\text{min}}^2 = 1^2 = 1
\]
7. The difference between the maximum and minimum values of \(x^2 + y^2\) is:
\[
M - m = 9 - 1 = 8
\]
The final answer is \(\boxed{8}\).
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Determine the largest integer $n$ such that $2^n$ divides the decimal representation given by some permutation of the digits $2$, $0$, $1$, and $5$. (For example, $2^1$ divides $2150$. It may start with $0$.)
|
1. **Identify the permutations of the digits 2, 0, 1, and 5:**
The possible permutations of the digits 2, 0, 1, and 5 are:
\[
\{2015, 2051, 2105, 2150, 2501, 2510, 1025, 1052, 1205, 1250, 1502, 1520, 5012, 5021, 5102, 5120, 5201, 5210, 0125, 0152, 0215, 0251, 0512, 0521\}
\]
2. **Check divisibility by powers of 2:**
We need to determine the highest power of 2 that divides any of these permutations. We will check each permutation to see how many trailing zeros (which indicate factors of 2) they have.
3. **Evaluate each permutation:**
- **2015:** Ends in 5, not divisible by 2.
- **2051:** Ends in 1, not divisible by 2.
- **2105:** Ends in 5, not divisible by 2.
- **2150:** Ends in 0, divisible by \(2^1\).
- **2501:** Ends in 1, not divisible by 2.
- **2510:** Ends in 0, divisible by \(2^1\).
- **1025:** Ends in 5, not divisible by 2.
- **1052:** Ends in 2, divisible by \(2^1\).
- **1205:** Ends in 5, not divisible by 2.
- **1250:** Ends in 0, divisible by \(2^1\).
- **1502:** Ends in 2, divisible by \(2^1\).
- **1520:** Ends in 0, divisible by \(2^1\).
- **5012:** Ends in 2, divisible by \(2^1\).
- **5021:** Ends in 1, not divisible by 2.
- **5102:** Ends in 2, divisible by \(2^1\).
- **5120:** Ends in 0, divisible by \(2^4\) (since \(5120 = 2^4 \times 320\)).
- **5201:** Ends in 1, not divisible by 2.
- **5210:** Ends in 0, divisible by \(2^1\).
- **0125:** Ends in 5, not divisible by 2.
- **0152:** Ends in 2, divisible by \(2^1\).
- **0215:** Ends in 5, not divisible by 2.
- **0251:** Ends in 1, not divisible by 2.
- **0512:** Ends in 2, divisible by \(2^1\).
- **0521:** Ends in 1, not divisible by 2.
4. **Determine the highest power of 2:**
From the above evaluations, the permutation **5120** is divisible by \(2^4\), which is the highest power of 2 among all permutations.
5. **Conclusion:**
The largest integer \(n\) such that \(2^n\) divides the decimal representation given by some permutation of the digits 2, 0, 1, and 5 is \(n = 4\).
The final answer is \(\boxed{4}\)
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose we list the decimal representations of the positive even numbers from left to right. Determine the $2015^{th}$ digit in the list.
|
To determine the $2015^{th}$ digit in the concatenation of the decimal representations of positive even numbers, we need to break down the problem into manageable parts by considering the number of digits contributed by different ranges of even numbers.
1. **One-digit even numbers:**
The one-digit even numbers are \(2, 4, 6, 8\). There are 4 such numbers, each contributing 1 digit.
\[
\text{Total digits from one-digit numbers} = 4 \times 1 = 4
\]
2. **Two-digit even numbers:**
The two-digit even numbers range from 10 to 98. There are 45 such numbers (since \( \frac{98 - 10}{2} + 1 = 45 \)), each contributing 2 digits.
\[
\text{Total digits from two-digit numbers} = 45 \times 2 = 90
\]
3. **Three-digit even numbers:**
The three-digit even numbers range from 100 to 998. There are 450 such numbers (since \( \frac{998 - 100}{2} + 1 = 450 \)), each contributing 3 digits.
\[
\text{Total digits from three-digit numbers} = 450 \times 3 = 1350
\]
4. **Four-digit even numbers:**
The four-digit even numbers start from 1000. We need to find how many more digits are required to reach the $2015^{th}$ digit.
\[
\text{Digits needed} = 2015 - (4 + 90 + 1350) = 2015 - 1444 = 571
\]
Each four-digit even number contributes 4 digits. To find how many four-digit even numbers are needed to contribute 571 digits:
\[
\frac{571}{4} = 142 \text{ remainder } 3
\]
This means we need 142 full four-digit even numbers and 3 additional digits from the next number.
5. **Identifying the specific number:**
The first four-digit even number is 1000. The $142^{nd}$ four-digit even number can be found as follows:
\[
a_n = 1000 + 2(n-1) = 1000 + 2 \times 141 = 1000 + 282 = 1282
\]
The next number after 1282 is 1284. Since we need 3 more digits, we look at the first three digits of 1284, which are "128".
6. **Conclusion:**
The $2015^{th}$ digit is the third digit of 1284, which is 8.
The final answer is \(\boxed{8}\).
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The number $2^{29}$ has a $9$-digit decimal representation that contains all but one of the $10$ (decimal) digits. Determine which digit is missing
|
1. **Determine the modulo 9 of \(2^{29}\):**
\[
2^{29} \equiv (2^6)^4 \cdot 2^5 \pmod{9}
\]
First, calculate \(2^6 \mod 9\):
\[
2^6 = 64 \quad \text{and} \quad 64 \div 9 = 7 \quad \text{remainder} \quad 1 \quad \Rightarrow \quad 64 \equiv 1 \pmod{9}
\]
Therefore,
\[
(2^6)^4 \equiv 1^4 \equiv 1 \pmod{9}
\]
Next, calculate \(2^5 \mod 9\):
\[
2^5 = 32 \quad \text{and} \quad 32 \div 9 = 3 \quad \text{remainder} \quad 5 \quad \Rightarrow \quad 32 \equiv 5 \pmod{9}
\]
Combining these results:
\[
2^{29} \equiv 1 \cdot 5 \equiv 5 \pmod{9}
\]
2. **Sum of digits of \(2^{29}\):**
Since \(2^{29}\) has a 9-digit decimal representation containing all but one of the 10 decimal digits, the sum of all digits from 0 to 9 is:
\[
0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
\]
Let \(m\) be the missing digit. The sum of the digits of \(2^{29}\) is:
\[
45 - m
\]
3. **Modulo 9 condition:**
The sum of the digits of a number is congruent to the number itself modulo 9. Therefore:
\[
45 - m \equiv 2^{29} \equiv 5 \pmod{9}
\]
Solving for \(m\):
\[
45 - m \equiv 5 \pmod{9}
\]
Subtract 5 from both sides:
\[
45 - m - 5 \equiv 0 \pmod{9}
\]
Simplify:
\[
40 - m \equiv 0 \pmod{9}
\]
Therefore:
\[
40 \equiv m \pmod{9}
\]
Since \(40 \div 9 = 4\) remainder \(4\):
\[
40 \equiv 4 \pmod{9}
\]
Thus, the missing digit \(m\) is:
\[
m = 4
\]
The final answer is \(\boxed{4}\).
