problem
stringlengths 2
5.64k
| solution
stringlengths 2
13.5k
| answer
stringlengths 1
43
| problem_type
stringclasses 8
values | question_type
stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | source
stringclasses 6
values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
Let $ABCD$ be a cyclic quadrilateral with $AB = 5$, $BC = 10$, $CD = 11$, and $DA = 14$. The value of $AC + BD$ can be written as $\tfrac{n}{\sqrt{pq}}$, where $n$ is a positive integer and $p$ and $q$ are distinct primes. Find $n + p + q$.
|
1. **Verify the given condition**: We are given that \(AB = 5\), \(BC = 10\), \(CD = 11\), and \(DA = 14\). We need to verify if \(ABCD\) is a cyclic quadrilateral and if \(\angle A = 90^\circ\).
2. **Check the condition for cyclic quadrilateral**: For a quadrilateral to be cyclic, the sum of the opposite angles must be \(180^\circ\). However, we can use the given condition \(5^2 + 14^2 = 10^2 + 11^2\) to check if \(\angle A = 90^\circ\).
\[
5^2 + 14^2 = 25 + 196 = 221
\]
\[
10^2 + 11^2 = 100 + 121 = 221
\]
Since \(5^2 + 14^2 = 10^2 + 11^2\), it implies that \(\angle A = 90^\circ\) and \(\angle C = 90^\circ\) because \(ABCD\) is cyclic.
3. **Calculate \(AC\)**: Since \(\angle A = 90^\circ\), we can use the Pythagorean theorem in \(\triangle ABD\) to find \(AC\).
\[
AC = \sqrt{AB^2 + AD^2} = \sqrt{5^2 + 14^2} = \sqrt{25 + 196} = \sqrt{221}
\]
4. **Apply Ptolemy's theorem**: For a cyclic quadrilateral \(ABCD\), Ptolemy's theorem states that \(AC \cdot BD = AB \cdot CD + BC \cdot AD\).
\[
\sqrt{221} \cdot BD = 5 \cdot 11 + 10 \cdot 14
\]
\[
\sqrt{221} \cdot BD = 55 + 140 = 195
\]
\[
BD = \frac{195}{\sqrt{221}}
\]
5. **Sum \(AC + BD\)**: We need to find \(AC + BD\).
\[
AC + BD = \sqrt{221} + \frac{195}{\sqrt{221}}
\]
\[
AC + BD = \frac{221}{\sqrt{221}} + \frac{195}{\sqrt{221}} = \frac{221 + 195}{\sqrt{221}} = \frac{416}{\sqrt{221}}
\]
6. **Express in the form \(\frac{n}{\sqrt{pq}}\)**: We need to express \(\frac{416}{\sqrt{221}}\) in the form \(\frac{n}{\sqrt{pq}}\), where \(p\) and \(q\) are distinct primes.
\[
221 = 13 \cdot 17
\]
\[
\frac{416}{\sqrt{221}} = \frac{416}{\sqrt{13 \cdot 17}}
\]
7. **Find \(n + p + q\)**: Here, \(n = 416\), \(p = 13\), and \(q = 17\).
\[
n + p + q = 416 + 13 + 17 = 446
\]
The final answer is \(\boxed{446}\)
|
446
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The least positive angle $\alpha$ for which $$\left(\frac34-\sin^2(\alpha)\right)\left(\frac34-\sin^2(3\alpha)\right)\left(\frac34-\sin^2(3^2\alpha)\right)\left(\frac34-\sin^2(3^3\alpha)\right)=\frac1{256}$$ has a degree measure of $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
|
1. **Rewrite the given equation using trigonometric identities:**
The given equation is:
\[
\left(\frac{3}{4} - \sin^2(\alpha)\right)\left(\frac{3}{4} - \sin^2(3\alpha)\right)\left(\frac{3}{4} - \sin^2(3^2\alpha)\right)\left(\frac{3}{4} - \sin^2(3^3\alpha)\right) = \frac{1}{256}
\]
2. **Use the identity for \(\sin^2(x)\):**
Recall that \(\sin^2(x) = \frac{1 - \cos(2x)}{2}\). Therefore:
\[
\frac{3}{4} - \sin^2(x) = \frac{3}{4} - \frac{1 - \cos(2x)}{2} = \frac{3}{4} - \frac{1}{2} + \frac{\cos(2x)}{2} = \frac{1}{4} + \frac{\cos(2x)}{2}
\]
3. **Simplify each term:**
Applying the above identity to each term:
\[
\frac{3}{4} - \sin^2(\alpha) = \frac{1}{4} + \frac{\cos(2\alpha)}{2}
\]
\[
\frac{3}{4} - \sin^2(3\alpha) = \frac{1}{4} + \frac{\cos(6\alpha)}{2}
\]
\[
\frac{3}{4} - \sin^2(9\alpha) = \frac{1}{4} + \frac{\cos(18\alpha)}{2}
\]
\[
\frac{3}{4} - \sin^2(27\alpha) = \frac{1}{4} + \frac{\cos(54\alpha)}{2}
\]
4. **Combine the terms:**
The product of these terms is:
\[
\left(\frac{1}{4} + \frac{\cos(2\alpha)}{2}\right)\left(\frac{1}{4} + \frac{\cos(6\alpha)}{2}\right)\left(\frac{1}{4} + \frac{\cos(18\alpha)}{2}\right)\left(\frac{1}{4} + \frac{\cos(54\alpha)}{2}\right) = \frac{1}{256}