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Define $ P(\tau ) = (\tau + 1)^3$ . If $x + y = 0$, what is the minimum possible value of $P(x) + P(y)$?
|
1. Given the function \( P(\tau) = (\tau + 1)^3 \), we need to find the minimum possible value of \( P(x) + P(y) \) given that \( x + y = 0 \).
2. Since \( x + y = 0 \), we can substitute \( y = -x \).
3. Therefore, we need to evaluate \( P(x) + P(-x) \):
\[
P(x) + P(-x) = (x + 1)^3 + (-x + 1)^3
\]
4. Expand both terms using the binomial theorem:
\[
(x + 1)^3 = x^3 + 3x^2 + 3x + 1
\]
\[
(-x + 1)^3 = (-x)^3 + 3(-x)^2 + 3(-x) + 1 = -x^3 + 3x^2 - 3x + 1
\]
5. Add the expanded forms:
\[
P(x) + P(-x) = (x^3 + 3x^2 + 3x + 1) + (-x^3 + 3x^2 - 3x + 1)
\]
6. Combine like terms:
\[
P(x) + P(-x) = x^3 - x^3 + 3x^2 + 3x^2 + 3x - 3x + 1 + 1
\]
\[
P(x) + P(-x) = 6x^2 + 2
\]
7. The expression \( 6x^2 + 2 \) is a quadratic function in terms of \( x \). The minimum value of a quadratic function \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). Here, \( a = 6 \) and \( b = 0 \), so the vertex is at:
\[
x = -\frac{0}{2 \cdot 6} = 0
\]
8. Substitute \( x = 0 \) into the quadratic expression to find the minimum value:
\[
6(0)^2 + 2 = 2
\]
9. Therefore, the minimum possible value of \( P(x) + P(y) \) is \( 2 \).
The final answer is \( \boxed{2} \).
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Consider the graph on $1000$ vertices $v_1, v_2, ...v_{1000}$ such that for all $1 \le i < j \le 1000$, $v_i$ is connected to $v_j$ if and only if $i$ divides $j$. Determine the minimum number of colors that must be used to color the vertices of this graph such that no two vertices sharing an edge are the same color.
|
1. **Understanding the Graph Construction**:
- We have a graph with vertices \( v_1, v_2, \ldots, v_{1000} \).
- An edge exists between \( v_i \) and \( v_j \) if and only if \( i \) divides \( j \) (i.e., \( i \mid j \)).
2. **Identifying Cliques**:
- A clique is a subset of vertices such that every two distinct vertices are adjacent.
- Consider the vertices \( v_1, v_2, v_4, \ldots, v_{512} \). These vertices form a clique because:
- \( 1 \mid 2 \), \( 1 \mid 4 \), \( 2 \mid 4 \), and so on.
- Each vertex \( v_{2^k} \) (where \( k \) ranges from 0 to 9) divides \( v_{2^m} \) for \( m > k \).
3. **Clique Size**:
- The vertices \( v_1, v_2, v_4, \ldots, v_{512} \) form a clique of size 10 because \( 2^9 = 512 \) and \( 2^{10} = 1024 \) which is greater than 1000.
- Therefore, we need at least 10 colors to color this clique since each vertex in a clique must have a unique color.
4. **Coloring Strategy**:
- We color the vertices based on the number of prime factors in their prime factorization (counted with multiplicity).
- For example, \( 12 = 2^2 \times 3 \) has 3 prime factors (2, 2, and 3).
5. **Prime Factorization and Coloring**:
- Since \( 1000 < 2^{10} \), each number \( v_i \) (where \( 1 \leq i \leq 1000 \)) has at most 9 prime factors.
- Therefore, we need 10 colors to ensure that no two adjacent vertices (where one divides the other) share the same color.
6. **Verification**:
- If \( i \mid j \), then the number of prime factors of \( i \) is less than or equal to the number of prime factors of \( j \).
- Thus, by coloring vertices based on the number of prime factors, we ensure that adjacent vertices have different colors.
Conclusion:
\[
\boxed{10}
\]
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose the side lengths of triangle $ABC$ are the roots of polynomial $x^3 - 27x^2 + 222x - 540$. What is the product of its inradius and circumradius?
|
1. **Identify the roots of the polynomial:**
The polynomial given is \(x^3 - 27x^2 + 222x - 540\). The roots of this polynomial are the side lengths \(a\), \(b\), and \(c\) of the triangle \(ABC\).
2. **Sum and product of the roots:**
By Vieta's formulas, for the polynomial \(x^3 - 27x^2 + 222x - 540\):
- The sum of the roots \(a + b + c = 27\).
- The product of the roots \(abc = 540\).
3. **Inradius and circumradius formulas:**
- The inradius \(r\) of a triangle is given by:
\[
r = \frac{\text{Area}}{s}
\]
where \(s\) is the semi-perimeter of the triangle, \(s = \frac{a + b + c}{2}\).
- The circumradius \(R\) of a triangle is given by:
\[
R = \frac{abc}{4 \cdot \text{Area}}
\]
4. **Product of inradius and circumradius:**
- The product of the inradius and circumradius is:
\[
r \cdot R = \left(\frac{\text{Area}}{s}\right) \cdot \left(\frac{abc}{4 \cdot \text{Area}}\right)
\]
Simplifying this expression:
\[
r \cdot R = \frac{\text{Area} \cdot abc}{s \cdot 4 \cdot \text{Area}} = \frac{abc}{4s}
\]
5. **Substitute the known values:**
- The semi-perimeter \(s\) is:
\[
s = \frac{a + b + c}{2} = \frac{27}{2}
\]
- The product of the roots \(abc = 540\).
6. **Calculate the product \(r \cdot R\):**
\[
r \cdot R = \frac{abc}{4s} = \frac{540}{4 \cdot \frac{27}{2}} = \frac{540}{54} = 10
\]
The final answer is \(\boxed{10}\).
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
$4$ equilateral triangles of side length $1$ are drawn on the interior of a unit square, each one of which shares a side with one of the $4$ sides of the unit square. What is the common area enclosed by all $4$ equilateral triangles?
|
1. **Define the square and the equilateral triangles:**
- Consider a unit square \(ABCD\) with vertices \(A(0,0)\), \(B(1,0)\), \(C(1,1)\), and \(D(0,1)\).
- Four equilateral triangles are drawn on the interior of the square, each sharing a side with one of the sides of the square.
2. **Determine the vertices of the equilateral triangles:**
- For the triangle on side \(AB\), the third vertex is \(E\left(\frac{1}{2}, 1 - \frac{\sqrt{3}}{2}\right)\).
- For the triangle on side \(BC\), the third vertex is \(F\left(1 - \frac{\sqrt{3}}{2}, \frac{1}{2}\right)\).
3. **Find the intersection point of the lines:**
- The line \(AF\) has the equation \(y = \frac{\sqrt{3}}{2}x\).
- The line \(DE\) has the equation \(y - 1 = -\sqrt{3}x\), which simplifies to \(y = 1 - \sqrt{3}x\).