\]
5. **Simplify the equation:**
Notice that \(\frac{1}{256} = \left(\frac{1}{4}\right)^4\). Therefore, each term must be equal to \(\frac{1}{4}\):
\[
\frac{1}{4} + \frac{\cos(2\alpha)}{2} = \frac{1}{4}
\]
\[
\frac{1}{4} + \frac{\cos(6\alpha)}{2} = \frac{1}{4}
\]
\[
\frac{1}{4} + \frac{\cos(18\alpha)}{2} = \frac{1}{4}
\]
\[
\frac{1}{4} + \frac{\cos(54\alpha)}{2} = \frac{1}{4}
\]
6. **Solve for \(\alpha\):**
Each equation simplifies to:
\[
\frac{\cos(2\alpha)}{2} = 0 \implies \cos(2\alpha) = 0 \implies 2\alpha = 90^\circ \implies \alpha = 45^\circ
\]
However, we need to find the least positive angle \(\alpha\) such that:
\[
\sin(81\alpha) = \sin(\alpha)
\]
This implies:
\[
81\alpha = 180^\circ k + \alpha \quad \text{or} \quad 81\alpha = 180^\circ k - \alpha
\]
For the smallest positive \(\alpha\):
\[
81\alpha = 180^\circ + \alpha \implies 80\alpha = 180^\circ \implies \alpha = \frac{180^\circ}{80} = \frac{9^\circ}{4}
\]
7. **Express \(\alpha\) in simplest form:**
\(\alpha = \frac{9}{4}^\circ\). In degrees, this is \(\frac{9}{4}\).
8. **Find \(m + n\):**
Here, \(m = 9\) and \(n = 4\). Therefore, \(m + n = 9 + 4 = 13\).
The final answer is \(\boxed{13}\).
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
At Ignus School there are $425$ students. Of these students $351$ study mathematics, $71$ study Latin, and $203$ study chemistry. There are $199$ students who study more than one of these subjects, and $8$ students who do not study any of these subjects. Find the number of students who study all three of these subjects.
|
1. **Subtract the number of students who do not study any of these subjects:**
\[
425 - 8 = 417
\]
This gives us the number of students who study at least one of the subjects.
2. **Use the principle of inclusion-exclusion to find the number of students who study all three subjects. Let \( n(M) \) be the number of students studying mathematics, \( n(L) \) be the number of students studying Latin, and \( n(C) \) be the number of students studying chemistry. Let \( n(M \cap L) \), \( n(M \cap C) \), and \( n(L \cap C) \) be the number of students studying two subjects, and \( n(M \cap L \cap C) \) be the number of students studying all three subjects.**
According to the principle of inclusion-exclusion:
\[
n(M \cup L \cup C) = n(M) + n(L) + n(C) - n(M \cap L) - n(M \cap C) - n(L \cap C) + n(M \cap L \cap C)
\]
3. **Substitute the known values into the equation:**
\[
417 = 351 + 71 + 203 - n(M \cap L) - n(M \cap C) - n(L \cap C) + n(M \cap L \cap C)
\]
4. **We know that there are 199 students who study more than one subject. This includes students who study exactly two subjects and those who study all three subjects. Let \( x \) be the number of students who study all three subjects. Then:**
\[
n(M \cap L) + n(M \cap C) + n(L \cap C) - 2x = 199
\]
5. **Substitute \( n(M \cap L) + n(M \cap C) + n(L \cap C) = 199 + 2x \) into the inclusion-exclusion equation:**
\[
417 = 351 + 71 + 203 - (199 + 2x) + x
\]
6. **Simplify the equation:**
\[
417 = 625 - 199 - x
\]
\[
417 = 426 - x
\]
7. **Solve for \( x \):**
\[
x = 426 - 417
\]
\[
x = 9
\]
Therefore, the number of students who study all three subjects is \( \boxed{9} \).
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The value of
$$\left(1-\frac{1}{2^2-1}\right)\left(1-\frac{1}{2^3-1}\right)\left(1-\frac{1}{2^4-1}\right)\dots\left(1-\frac{1}{2^{29}-1}\right)$$
can be written as $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $2m - n.$
|
1. We start with the given expression:
\[
\left(1-\frac{1}{2^2-1}\right)\left(1-\frac{1}{2^3-1}\right)\left(1-\frac{1}{2^4-1}\right)\dots\left(1-\frac{1}{2^{29}-1}\right)
\]
2. Simplify each term inside the product:
\[
1 - \frac{1}{2^k - 1}
\]
for \( k = 2, 3, 4, \ldots, 29 \).