- To find the intersection point \(G\), solve the system of equations:
\[
\frac{\sqrt{3}}{2}x = 1 - \sqrt{3}x
\]
\[
\frac{\sqrt{3}}{2}x + \sqrt{3}x = 1
\]
\[
\frac{3\sqrt{3}}{2}x = 1
\]
\[
x = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}
\]
\[
y = \frac{\sqrt{3}}{2} \cdot \frac{2\sqrt{3}}{9} = \frac{3}{9} = \frac{1}{3}
\]
Thus, the intersection point \(G\) is \(\left(\frac{2\sqrt{3}}{9}, \frac{1}{3}\right)\).
4. **Calculate the area of the triangle \(AGG'\):**
- The coordinates of \(G'\) are \(\left(\frac{2\sqrt{3}}{9}, 0\right)\).
- The area of \(\triangle AGG'\) is:
\[
\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot \frac{2\sqrt{3}}{9} \cdot \frac{1}{3} = \frac{\sqrt{3}}{27}
\]
5. **Calculate the area of the trapezoid \(GG'E'E\):**
- The height of the trapezoid is \(\frac{1}{3}\).
- The lengths of the parallel sides are \(\frac{1}{2} - \frac{2\sqrt{3}}{9}\) and \(1 - \frac{\sqrt{3}}{2}\).
- The area of the trapezoid is:
\[
\text{Area} = \frac{1}{2} \cdot \left(\frac{1}{2} - \frac{2\sqrt{3}}{9} + 1 - \frac{\sqrt{3}}{2}\right) \cdot \frac{1}{3}
\]
\[
= \frac{1}{2} \cdot \left(\frac{1}{2} + 1 - \frac{\sqrt{3}}{2} - \frac{2\sqrt{3}}{9}\right) \cdot \frac{1}{3}
\]
\[
= \frac{1}{2} \cdot \left(\frac{3}{2} - \frac{2\sqrt{3}}{9} - \frac{\sqrt{3}}{2}\right) \cdot \frac{1}{3}
\]
\[
= \frac{1}{2} \cdot \left(\frac{3}{2} - \frac{4\sqrt{3}}{9}\right) \cdot \frac{1}{3}
\]
\[
= \frac{1}{2} \cdot \left(\frac{27 - 4\sqrt{3}}{18}\right) \cdot \frac{1}{3}
\]
\[
= \frac{1}{2} \cdot \frac{27 - 4\sqrt{3}}{54}
\]
\[
= \frac{27 - 4\sqrt{3}}{108}
\]
6. **Calculate the total common area enclosed by all four equilateral triangles:**
- The total area is:
\[
1 - 8 \cdot \left(\frac{\sqrt{3}}{27} + \frac{27 - 4\sqrt{3}}{108}\right)
\]
\[
= 1 - 8 \cdot \left(\frac{4\sqrt{3}}{108} + \frac{27 - 4\sqrt{3}}{108}\right)
\]
\[
= 1 - 8 \cdot \frac{27}{108}
\]
\[
= 1 - 8 \cdot \frac{1}{4}
\]
\[
= 1 - 2
\]
\[
= -1
\]
The final answer is \( \boxed{ -1 } \)
|
-1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
In three years, Xingyou’s age in years will be twice his current height in feet. If Xingyou’s current age in years is also his current height in feet, what is Xingyou’s age in years right now?
|
1. Let \( x \) be Xingyou’s current age in years, which is also his current height in feet.
2. According to the problem, in three years, Xingyou’s age will be twice his current height. This can be expressed as:
\[
x + 3 = 2x
\]
3. To solve for \( x \), we subtract \( x \) from both sides of the equation:
\[
x + 3 - x = 2x - x
\]
Simplifying this, we get:
\[
3 = x
\]
The final answer is \( \boxed{3} \).
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the product of all values of $d$ such that $x^{3} +2x^{2} +3x +4 = 0$ and $x^{2} +dx +3 = 0$ have a common root.
|
1. Let \( \alpha \) be the common root of the polynomials \( x^3 + 2x^2 + 3x + 4 = 0 \) and \( x^2 + dx + 3 = 0 \).
2. Since \( \alpha \) is a root of \( x^2 + dx + 3 = 0 \), we have:
\[
\alpha^2 + d\alpha + 3 = 0 \quad \text{(1)}
\]
3. Similarly, since \( \alpha \) is a root of \( x^3 + 2x^2 + 3x + 4 = 0 \), we have:
\[
\alpha^3 + 2\alpha^2 + 3\alpha + 4 = 0 \quad \text{(2)}
\]
4. From equation (1), solve for \( \alpha^2 \):
\[
\alpha^2 = -d\alpha - 3
\]
5. Substitute \( \alpha^2 = -d\alpha - 3 \) into equation (2):
\[
\alpha^3 + 2(-d\alpha - 3) + 3\alpha + 4 = 0
\]
Simplify:
\[
\alpha^3 - 2d\alpha - 6 + 3\alpha + 4 = 0
\]
\[
\alpha^3 + (3 - 2d)\alpha - 2 = 0 \quad \text{(3)}
\]
6. From equation (1), multiply both sides by \( \alpha \):
\[
\alpha^3 + d\alpha^2 + 3\alpha = 0
\]
Substitute \( \alpha^2 = -d\alpha - 3 \):
\[
\alpha^3 + d(-d\alpha - 3) + 3\alpha = 0
\]
Simplify:
\[
\alpha^3 - d^2\alpha - 3d + 3\alpha = 0
\]
\[
\alpha^3 + (3 - d^2)\alpha - 3d = 0 \quad \text{(4)}
\]
7. Equate the coefficients of \( \alpha \) from equations (3) and (4):
\[
3 - 2d = 3 - d^2
\]
Simplify:
\[
-2d = -d^2
\]
\[
d^2 - 2d = 0
\]
Factorize:
\[
d(d - 2) = 0
\]
Thus, \( d = 0 \) or \( d = 2 \).
8. The product of all values of \( d \) is:
\[
0 \times 2 = 0
\]
The final answer is \(\boxed{0}\).
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
If there is only $1$ complex solution to the equation $8x^3 + 12x^2 + kx + 1 = 0$, what is $k$?
|
1. Given the cubic equation \(8x^3 + 12x^2 + kx + 1 = 0\), we need to find the value of \(k\) such that there is exactly one complex solution.
2. For a cubic equation to have exactly one complex solution, it must have a triple root. This means the equation can be written in the form \(8(x - \alpha)^3 = 0\) for some complex number \(\alpha\).
3. Expanding \(8(x - \alpha)^3\), we get:
\[
8(x - \alpha)^3 = 8(x^3 - 3\alpha x^2 + 3\alpha^2 x - \alpha^3)
\]
Simplifying, we have:
\[
8(x - \alpha)^3 = 8x^3 - 24\alpha x^2 + 24\alpha^2 x - 8\alpha^3
\]
4. Comparing this with the given equation \(8x^3 + 12x^2 + kx + 1 = 0\), we equate the coefficients:
\[
8x^3 - 24\alpha x^2 + 24\alpha^2 x - 8\alpha^3 = 8x^3 + 12x^2 + kx + 1
\]
5. From the coefficient of \(x^2\), we get:
\[
-24\alpha = 12 \implies \alpha = -\frac{1}{2}
\]
6. Substituting \(\alpha = -\frac{1}{2}\) into the coefficient of \(x\), we get:
\[
24\alpha^2 = k \implies 24\left(-\frac{1}{2}\right)^2 = k \implies 24 \cdot \frac{1}{4} = k \implies k = 6
\]
7. Finally, we check the constant term:
\[
-8\alpha^3 = 1 \implies -8\left(-\frac{1}{2}\right)^3 = 1 \implies -8 \cdot -\frac{1}{8} = 1 \implies 1 = 1
\]
This confirms that our value of \(\alpha\) and \(k\) are correct.
The final answer is \(\boxed{6}\).
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
If $f$ is a polynomial, and $f(-2)=3$, $f(-1)=-3=f(1)$, $f(2)=6$, and $f(3)=5$, then what is the minimum possible degree of $f$?
|
1. **Given Points and Polynomial Degree**:
We are given the points $f(-2)=3$, $f(-1)=-3$, $f(1)=-3$, $f(2)=6$, and $f(3)=5$. To determine the minimum possible degree of the polynomial $f$, we start by noting that a polynomial of degree $n$ is uniquely determined by $n+1$ points. Since we have 5 points, the polynomial could be of degree at most 4.
2. **Checking for Degree 3**:
To check if a polynomial of degree 3 could fit these points, we assume $f(x)$ is a cubic polynomial:
\[
f(x) = ax^3 + bx^2 + cx + d
\]
We need to determine if there exist coefficients $a$, $b$, $c$, and $d$ such that the polynomial passes through the given points.
3. **System of Equations**:
Substituting the given points into the polynomial, we get the following system of equations:
\[
\begin{cases}
a(-2)^3 + b(-2)^2 + c(-2) + d = 3 \\
a(-1)^3 + b(-1)^2 + c(-1) + d = -3 \\
a(1)^3 + b(1)^2 + c(1) + d = -3 \\
a(2)^3 + b(2)^2 + c(2) + d = 6 \\
a(3)^3 + b(3)^2 + c(3) + d = 5
\end{cases}
\]
Simplifying these equations, we get:
\[
\begin{cases}
-8a + 4b - 2c + d = 3 \\
-a + b - c + d = -3 \\
a + b + c + d = -3 \\
8a + 4b + 2c + d = 6 \\
27a + 9b + 3c + d = 5
\end{cases}
\]
4. **Solving the System**:
We solve this system of linear equations to find the coefficients $a$, $b$, $c$, and $d$. However, solving this system shows that there is no consistent solution for $a$, $b$, $c$, and $d$ that satisfies all five equations simultaneously. This indicates that a cubic polynomial cannot fit all the given points.
5. **Conclusion**:
Since a cubic polynomial (degree 3) cannot fit the given points, the minimum possible degree of the polynomial $f$ must be 4.
The final answer is $\boxed{4}$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Define $ f(n) = \dfrac{n^2 + n}{2} $. Compute the number of positive integers $ n $ such that $ f(n) \leq 1000 $ and $ f(n) $ is the product of two prime numbers.
|
To solve the problem, we need to find the number of positive integers \( n \) such that \( f(n) \leq 1000 \) and \( f(n) \) is the product of exactly two prime numbers. The function \( f(n) \) is defined as:
\[ f(n) = \frac{n^2 + n}{2} = \frac{n(n+1)}{2} \]
We need to consider two cases: when \( n \) is a prime number and when \( n \) is an even number.
1. **Determine the upper bound for \( n \):**
\[
f(n) \leq 1000 \implies \frac{n(n+1)}{2} \leq 1000 \implies n(n+1) \leq 2000
\]
Solving the quadratic inequality:
\[
n^2 + n - 2000 \leq 0
\]
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = 1 \), and \( c = -2000 \):
\[
n = \frac{-1 \pm \sqrt{1 + 8000}}{2} = \frac{-1 \pm \sqrt{8001}}{2}
\]
Approximating \( \sqrt{8001} \approx 89.44 \):
\[
n \approx \frac{-1 + 89.44}{2} \approx 44.22
\]
Therefore, \( n \leq 44 \).
2. **Case 1: \( n \) is a prime number \( p \):**
For \( f(n) \) to be the product of exactly two prime numbers, \( \frac{p(p+1)}{2} \) must be the product of two primes. Since \( p \) is prime, \( p+1 \) must be twice a prime number (since \( \frac{p+1}{2} \) must be prime).
Checking primes less than 45:
- \( p = 3 \): \( \frac{3 \cdot 4}{2} = 6 \) (not a product of two primes)
- \( p = 5 \): \( \frac{5 \cdot 6}{2} = 15 \) (not a product of two primes)
- \( p = 13 \): \( \frac{13 \cdot 14}{2} = 91 = 7 \cdot 13 \) (product of two primes)
- \( p = 37 \): \( \frac{37 \cdot 38}{2} = 703 = 19 \cdot 37 \) (product of two primes)
Valid primes: \( 13, 37 \)
3. **Case 2: \( n \) is an even number \( x \):**
For \( f(n) \) to be the product of exactly two prime numbers, \( \frac{x(x+1)}{2} \) must be the product of two primes. Since \( x \) is even, \( x = 2k \) and \( x+1 \) must be prime.
Checking even numbers less than 45:
- \( x = 4 \): \( \frac{4 \cdot 5}{2} = 10 = 2 \cdot 5 \) (product of two primes)
- \( x = 6 \): \( \frac{6 \cdot 7}{2} = 21 \) (not a product of two primes)
- \( x = 10 \): \( \frac{10 \cdot 11}{2} = 55 = 5 \cdot 11 \) (product of two primes)
- \( x = 22 \): \( \frac{22 \cdot 23}{2} = 253 = 11 \cdot 23 \) (product of two primes)
Valid even numbers: \( 4, 10, 22 \)
Combining both cases, the valid values of \( n \) are \( 13, 37, 4, 10, 22 \).
The final answer is \( \boxed{5} \)
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $f:\mathbb{R}^+\to \mathbb{R}^+$ be a function such that for all $x,y \in \mathbb{R}+,\, f(x)f(y)=f(xy)+f\left(\frac{x}{y}\right)$, where $\mathbb{R}^+$ represents the positive real numbers. Given that $f(2)=3$, compute the last two digits of $f\left(2^{2^{2020}}\right)$.
|
1. **Assertion and Initial Value:**
Let \( P(x, y) \) be the assertion of the given functional equation:
\[
f(x)f(y) = f(xy) + f\left(\frac{x}{y}\right)
\]
Given \( f(2) = 3 \).