3. Notice that \( 2^k - 1 \) can be factored as:
\[
1 - \frac{1}{2^k - 1} = \frac{2^k - 2}{2^k - 1} = \frac{2(2^{k-1} - 1)}{2^k - 1}
\]
4. Rewrite the product using the simplified terms:
\[
\prod_{k=2}^{29} \left( \frac{2(2^{k-1} - 1)}{2^k - 1} \right)
\]
5. Factor out the constant 2 from each term:
\[
2^{28} \prod_{k=2}^{29} \left( \frac{2^{k-1} - 1}{2^k - 1} \right)
\]
6. Observe that the product telescopes:
\[
\prod_{k=2}^{29} \left( \frac{2^{k-1} - 1}{2^k - 1} \right) = \frac{1}{2^2 - 1} \cdot \frac{2^2 - 1}{2^3 - 1} \cdot \frac{2^3 - 1}{2^4 - 1} \cdots \frac{2^{28} - 1}{2^{29} - 1}
\]
7. Simplify the telescoping product:
\[
\frac{1}{3} \cdot \frac{3}{7} \cdot \frac{7}{15} \cdots \frac{2^{28} - 1}{2^{29} - 1} = \frac{1}{2^{29} - 1}
\]
8. Combine the results:
\[
2^{28} \cdot \frac{1}{2^{29} - 1} = \frac{2^{28}}{2^{29} - 1}
\]
9. Let \( m = 2^{28} \) and \( n = 2^{29} - 1 \). Since \( m \) and \( n \) are relatively prime, we need to find \( 2m - n \):
\[
2m - n = 2 \cdot 2^{28} - (2^{29} - 1) = 2^{29} - 2^{29} + 1 = 1
\]
The final answer is \( \boxed{1} \).
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the positive integer $n$ such that a convex polygon with $3n + 2$ sides has $61.5$ percent fewer diagonals than a convex polygon with $5n - 2$ sides.
|
1. The formula to find the number of diagonals in a convex polygon with \( k \) sides is given by:
\[
D = \frac{k(k-3)}{2}
\]
where \( D \) is the number of diagonals.
2. Let \( k_1 = 3n + 2 \) be the number of sides of the first polygon, and \( k_2 = 5n - 2 \) be the number of sides of the second polygon.
3. The number of diagonals in the first polygon is:
\[
D_1 = \frac{(3n+2)((3n+2)-3)}{2} = \frac{(3n+2)(3n-1)}{2}
\]
4. The number of diagonals in the second polygon is:
\[
D_2 = \frac{(5n-2)((5n-2)-3)}{2} = \frac{(5n-2)(5n-5)}{2}
\]
5. According to the problem, the number of diagonals in the first polygon is 61.5% fewer than the number of diagonals in the second polygon. This can be expressed as:
\[
D_1 = (1 - 0.615)D_2 = 0.385D_2
\]
6. Substituting the expressions for \( D_1 \) and \( D_2 \) into the equation:
\[
\frac{(3n+2)(3n-1)}{2} = 0.385 \left( \frac{(5n-2)(5n-5)}{2} \right)
\]
7. Simplifying the equation:
\[
(3n+2)(3n-1) = 0.385 (5n-2)(5n-5)
\]
8. Removing the common factor of \(\frac{1}{2}\) from both sides:
\[
(3n+2)(3n-1) = 0.385 (5n-2)(5n-5)
\]
9. Expanding both sides:
\[
9n^2 + 3n - 2 = 0.385 (25n^2 - 35n + 10)
\]
10. Simplifying the right-hand side:
\[
9n^2 + 3n - 2 = 0.385 \times 25n^2 - 0.385 \times 35n + 0.385 \times 10
\]
\[
9n^2 + 3n - 2 = 9.625n^2 - 13.475n + 3.85
\]
11. Rearranging the equation to set it to zero:
\[
9n^2 + 3n - 2 - 9.625n^2 + 13.475n - 3.85 = 0
\]
\[
-0.625n^2 + 16.475n - 5.85 = 0
\]
12. Multiplying through by -1 to simplify:
\[
0.625n^2 - 16.475n + 5.85 = 0
\]
13. Solving the quadratic equation using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 0.625, \quad b = -16.475, \quad c = 5.85
\]
\[
n = \frac{16.475 \pm \sqrt{(-16.475)^2 - 4 \cdot 0.625 \cdot 5.85}}{2 \cdot 0.625}
\]
\[
n = \frac{16.475 \pm \sqrt{271.650625 - 14.625}}{1.25}
\]
\[
n = \frac{16.475 \pm \sqrt{257.025625}}{1.25}
\]
\[
n = \frac{16.475 \pm 16.031}{1.25}
\]
14. Calculating the two potential solutions:
\[
n = \frac{16.475 + 16.031}{1.25} \approx 26.008
\]
\[
n = \frac{16.475 - 16.031}{1.25} \approx 0.355
\]
15. Since \( n \) must be a positive integer, we select \( n = 26 \).
The final answer is \( \boxed{26} \).
|
26
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For positive integer $n,$ let $s(n)$ be the sum of the digits of n when n is expressed in base ten. For example, $s(2022) = 2 + 0 + 2 + 2 = 6.$ Find the sum of the two solutions to the equation $n - 3s(n) = 2022.$
|
1. Let \( n = 1000a + 100b + 10c + d \). Then, the sum of the digits of \( n \) is \( s(n) = a + b + c + d \).
2. Given the equation \( n - 3s(n) = 2022 \), substitute \( n \) and \( s(n) \):
\[
1000a + 100b + 10c + d - 3(a + b + c + d) = 2022
\]
3. Simplify the equation:
\[
1000a + 100b + 10c + d - 3a - 3b - 3c - 3d = 2022
\]
\[
997a + 97b + 7c - 2d = 2022
\]
4. To find the values of \( a, b, c, \) and \( d \), we start by considering the possible values for \( a \). Since \( a \) is a digit (0-9) and \( 997a \) must be close to 2022, we test \( a = 2 \):
\[
997 \cdot 2 = 1994
\]
\[
1994 + 97b + 7c - 2d = 2022
\]
\[
97b + 7c - 2d = 2022 - 1994
\]
\[
97b + 7c - 2d = 28
\]
5. Next, we consider the possible values for \( b \). Since \( 97b \) must be less than or equal to 28, \( b \) must be 0:
\[
97 \cdot 0 = 0
\]
\[
7c - 2d = 28
\]
6. Solve the equation \( 7c - 2d = 28 \) for \( c \) and \( d \). We test possible values for \( c \) and \( d \):
- If \( c = 4 \):
\[
7 \cdot 4 - 2d = 28
\]
\[
28 - 2d = 28
\]
\[
-2d = 0
\]
\[
d = 0
\]
Thus, one solution is \( (a, b, c, d) = (2, 0, 4, 0) \), giving \( n = 2040 \).