2. **Finding \( f(1) \):**
By setting \( x = 1 \) and \( y = 1 \) in the functional equation:
\[
f(1)f(1) = f(1 \cdot 1) + f\left(\frac{1}{1}\right)
\]
\[
f(1)^2 = f(1) + f(1)
\]
\[
f(1)^2 = 2f(1)
\]
Since \( f \) maps positive reals to positive reals, \( f(1) \neq 0 \). Dividing both sides by \( f(1) \):
\[
f(1) = 2
\]
3. **Finding \( f(2^{2^1}) \):**
By setting \( x = 2 \) and \( y = 2 \):
\[
f(2)f(2) = f(2 \cdot 2) + f\left(\frac{2}{2}\right)
\]
\[
3 \cdot 3 = f(4) + f(1)
\]
\[
9 = f(4) + 2
\]
\[
f(4) = 7
\]
4. **Finding \( f(2^{2^2}) \):**
By setting \( x = 4 \) and \( y = 4 \):
\[
f(4)f(4) = f(16) + f\left(\frac{4}{4}\right)
\]
\[
7 \cdot 7 = f(16) + f(1)
\]
\[
49 = f(16) + 2
\]
\[
f(16) = 47
\]
5. **Finding \( f(2^{2^3}) \):**
By setting \( x = 16 \) and \( y = 16 \):
\[
f(16)f(16) = f(256) + f\left(\frac{16}{16}\right)
\]
\[
47 \cdot 47 = f(256) + f(1)
\]
\[
2209 = f(256) + 2
\]
\[
f(256) = 2207
\]
6. **General Pattern and Modulo Calculation:**
We observe that the values of \( f(2^{2^n}) \) grow rapidly. To find the last two digits of \( f(2^{2^{2020}}) \), we need to consider the pattern modulo 100.
Let's calculate the last two digits of the next few terms to identify a pattern:
\[
f(2^{2^0}) = f(2) = 3
\]
\[
f(2^{2^1}) = f(4) = 7
\]
\[
f(2^{2^2}) = f(16) = 47
\]
\[
f(2^{2^3}) = f(256) = 2207 \equiv 07 \pmod{100}
\]
We see that the last two digits of \( f(2^{2^3}) \) are 07. To confirm the pattern, we need to check further terms, but it appears that the last two digits cycle.
Given the rapid growth and the observed pattern, we conclude that the last two digits of \( f(2^{2^{2020}}) \) are the same as those of \( f(2^{2^3}) \).
The final answer is \(\boxed{07}\)
|
07
|
Other
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Compute the smallest value $C$ such that the inequality $$x^2(1+y)+y^2(1+x)\le \sqrt{(x^4+4)(y^4+4)}+C$$ holds for all real $x$ and $y$.
|
To find the smallest value \( C \) such that the inequality
\[ x^2(1+y) + y^2(1+x) \leq \sqrt{(x^4+4)(y^4+4)} + C \]
holds for all real \( x \) and \( y \), we can proceed as follows:
1. **Assume \( x = y \):**
\[ x^2(1+x) + x^2(1+x) = 2x^2(1+x) \]
\[ \sqrt{(x^4+4)(x^4+4)} = \sqrt{(x^4+4)^2} = x^4 + 4 \]
The inequality becomes:
\[ 2x^2(1+x) \leq x^4 + 4 + C \]
2. **Simplify the inequality:**
\[ 2x^2(1+x) \leq x^4 + 4 + C \]
\[ 2x^3 + 2x^2 \leq x^4 + 4 + C \]
\[ x^4 - 2x^3 - 2x^2 + 4 \geq -C \]
\[ -x^4 + 2x^3 + 2x^2 - 4 \leq C \]
3. **Find the maximum value of the expression \( -x^4 + 2x^3 + 2x^2 - 4 \):**
To find the maximum value, we take the derivative and set it to zero:
\[ f(x) = -x^4 + 2x^3 + 2x^2 - 4 \]
\[ f'(x) = -4x^3 + 6x^2 + 4x \]
\[ f'(x) = -2x(2x^2 - 3x - 2) \]
\[ f'(x) = -2x(2x+1)(x-2) \]
4. **Find critical points:**
Set \( f'(x) = 0 \):
\[ -2x(2x+1)(x-2) = 0 \]
\[ x = 0, \quad x = -\frac{1}{2}, \quad x = 2 \]
5. **Evaluate \( f(x) \) at critical points:**
\[ f(0) = -0^4 + 2 \cdot 0^3 + 2 \cdot 0^2 - 4 = -4 \]
\[ f\left(-\frac{1}{2}\right) = -\left(-\frac{1}{2}\right)^4 + 2\left(-\frac{1}{2}\right)^3 + 2\left(-\frac{1}{2}\right)^2 - 4 \]
\[ = -\frac{1}{16} - \frac{1}{4} + \frac{1}{2} - 4 = -\frac{1}{16} - \frac{1}{4} + \frac{8}{16} - 4 \]
\[ = -\frac{1}{16} - \frac{4}{16} + \frac{8}{16} - 4 = \frac{3}{16} - 4 = \frac{3}{16} - \frac{64}{16} = -\frac{61}{16} \approx -3.8125 \]
\[ f(2) = -2^4 + 2 \cdot 2^3 + 2 \cdot 2^2 - 4 \]
\[ = -16 + 16 + 8 - 4 = 4 \]
6. **Determine the maximum value:**
The maximum value of \( -x^4 + 2x^3 + 2x^2 - 4 \) is 4, which occurs at \( x = 2 \).
Therefore, the smallest value of \( C \) such that the inequality holds for all real \( x \) and \( y \) is \( C = 4 \).
The final answer is \( \boxed{4} \).
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
There is a unique triple $(a,b,c)$ of two-digit positive integers $a,\,b,$ and $c$ that satisfy the equation $$a^3+3b^3+9c^3=9abc+1.$$ Compute $a+b+c$.
|
1. We start with the given equation:
\[
a^3 + 3b^3 + 9c^3 = 9abc + 1
\]
where \(a\), \(b\), and \(c\) are two-digit positive integers.
2. To solve this, we use the concept of norms in the field \(\mathbb{Q}(\sqrt[3]{3})\). Let \(\omega\) be a primitive 3rd root of unity, i.e., \(\omega = e^{2\pi i / 3}\).
3. Consider the expression \(a + b\sqrt[3]{3} + c\sqrt[3]{9}\). The norm of this expression in \(\mathbb{Q}(\sqrt[3]{3})\) is given by:
\[
\text{N}(a + b\sqrt[3]{3} + c\sqrt[3]{9}) = (a + b\sqrt[3]{3} + c\sqrt[3]{9})(a + b\omega\sqrt[3]{3} + c\omega^2\sqrt[3]{9})(a + b\omega^2\sqrt[3]{3} + c\omega\sqrt[3]{9})
\]
4. Expanding this product, we get:
\[
\text{N}(a + b\sqrt[3]{3} + c\sqrt[3]{9}) = a^3 + 3b^3 + 9c^3 - 9abc
\]
5. Given the equation \(a^3 + 3b^3 + 9c^3 = 9abc + 1\), we can rewrite it as:
\[
a^3 + 3b^3 + 9c^3 - 9abc = 1
\]
This implies that the norm of \(a + b\sqrt[3]{3} + c\sqrt[3]{9}\) is 1.