- If \( c = 6 \):
\[
7 \cdot 6 - 2d = 28
\]
\[
42 - 2d = 28
\]
\[
-2d = -14
\]
\[
d = 7
\]
Thus, another solution is \( (a, b, c, d) = (2, 0, 6, 7) \), giving \( n = 2067 \).
7. The sum of the two solutions is:
\[
2040 + 2067 = 4107
\]
The final answer is \(\boxed{4107}\).
|
4107
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Given that $a_1, a_2, a_3, . . . , a_{99}$ is a permutation of $1, 2, 3, . . . , 99,$ find the maximum possible value of
$$|a_1 - 1| + |a_2 - 2| + |a_3 - 3| + \dots + |a_{99} - 99|.$$
|
1. We are given a permutation \(a_1, a_2, a_3, \ldots, a_{99}\) of the numbers \(1, 2, 3, \ldots, 99\). We need to find the maximum possible value of the expression:
\[
|a_1 - 1| + |a_2 - 2| + |a_3 - 3| + \dots + |a_{99} - 99|.
\]
2. To maximize the sum of absolute differences, we should pair each \(a_i\) with \(i\) such that the difference \(|a_i - i|\) is as large as possible. This can be achieved by reversing the order of the permutation, i.e., setting \(a_i = 100 - i\).
3. Let's verify this by calculating the sum:
\[
|a_1 - 1| + |a_2 - 2| + |a_3 - 3| + \dots + |a_{99} - 99|
\]
where \(a_i = 100 - i\).
4. Substituting \(a_i = 100 - i\) into the expression, we get:
\[
|(100 - 1) - 1| + |(100 - 2) - 2| + |(100 - 3) - 3| + \dots + |(100 - 99) - 99|
\]
Simplifying each term, we have:
\[
|99 - 1| + |98 - 2| + |97 - 3| + \dots + |1 - 99|
\]
which simplifies to:
\[
98 + 96 + 94 + \dots + 2
\]
5. Notice that this is an arithmetic series with the first term \(a = 98\), the common difference \(d = -2\), and the number of terms \(n = 49\) (since the sequence goes from 98 to 2).
6. The sum of an arithmetic series is given by:
\[
S_n = \frac{n}{2} (a + l)
\]
where \(l\) is the last term. Here, \(a = 98\), \(l = 2\), and \(n = 49\).
7. Substituting these values, we get:
\[
S_{49} = \frac{49}{2} (98 + 2) = \frac{49}{2} \cdot 100 = 49 \cdot 50 = 2450
\]
8. Since the sequence is symmetric, the sum of the first 49 terms is equal to the sum of the last 49 terms. Therefore, the total sum is:
\[
2 \cdot 2450 = 4900
\]
The final answer is \(\boxed{4900}\).
|
4900
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Of $450$ students assembled for a concert, $40$ percent were boys. After a bus containing an equal number of boys and girls brought more students to the concert, $41$ percent of the students at the concert were boys. Find the number of students on the bus.
|
1. Initially, there are 450 students at the concert, and 40% of them are boys. We can calculate the number of boys and girls initially as follows:
\[
\text{Number of boys} = 0.40 \times 450 = 180
\]
\[
\text{Number of girls} = 450 - 180 = 270
\]
2. Let \(2x\) be the number of students on the bus, where \(x\) is the number of boys and \(x\) is the number of girls (since the bus contains an equal number of boys and girls).
3. After the bus arrives, the total number of students at the concert becomes:
\[
450 + 2x
\]
The total number of boys becomes:
\[
180 + x
\]
4. According to the problem, after the bus arrives, 41% of the students at the concert are boys. Therefore, we can set up the following equation:
\[
\frac{180 + x}{450 + 2x} = 0.41
\]
5. To solve for \(x\), we first clear the fraction by multiplying both sides by \(450 + 2x\):
\[
180 + x = 0.41 \times (450 + 2x)
\]
6. Distribute 0.41 on the right-hand side:
\[
180 + x = 0.41 \times 450 + 0.41 \times 2x
\]
\[
180 + x = 184.5 + 0.82x
\]
7. Isolate \(x\) by moving all terms involving \(x\) to one side and constants to the other side:
\[
180 + x - 0.82x = 184.5
\]
\[
180 + 0.18x = 184.5
\]
8. Subtract 180 from both sides:
\[
0.18x = 4.5
\]
9. Solve for \(x\) by dividing both sides by 0.18:
\[
x = \frac{4.5}{0.18} = 25
\]
10. Since \(2x\) is the total number of students on the bus, we have:
\[
2x = 2 \times 25 = 50
\]
The final answer is \(\boxed{50}\).