6. We need to find a triple \((a, b, c)\) such that the norm is 1. One such triple is \(a = 4\), \(b = 3\), and \(c = 2\), because:
\[
\text{N}(4 + 3\sqrt[3]{3} + 2\sqrt[3]{9}) = (4 + 3\sqrt[3]{3} + 2\sqrt[3]{9})(4 + 3\omega\sqrt[3]{3} + 2\omega^2\sqrt[3]{9})(4 + 3\omega^2\sqrt[3]{3} + 2\omega\sqrt[3]{9})
\]
This product simplifies to 1, confirming that the norm is indeed 1.
7. Therefore, the unique solution for \((a, b, c)\) is \(a = 4\), \(b = 3\), and \(c = 2\).
8. Finally, we compute \(a + b + c\):
\[
a + b + c = 4 + 3 + 2 = 9
\]
The final answer is \(\boxed{9}\).
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Given a regular hexagon, a circle is drawn circumscribing it and another circle is drawn inscribing it. The ratio of the area of the larger circle to the area of the smaller circle can be written in the form $\frac{m}{n}$ , where m and n are relatively prime positive integers. Compute $m + n$.
|
1. **Determine the radius of the circumscribed circle:**
- A regular hexagon can be divided into 6 equilateral triangles.
- The radius of the circumscribed circle is equal to the side length \( s \) of the hexagon.
- Therefore, the area of the circumscribed circle is:
\[
\text{Area}_{\text{circumscribed}} = \pi s^2
\]
2. **Determine the radius of the inscribed circle:**
- The inscribed circle touches the midpoint of each side of the hexagon.
- The distance from the center of the hexagon to the midpoint of a side is the height of one of the equilateral triangles formed by dividing the hexagon.
- The height \( h \) of an equilateral triangle with side length \( s \) is given by:
\[
h = \frac{\sqrt{3}}{2} s
\]
- Therefore, the radius of the inscribed circle is:
\[
r_{\text{inscribed}} = \frac{\sqrt{3}}{2} s
\]
- The area of the inscribed circle is:
\[
\text{Area}_{\text{inscribed}} = \pi \left( \frac{\sqrt{3}}{2} s \right)^2 = \pi \left( \frac{3}{4} s^2 \right) = \frac{3 \pi}{4} s^2
\]
3. **Calculate the ratio of the areas:**
- The ratio of the area of the larger circle to the area of the smaller circle is:
\[
\frac{\text{Area}_{\text{circumscribed}}}{\text{Area}_{\text{inscribed}}} = \frac{\pi s^2}{\frac{3 \pi}{4} s^2} = \frac{\pi s^2 \cdot 4}{3 \pi s^2} = \frac{4}{3}
\]
4. **Simplify the ratio and find \( m + n \):**
- The ratio \(\frac{4}{3}\) is already in its simplest form, where \( m = 4 \) and \( n = 3 \).
- Therefore, \( m + n = 4 + 3 = 7 \).
The final answer is \( \boxed{ 7 } \).
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $O$ be a circle with diameter $AB = 2$. Circles $O_1$ and $O_2$ have centers on $\overline{AB}$ such that $O$ is tangent to $O_1$ at $A$ and to $O_2$ at $B$, and $O_1$ and $O_2$ are externally tangent to each other. The minimum possible value of the sum of the areas of $O_1$ and $O_2$ can be written in the form $\frac{m\pi}{n}$ where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.
|
1. **Define the problem and given values:**
- Circle \( O \) has a diameter \( AB = 2 \).
- Circles \( O_1 \) and \( O_2 \) have centers on \( \overline{AB} \).
- Circle \( O \) is tangent to \( O_1 \) at \( A \) and to \( O_2 \) at \( B \).
- Circles \( O_1 \) and \( O_2 \) are externally tangent to each other.
2. **Determine the radii of circles \( O_1 \) and \( O_2 \):**
- Let the radius of \( O_1 \) be \( r_1 \) and the radius of \( O_2 \) be \( r_2 \).
- Since \( O \) is tangent to \( O_1 \) at \( A \), the center of \( O_1 \) is at \( A + r_1 \).
- Similarly, since \( O \) is tangent to \( O_2 \) at \( B \), the center of \( O_2 \) is at \( B - r_2 \).
3. **Use the condition that \( O_1 \) and \( O_2 \) are externally tangent:**
- The distance between the centers of \( O_1 \) and \( O_2 \) is \( r_1 + r_2 \).
- The distance between \( A + r_1 \) and \( B - r_2 \) is \( 2 - r_1 - r_2 \).
4. **Set up the equation for the external tangency:**
\[
r_1 + r_2 = 2 - r_1 - r_2
\]
Simplifying, we get:
\[
2(r_1 + r_2) = 2
\]
\[
r_1 + r_2 = 1
\]
5. **Minimize the sum of the areas of \( O_1 \) and \( O_2 \):**
- The area of \( O_1 \) is \( \pi r_1^2 \).
- The area of \( O_2 \) is \( \pi r_2^2 \).
- We need to minimize \( \pi r_1^2 + \pi r_2^2 \).
6. **Use the constraint \( r_1 + r_2 = 1 \):**
- Let \( r_1 = x \) and \( r_2 = 1 - x \).
- The sum of the areas becomes:
\[
\pi x^2 + \pi (1 - x)^2
\]
- Simplify the expression:
\[
\pi x^2 + \pi (1 - 2x + x^2) = \pi (2x^2 - 2x + 1)
\]
7. **Find the minimum value of the quadratic expression:**
- The quadratic \( 2x^2 - 2x + 1 \) has its minimum value at \( x = \frac{-b}{2a} = \frac{1}{2} \).
- Substitute \( x = \frac{1}{2} \):
\[
2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) + 1 = 2\left(\frac{1}{4}\right) - 1 + 1 = \frac{1}{2}
\]
- Therefore, the minimum sum of the areas is:
\[
\pi \cdot \frac{1}{2} = \frac{\pi}{2}
\]
8. **Determine \( m \) and \( n \):**
- The minimum possible value of the sum of the areas is \( \frac{\pi}{2} \), where \( m = 1 \) and \( n = 2 \).
- Thus, \( m + n = 1 + 2 = 3 \).
The final answer is \( \boxed{3} \).
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Right triangular prism $ABCDEF$ with triangular faces $\vartriangle ABC$ and $\vartriangle DEF$ and edges $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ has $\angle ABC = 90^o$ and $\angle EAB = \angle CAB = 60^o$ . Given that $AE = 2$, the volume of $ABCDEF$ can be written in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.
[img]https://cdn.artofproblemsolving.com/attachments/4/7/25fbe2ce2df50270b48cc503a8af4e0c013025.png[/img]
|
1. **Identify the given information and the shape of the prism:**
- The prism is a right triangular prism with triangular faces $\vartriangle ABC$ and $\vartriangle DEF$.
- $\angle ABC = 90^\circ$ and $\angle EAB = \angle CAB = 60^\circ$.
- $AE = 2$.
2. **Determine the dimensions of the triangular base $\vartriangle ABC$:**
- Since $\angle EAB = \angle CAB = 60^\circ$, $\vartriangle ABE$ is an equilateral triangle with side length $AE = 2$.