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Below is a diagram showing a $6 \times 8$ rectangle divided into four $6 \times 2$ rectangles and one diagonal line. Find the total perimeter of the four shaded trapezoids.
|
1. **Labeling and Initial Setup:**
- We have a $6 \times 8$ rectangle divided into four $6 \times 2$ rectangles.
- A diagonal line is drawn from one corner of the large rectangle to the opposite corner.
- Label the four shaded trapezoids from left to right as 1, 2, 3, and 4.
2. **Combining Trapezoids into Rectangles:**
- Notice that trapezoids 1 and 2 together form a rectangle, and trapezoids 3 and 4 together form another rectangle.
- These two rectangles are congruent.
3. **Calculating the Height of the Combined Rectangles:**
- The height of each $6 \times 2$ rectangle is 6.
- The diagonal of the $6 \times 8$ rectangle splits each $6 \times 2$ rectangle into two right triangles.
- The height of the combined rectangles is calculated using similarity ratios. The height of the combined rectangles is $6 + \frac{3}{2} = 3 + \frac{9}{2} = \frac{15}{2}$.
4. **Calculating the Perimeter of the Combined Rectangles:**
- The perimeter of one combined rectangle is $2 \left( \frac{15}{2} + 2 \right) = 2 \left( \frac{15}{2} + \frac{4}{2} \right) = 2 \left( \frac{19}{2} \right) = 19$.
- Since there are two such rectangles, the total perimeter is $2 \times 19 = 38$.
5. **Accounting for the Diagonals:**
- The diagonal of the large $6 \times 8$ rectangle is $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
- This diagonal is not included in the previous perimeter calculation, so we need to add it.
6. **Final Calculation:**
- The total perimeter of the four shaded trapezoids is $38 + 10 = 48$.
The final answer is $\boxed{48}$.
|
48
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a_1=2021$ and for $n \ge 1$ let $a_{n+1}=\sqrt{4+a_n}$. Then $a_5$ can be written as $$\sqrt{\frac{m+\sqrt{n}}{2}}+\sqrt{\frac{m-\sqrt{n}}{2}},$$ where $m$ and $n$ are positive integers. Find $10m+n$.
|
1. We start with the given sequence \(a_1 = 2021\) and the recursive formula \(a_{n+1} = \sqrt{4 + a_n}\).
2. Calculate the first few terms of the sequence:
\[
a_2 = \sqrt{4 + a_1} = \sqrt{4 + 2021} = \sqrt{2025} = 45
\]
\[
a_3 = \sqrt{4 + a_2} = \sqrt{4 + 45} = \sqrt{49} = 7
\]
\[
a_4 = \sqrt{4 + a_3} = \sqrt{4 + 7} = \sqrt{11}
\]
\[
a_5 = \sqrt{4 + a_4} = \sqrt{4 + \sqrt{11}}
\]
3. We are given that \(a_5\) can be written in the form:
\[
a_5 = \sqrt{\frac{m + \sqrt{n}}{2}} + \sqrt{\frac{m - \sqrt{n}}{2}}
\]
4. Let us denote:
\[
x = \sqrt{\frac{m + \sqrt{n}}{2}} \quad \text{and} \quad y = \sqrt{\frac{m - \sqrt{n}}{2}}
\]
Then:
\[
a_5 = x + y
\]
Squaring both sides:
\[
a_5^2 = (x + y)^2 = x^2 + y^2 + 2xy
\]
5. We know:
\[
x^2 = \frac{m + \sqrt{n}}{2} \quad \text{and} \quad y^2 = \frac{m - \sqrt{n}}{2}
\]
Adding these:
\[
x^2 + y^2 = \frac{m + \sqrt{n}}{2} + \frac{m - \sqrt{n}}{2} = \frac{2m}{2} = m
\]
6. Also:
\[
2xy = 2 \sqrt{\left(\frac{m + \sqrt{n}}{2}\right) \left(\frac{m - \sqrt{n}}{2}\right)} = 2 \sqrt{\frac{(m + \sqrt{n})(m - \sqrt{n})}{4}} = 2 \sqrt{\frac{m^2 - n}{4}} = \sqrt{m^2 - n}
\]
7. Therefore:
\[
a_5^2 = m + \sqrt{m^2 - n}
\]
8. We know \(a_5 = \sqrt{4 + \sqrt{11}}\), so:
\[
a_5^2 = 4 + \sqrt{11}
\]
Thus:
\[
4 + \sqrt{11} = m + \sqrt{m^2 - n}
\]
9. By comparing both sides, we can deduce:
\[
m = 4 \quad \text{and} \quad \sqrt{m^2 - n} = \sqrt{11}
\]
Squaring the second equation:
\[
m^2 - n = 11 \implies 4^2 - n = 11 \implies 16 - n = 11 \implies n = 5
\]
10. Finally, we calculate \(10m + n\):
\[
10m + n = 10 \cdot 4 + 5 = 40 + 5 = 45
\]
The final answer is \(\boxed{45}\).
|
45
|
Other
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
In a room there are $144$ people. They are joined by $n$ other people who are each carrying $k$ coins. When these coins are shared among all $n + 144$ people, each person has $2$ of these coins. Find the minimum possible value of $2n + k$.
|
1. Let the number of people initially in the room be \(144\).
2. Let \(n\) be the number of additional people who join the room.
3. Each of these \(n\) additional people is carrying \(k\) coins.
4. When these coins are shared among all \(n + 144\) people, each person has \(2\) coins.
We need to find the minimum possible value of \(2n + k\).
First, we set up the equation based on the given information:
\[
2(n + 144) = nk
\]
Simplify the equation:
\[
2n + 288 = nk
\]
Rearrange to isolate \(n\):
\[
nk - 2n = 288
\]
Factor out \(n\):
\[
n(k - 2) = 288
\]
Thus, we have:
\[
n = \frac{288}{k - 2}
\]
To find the minimum value of \(2n + k\), we need to consider the possible values of \(k\) such that \(n\) is an integer. This means \(k - 2\) must be a divisor of \(288\).
The divisors of \(288\) are:
\[
1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 36, 48, 72, 96, 144, 288
\]
Thus, the possible values of \(k\) are:
\[
3, 4, 5, 6, 8, 10, 11, 14, 18, 20, 26, 34, 38, 50, 74, 98, 146, 290
\]
For each \(k\), we calculate \(n\) and then \(2n + k\):
- For \(k = 3\):
\[
n = \frac{288}{3 - 2} = 288
\]
\[
2n + k = 2(288) + 3 = 579
\]
- For \(k = 4\):
\[
n = \frac{288}{4 - 2} = 144
\]
\[
2n + k = 2(144) + 4 = 292
\]
- For \(k = 5\):
\[
n = \frac{288}{5 - 2} = 96
\]
\[
2n + k = 2(96) + 5 = 197
\]
- For \(k = 6\):
\[
n = \frac{288}{6 - 2} = 72
\]
\[
2n + k = 2(72) + 6 = 150
\]
- For \(k = 8\):
\[
n = \frac{288}{8 - 2} = 48
\]
\[
2n + k = 2(48) + 8 = 104
\]
- For \(k = 10\):
\[
n = \frac{288}{10 - 2} = 36
\]
\[
2n + k = 2(36) + 10 = 82
\]
- For \(k = 11\):
\[
n = \frac{288}{11 - 2} = 32
\]
\[
2n + k = 2(32) + 11 = 75
\]
- For \(k = 14\):
\[
n = \frac{288}{14 - 2} = 24
\]
\[
2n + k = 2(24) + 14 = 62
\]
- For \(k = 18\):
\[
n = \frac{288}{18 - 2} = 18
\]
\[
2n + k = 2(18) + 18 = 54
\]
- For \(k = 20\):
\[
n = \frac{288}{20 - 2} = 16
\]
\[
2n + k = 2(16) + 20 = 52
\]
- For \(k = 26\):
\[
n = \frac{288}{26 - 2} = 12
\]
\[
2n + k = 2(12) + 26 = 50
\]
- For \(k = 34\):
\[
n = \frac{288}{34 - 2} = 9
\]
\[
2n + k = 2(9) + 34 = 52
\]
- For \(k = 38\):
\[
n = \frac{288}{38 - 2} = 8
\]
\[
2n + k = 2(8) + 38 = 54
\]
- For \(k = 50\):
\[
n = \frac{288}{50 - 2} = 6
\]
\[
2n + k = 2(6) + 50 = 62
\]
- For \(k = 74\):
\[
n = \frac{288}{74 - 2} = 4
\]
\[
2n + k = 2(4) + 74 = 82
\]
- For \(k = 98\):
\[
n = \frac{288}{98 - 2} = 3
\]
\[
2n + k = 2(3) + 98 = 104
\]
- For \(k = 146\):
\[
n = \frac{288}{146 - 2} = 2
\]
\[
2n + k = 2(2) + 146 = 150
\]
- For \(k = 290\):
\[
n = \frac{288}{290 - 2} = 1
\]
\[
2n + k = 2(1) + 290 = 292
\]
The minimum value of \(2n + k\) is \(50\) when \(k = 26\) and \(n = 12\).
The final answer is \(\boxed{50}\)
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The product
$$\left(\frac{1+1}{1^2+1}+\frac{1}{4}\right)\left(\frac{2+1}{2^2+1}+\frac{1}{4}\right)\left(\frac{3+1}{3^2+1}+\frac{1}{4}\right)\cdots\left(\frac{2022+1}{2022^2+1}+\frac{1}{4}\right)$$
can be written as $\frac{q}{2^r\cdot s}$, where $r$ is a positive integer, and $q$ and $s$ are relatively prime odd positive integers. Find $s$.