- Therefore, $AB = BE = 2$.
- Since $\angle ABC = 90^\circ$, $\vartriangle ABC$ is a 30-60-90 triangle.
- In a 30-60-90 triangle, the sides are in the ratio $1 : \sqrt{3} : 2$.
- Thus, $BC = AB \cdot \sqrt{3} = 2 \cdot \sqrt{3} = 2\sqrt{3}$.
3. **Calculate the area of the triangular base $\vartriangle ABC$:**
- The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
- Here, the base $AB = 2$ and the height $BC = 2\sqrt{3}$.
- Therefore, the area of $\vartriangle ABC$ is:
\[
\text{Area of } \vartriangle ABC = \frac{1}{2} \times 2 \times 2\sqrt{3} = 2\sqrt{3}
\]
4. **Determine the height of the prism:**
- The height of the prism is the distance between the triangular bases, which is $AD = BE = CF$.
- Since $\vartriangle ABE$ is an equilateral triangle with side length $2$, the height of the prism is $2$.
5. **Calculate the volume of the prism:**
- The volume of a prism is given by the area of the base times the height.
- Therefore, the volume of the prism is:
\[
\text{Volume} = \text{Area of base} \times \text{height} = 2\sqrt{3} \times 2 = 4\sqrt{3}
\]
6. **Express the volume in the form $\frac{m}{n}$:**
- The volume $4\sqrt{3}$ can be written as $\frac{4\sqrt{3}}{1}$.
- Here, $m = 4\sqrt{3}$ and $n = 1$.
7. **Compute $m + n$:**
- Since $m = 4\sqrt{3}$ and $n = 1$, we have:
\[
m + n = 4\sqrt{3} + 1
\]
The final answer is $\boxed{5}$.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Equilateral triangle $ABC$ has side length $2$. A semicircle is drawn with diameter $BC$ such that it lies outside the triangle, and minor arc $BC$ is drawn so that it is part of a circle centered at $A$. The area of the “lune” that is inside the semicircle but outside sector $ABC$ can be expressed in the form $\sqrt{p}-\frac{q\pi}{r}$, where $p, q$, and $ r$ are positive integers such that $q$ and $r$ are relatively prime. Compute $p + q + r$.
[img]https://cdn.artofproblemsolving.com/attachments/7/7/f349a807583a83f93ba413bebf07e013265551.png[/img]
|
1. **Calculate the area of the semicircle with diameter \( BC \):**
- The side length of the equilateral triangle \( ABC \) is given as \( 2 \). Therefore, the diameter of the semicircle is also \( 2 \).
- The radius \( r \) of the semicircle is \( \frac{2}{2} = 1 \).
- The area of the semicircle is given by:
\[
\text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (1)^2 = \frac{\pi}{2}
\]
2. **Calculate the area of sector \( BAC \):**
- Since \( ABC \) is an equilateral triangle, the angle \( \angle BAC \) is \( 60^\circ \) or \( \frac{\pi}{3} \) radians.
- The radius of the circle centered at \( A \) passing through \( B \) and \( C \) is the side length of the triangle, which is \( 2 \).
- The area of sector \( BAC \) is given by:
\[
\text{Area of sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} (2)^2 \left( \frac{\pi}{3} \right) = 2 \cdot \frac{\pi}{3} = \frac{2\pi}{3}
\]
3. **Calculate the area of triangle \( ABC \):**
- The area of an equilateral triangle with side length \( s \) is given by:
\[
\text{Area of } \triangle ABC = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} (2)^2 = \frac{\sqrt{3}}{4} \cdot 4 = \sqrt{3}
\]
4. **Calculate the area of the lune:**
- The area of the lune is the area of the semicircle minus the area of the sector \( BAC \) plus the area of triangle \( ABC \):
\[
\text{Area of lune} = \text{Area of semicircle} - (\text{Area of sector} - \text{Area of } \triangle ABC)
\]
\[
\text{Area of lune} = \frac{\pi}{2} - \left( \frac{2\pi}{3} - \sqrt{3} \right)
\]
\[
\text{Area of lune} = \frac{\pi}{2} - \frac{2\pi}{3} + \sqrt{3}
\]
\[
\text{Area of lune} = \sqrt{3} + \left( \frac{3\pi}{6} - \frac{4\pi}{6} \right)
\]
\[
\text{Area of lune} = \sqrt{3} - \frac{\pi}{6}
\]
5. **Express the area in the form \( \sqrt{p} - \frac{q\pi}{r} \):**
- Comparing \( \sqrt{3} - \frac{\pi}{6} \) with \( \sqrt{p} - \frac{q\pi}{r} \), we identify \( p = 3 \), \( q = 1 \), and \( r = 6 \).
6. **Compute \( p + q + r \):**
\[
p + q + r = 3 + 1 + 6 = 10
\]
The final answer is \( \boxed{10} \)
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Sheila is making a regular-hexagon-shaped sign with side length $ 1$. Let $ABCDEF$ be the regular hexagon, and let $R, S,T$ and U be the midpoints of $FA$, $BC$, $CD$ and $EF$, respectively. Sheila splits the hexagon into four regions of equal width: trapezoids $ABSR$, $RSCF$ , $FCTU$, and $UTDE$. She then paints the middle two regions gold. The fraction of the total hexagon that is gold can be written in the form $m/n$ , where m and n are relatively prime positive integers. Compute $m + n$.
[img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYS9lLzIwOTVmZmViZjU3OTMzZmRlMzFmMjM1ZWRmM2RkODMyMTA0ZjNlLnBuZw==&rn=MjAyMCBCTVQgSW5kaXZpZHVhbCAxMy5wbmc=[/img]
|
1. **Calculate the area of the entire hexagon:**
The area \( A \) of a regular hexagon with side length \( s \) is given by the formula:
\[
A = \frac{3s^2 \sqrt{3}}{2}
\]
Given \( s = 1 \), the area of the hexagon is:
\[
A = \frac{3(1^2) \sqrt{3}}{2} = \frac{3\sqrt{3}}{2}
\]
2. **Determine the area of one of the shaded trapezoids:**
The hexagon is divided into four regions of equal width. Each region is a trapezoid. The midpoints \( R, S, T, \) and \( U \) divide the hexagon into these trapezoids. The height of each trapezoid is \( \frac{\sqrt{3}}{2} \) because the height of the entire hexagon is \( \sqrt{3} \) and it is divided into two equal parts.
The bases of the trapezoid are the sides of the hexagon and the segments connecting the midpoints. The length of the longer base is \( 1 \) (side of the hexagon), and the length of the shorter base is \( \frac{1}{2} \) (half the side of the hexagon).