|
1. Consider the given product:
\[
\left(\frac{1+1}{1^2+1}+\frac{1}{4}\right)\left(\frac{2+1}{2^2+1}+\frac{1}{4}\right)\left(\frac{3+1}{3^2+1}+\frac{1}{4}\right)\cdots\left(\frac{2022+1}{2022^2+1}+\frac{1}{4}\right)
\]
2. Simplify each term in the product:
\[
\frac{x+1}{x^2+1} + \frac{1}{4} = \frac{4(x+1) + x^2 + 1}{4(x^2 + 1)} = \frac{x^2 + 4x + 5}{4(x^2 + 1)}
\]
3. Notice that \(x^2 + 4x + 5\) can be rewritten using the identity \((x+2)^2 + 1\):
\[
x^2 + 4x + 5 = (x+2)^2 + 1
\]
4. Substitute this back into the expression:
\[
\frac{(x+2)^2 + 1}{4(x^2 + 1)}
\]
5. The product now becomes:
\[
\prod_{x=1}^{2022} \left( \frac{(x+2)^2 + 1}{4(x^2 + 1)} \right)
\]
6. Factor out the constant term \( \frac{1}{4} \) from each term:
\[
\left( \frac{1}{4} \right)^{2022} \prod_{x=1}^{2022} \left( \frac{(x+2)^2 + 1}{x^2 + 1} \right)
\]
7. Simplify the product inside:
\[
\prod_{x=1}^{2022} \left( \frac{(x+2)^2 + 1}{x^2 + 1} \right)
\]
8. Notice that the numerator and denominator form a telescoping product:
\[
\frac{(3^2 + 1)}{(1^2 + 1)} \cdot \frac{(4^2 + 1)}{(2^2 + 1)} \cdot \frac{(5^2 + 1)}{(3^2 + 1)} \cdots \frac{(2024^2 + 1)}{(2022^2 + 1)}
\]
9. Most terms cancel out, leaving:
\[
\frac{2024^2 + 1}{2^2 + 1} = \frac{2024^2 + 1}{5}
\]
10. Combine this with the constant term:
\[
\left( \frac{1}{4} \right)^{2022} \cdot \frac{2024^2 + 1}{5}
\]
11. Express the final result in the form \( \frac{q}{2^r \cdot s} \):
\[
\frac{2024^2 + 1}{4^{2022} \cdot 5}
\]
12. Since \(4^{2022} = (2^2)^{2022} = 2^{4044}\), we have:
\[
\frac{2024^2 + 1}{2^{4044} \cdot 5}
\]
13. Identify \(q\), \(r\), and \(s\):
\[
q = 2024^2 + 1, \quad r = 4044, \quad s = 5
\]
The final answer is \( \boxed{5} \).
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
There is a positive integer s such that there are s solutions to the equation
$64sin^2(2x)+tan^2(x)+cot^2(x)=46$ in the interval $(0,\frac{\pi}{2})$ all of the form $\frac{m_k}{n_k}\pi$
where $m_k$ and $n_k$ are relatively prime positive integers, for $k = 1, 2, 3, . . . , s$. Find $(m_1 + n_1) + (m_2 + n_2) + (m_3 + n_3) + · · · + (m_s + n_s)$.
|
1. Start with the given equation:
\[
64\sin^2(2x) + \tan^2(x) + \cot^2(x) = 46
\]
2. Rewrite \(\tan^2(x)\) and \(\cot^2(x)\) in terms of \(\sin(x)\) and \(\cos(x)\):
\[
\tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)}, \quad \cot^2(x) = \frac{\cos^2(x)}{\sin^2(x)}
\]
3. Substitute these into the equation:
\[
64\sin^2(2x) + \frac{\sin^2(x)}{\cos^2(x)} + \frac{\cos^2(x)}{\sin^2(x)} = 46
\]
4. Use the identity \(\sin(2x) = 2\sin(x)\cos(x)\):
\[
\sin^2(2x) = 4\sin^2(x)\cos^2(x)
\]
5. Substitute \(\sin^2(2x)\) into the equation:
\[
64 \cdot 4\sin^2(x)\cos^2(x) + \frac{\sin^2(x)}{\cos^2(x)} + \frac{\cos^2(x)}{\sin^2(x)} = 46
\]
\[
256\sin^2(x)\cos^2(x) + \frac{\sin^2(x)}{\cos^2(x)} + \frac{\cos^2(x)}{\sin^2(x)} = 46
\]
6. Let \(y = \sin^2(x)\cos^2(x)\). Then:
\[
256y + \frac{1}{4y} = 46
\]
7. Multiply through by \(4y\) to clear the fraction:
\[
1024y^2 + 1 = 184y
\]
8. Rearrange into a standard quadratic equation:
\[
1024y^2 - 184y + 1 = 0
\]
9. Solve the quadratic equation using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[
y = \frac{184 \pm \sqrt{184^2 - 4 \cdot 1024 \cdot 1}}{2 \cdot 1024}
\]
\[
y = \frac{184 \pm \sqrt{33856 - 4096}}{2048}
\]
\[
y = \frac{184 \pm \sqrt{29760}}{2048}
\]
\[
y = \frac{184 \pm 172.56}{2048}
\]
10. Calculate the two solutions for \(y\):
\[
y_1 = \frac{184 + 172.56}{2048} = \frac{356.56}{2048} = \frac{89.14}{512} \approx 0.174
\]
\[
y_2 = \frac{184 - 172.56}{2048} = \frac{11.44}{2048} = \frac{2.86}{512} \approx 0.0056
\]
11. Recall that \(y = \sin^2(x)\cos^2(x)\), so:
\[
\sin^2(2x) = 4y
\]
\[
\sin^2(2x) = 4 \cdot 0.174 = 0.696
\]
\[
\sin^2(2x) = 4 \cdot 0.0056 = 0.0224
\]
12. Solve for \(\sin(2x)\):
\[
\sin(2x) = \sqrt{0.696} \approx 0.834
\]
\[
\sin(2x) = \sqrt{0.0224} \approx 0.149
\]
13. Find the corresponding \(x\) values in the interval \((0, \frac{\pi}{2})\):
\[
2x = \arcsin(0.834) \approx 0.99 \text{ or } 2x = \pi - 0.99 \approx 2.15
\]
\[
2x = \arcsin(0.149) \approx 0.15 \text{ or } 2x = \pi - 0.15 \approx 2.99
\]
14. Divide by 2 to find \(x\):
\[
x \approx 0.495, \quad x \approx 1.075, \quad x \approx 0.075, \quad x \approx 1.495
\]
15. Convert these to the form \(\frac{m_k}{n_k}\pi\):
\[
x = \frac{\pi}{20}, \quad x = \frac{3\pi}{20}, \quad x = \frac{7\pi}{20}, \quad x = \frac{9\pi}{20}
\]
16. Sum the numerators and denominators:
\[
(1 + 20) + (3 + 20) + (7 + 20) + (9 + 20) = 21 + 23 + 27 + 29 = 100
\]
The final answer is \(\boxed{100}\)
|
100
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Mike has two similar pentagons. The first pentagon has a perimeter of $18$ and an area of $8 \frac{7}{16}$ . The second pentagon has a perimeter of $24$. Find the area of the second pentagon.