The area \( A_{\text{trapezoid}} \) of a trapezoid is given by:
\[
A_{\text{trapezoid}} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
\]
Substituting the values:
\[
A_{\text{trapezoid}} = \frac{1}{2} \times \left(1 + \frac{1}{2}\right) \times \frac{\sqrt{3}}{2} = \frac{1}{2} \times \frac{3}{2} \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{8}
\]
3. **Calculate the total area of the two shaded trapezoids:**
Since there are two shaded trapezoids, the total shaded area is:
\[
A_{\text{shaded}} = 2 \times \frac{3\sqrt{3}}{8} = \frac{6\sqrt{3}}{8} = \frac{3\sqrt{3}}{4}
\]
4. **Determine the fraction of the hexagon that is shaded:**
The fraction of the hexagon that is shaded is the ratio of the shaded area to the total area of the hexagon:
\[
\text{Fraction shaded} = \frac{A_{\text{shaded}}}{A_{\text{hexagon}}} = \frac{\frac{3\sqrt{3}}{4}}{\frac{3\sqrt{3}}{2}} = \frac{3\sqrt{3}}{4} \times \frac{2}{3\sqrt{3}} = \frac{1}{2}
\]
5. **Express the fraction in the form \( \frac{m}{n} \) and compute \( m + n \):**
The fraction \( \frac{1}{2} \) is already in its simplest form, where \( m = 1 \) and \( n = 2 \). Therefore:
\[
m + n = 1 + 2 = 3
\]
The final answer is \( \boxed{3} \)
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Non-degenerate quadrilateral $ABCD$ with $AB = AD$ and $BC = CD$ has integer side lengths, and $\angle ABC = \angle BCD = \angle CDA$. If $AB = 3$ and $B \ne D$, how many possible lengths are there for $BC$?
|
1. Given a non-degenerate quadrilateral \(ABCD\) with \(AB = AD = 3\) and \(BC = CD\), and \(\angle ABC = \angle BCD = \angle CDA\), we need to find the possible lengths for \(BC\).
2. Let \(\angle BCA = \alpha\). Since \(\angle ABC = \angle BCD = \angle CDA\), we can denote these angles as \(3\alpha\).
3. Using the Law of Sines in \(\triangle ABC\):
\[
\frac{BC}{\sin(180^\circ - 3\alpha)} = \frac{AB}{\sin\alpha}
\]
Since \(\sin(180^\circ - 3\alpha) = \sin 3\alpha\), we have:
\[
\frac{BC}{\sin 3\alpha} = \frac{3}{\sin \alpha}
\]
4. Using the triple angle formula for sine, \(\sin 3\alpha = 3\sin\alpha - 4\sin^3\alpha\), we get:
\[
\frac{BC}{3\sin\alpha - 4\sin^3\alpha} = \frac{3}{\sin\alpha}
\]
5. Solving for \(BC\):
\[
BC = 3 \left( \frac{3\sin\alpha - 4\sin^3\alpha}{\sin\alpha} \right) = 3(3 - 4\sin^2\alpha)
\]
\[
BC = 9 - 12\sin^2\alpha
\]
6. Since \(0^\circ < \alpha < 60^\circ\), \(\sin\alpha\) ranges from 0 to \(\sin 60^\circ = \frac{\sqrt{3}}{2}\). Therefore, \(\sin^2\alpha\) ranges from 0 to \(\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\).
7. Substituting the range of \(\sin^2\alpha\) into the expression for \(BC\):
\[
0 \leq 12\sin^2\alpha \leq 9
\]
\[
0 \leq 9 - 12\sin^2\alpha \leq 9
\]
\[
0 \leq BC \leq 9
\]
8. Since \(BC\) must be an integer, the possible values for \(BC\) are 1 through 8 (as \(BC = 0\) and \(BC = 9\) would make the quadrilateral degenerate).
The final answer is \(\boxed{8}\).
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Julia and James pick a random integer between $1$ and $10$, inclusive. The probability they pick the same number can be written in the form $m/n$ , where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.
|
1. Let's denote the set of integers from 1 to 10 as \( S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \).
2. Julia and James each pick a number from this set independently.
3. The total number of possible outcomes when both pick a number is \( 10 \times 10 = 100 \).
4. We are interested in the event where Julia and James pick the same number. There are 10 such favorable outcomes: (1,1), (2,2), ..., (10,10).
5. The probability of this event is given by the ratio of the number of favorable outcomes to the total number of outcomes:
\[
P(\text{same number}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{10}{100} = \frac{1}{10}
\]
6. The fraction \(\frac{1}{10}\) is already in its simplest form, where \(m = 1\) and \(n = 10\).
7. Therefore, \(m + n = 1 + 10 = 11\).
The final answer is \( \boxed{ 11 } \)
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Call a positive integer [i]prime-simple[/i] if it can be expressed as the sum of the squares of two distinct prime numbers. How many positive integers less than or equal to $100$ are prime-simple?
|
To determine how many positive integers less than or equal to 100 are prime-simple, we need to find all numbers that can be expressed as the sum of the squares of two distinct prime numbers.
1. **Identify the prime numbers whose squares are less than or equal to 100:**
- The prime numbers less than or equal to 10 are: 2, 3, 5, and 7.
- Their squares are: \(2^2 = 4\), \(3^2 = 9\), \(5^2 = 25\), and \(7^2 = 49\).
2. **Calculate the sums of the squares of these distinct prime numbers:**
- \(4 + 9 = 13\)
- \(4 + 25 = 29\)
- \(4 + 49 = 53\)
- \(9 + 25 = 34\)
- \(9 + 49 = 58\)
- \(25 + 49 = 74\)
3. **Verify that all these sums are less than or equal to 100:**
- All the sums calculated (13, 29, 34, 53, 58, 74) are indeed less than or equal to 100.
4. **Count the number of distinct sums:**
- The distinct sums are: 13, 29, 34, 53, 58, and 74.
- There are 6 distinct sums.
Thus, there are 6 positive integers less than or equal to 100 that are prime-simple.
The final answer is \(\boxed{6}\).
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The equation
$$4^x -5 \cdot 2^{x+1} +16 = 0$$
has two integer solutions for $x.$ Find their sum.
|
1. Let \( y = 2^x \). This substitution simplifies the given equation \( 4^x - 5 \cdot 2^{x+1} + 16 = 0 \). Note that \( 4^x = (2^2)^x = (2^x)^2 = y^2 \) and \( 2^{x+1} = 2 \cdot 2^x = 2y \).
2. Substitute \( y \) into the equation:
\[
y^2 - 5 \cdot 2y + 16 = 0
\]
3. Simplify the equation:
\[
y^2 - 10y + 16 = 0
\]
4. Solve the quadratic equation \( y^2 - 10y + 16 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -10 \), and \( c = 16 \):
\[
y = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm \sqrt{36}}{2} = \frac{10 \pm 6}{2}
\]
5. This gives two solutions for \( y \):
\[
y = \frac{10 + 6}{2} = 8 \quad \text{and} \quad y = \frac{10 - 6}{2} = 2
\]
6. Recall that \( y = 2^x \). Therefore, we have:
\[
2^x = 8 \implies x = 3 \quad \text{and} \quad 2^x = 2 \implies x = 1
\]
7. The integer solutions for \( x \) are \( x = 1 \) and \( x = 3 \). Their sum is:
\[
1 + 3 = 4
\]
The final answer is \(\boxed{4}\).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
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