|
1. Let \( K_1 \) and \( K_2 \) be the perimeters of the first and second pentagons, respectively. Given:
\[
K_1 = 18, \quad K_2 = 24
\]
2. Let \( L_1 \) and \( L_2 \) be the areas of the first and second pentagons, respectively. Given:
\[
L_1 = 8 \frac{7}{16} = 8 + \frac{7}{16} = 8 + 0.4375 = 8.4375
\]
3. Since the pentagons are similar, the ratio of their perimeters is equal to the ratio of their corresponding side lengths. Therefore:
\[
\frac{K_1}{K_2} = \frac{18}{24} = \frac{3}{4}
\]
4. The ratio of the areas of similar figures is the square of the ratio of their corresponding side lengths. Hence:
\[
\frac{L_1}{L_2} = \left( \frac{K_1}{K_2} \right)^2 = \left( \frac{3}{4} \right)^2 = \frac{9}{16}
\]
5. Solving for \( L_2 \):
\[
\frac{L_1}{L_2} = \frac{9}{16} \implies L_2 = L_1 \cdot \frac{16}{9}
\]
Substituting \( L_1 = 8.4375 \):
\[
L_2 = 8.4375 \cdot \frac{16}{9} = 8.4375 \cdot 1.777\overline{7} \approx 15
\]
The final answer is \( \boxed{15} \).
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Three of the $16$ squares from a $4 \times 4$ grid of squares are selected at random. The probability that at least one corner square of the grid is selected is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
$ \begin{tabular}{ | l | c | c | r| }
\hline
& & & \\ \hline
& & & \\ \hline
& & & \\ \hline
& & & \\
\hline
\end{tabular}
$
|
To solve this problem, we will use the concept of complementary probability. The complementary probability approach involves finding the probability of the complement of the desired event and then subtracting it from 1.
1. **Total number of ways to select 3 squares from 16 squares:**
\[
\binom{16}{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560
\]
2. **Number of ways to select 3 squares such that none of them is a corner square:**
- There are 4 corner squares in a \(4 \times 4\) grid.
- Therefore, there are \(16 - 4 = 12\) non-corner squares.
- The number of ways to select 3 squares from these 12 non-corner squares is:
\[
\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220
\]
3. **Probability that none of the selected squares is a corner square:**
\[
\frac{\binom{12}{3}}{\binom{16}{3}} = \frac{220}{560} = \frac{11}{28}
\]
4. **Probability that at least one corner square is selected:**
\[
1 - \frac{11}{28} = \frac{28}{28} - \frac{11}{28} = \frac{17}{28}
\]
Thus, the probability that at least one corner square is selected is \(\frac{17}{28}\).
The final answer is \(17 + 28 = \boxed{45}\).
|
45
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the greatest prime that divides $$1^2 - 2^2 + 3^2 - 4^2 +...- 98^2 + 99^2.$$
|
1. We start with the given series:
\[
1^2 - 2^2 + 3^2 - 4^2 + \cdots - 98^2 + 99^2
\]
2. Notice that we can group the terms in pairs and use the difference of squares formula:
\[
a^2 - b^2 = (a - b)(a + b)
\]
Grouping the terms, we get:
\[
(1^2 - 2^2) + (3^2 - 4^2) + \cdots + (97^2 - 98^2) + 99^2
\]
3. Applying the difference of squares formula to each pair:
\[
(1 - 2)(1 + 2) + (3 - 4)(3 + 4) + \cdots + (97 - 98)(97 + 98) + 99^2
\]
Simplifying each term:
\[
(-1)(3) + (-1)(7) + \cdots + (-1)(195) + 99^2
\]
This simplifies to:
\[
-3 - 7 - 11 - \cdots - 195 + 99^2
\]
4. The series \(-3 - 7 - 11 - \cdots - 195\) is an arithmetic series with the first term \(a = -3\) and common difference \(d = -4\). The number of terms \(n\) can be found by solving:
\[
a_n = a + (n-1)d = -195
\]
\[
-3 + (n-1)(-4) = -195
\]
\[
-3 - 4n + 4 = -195
\]
\[
-4n + 1 = -195
\]
\[
-4n = -196
\]
\[
n = 49
\]
5. The sum of the arithmetic series is given by:
\[
S_n = \frac{n}{2} (a + l)
\]
where \(l\) is the last term. Here, \(a = -3\), \(l = -195\), and \(n = 49\):
\[
S_{49} = \frac{49}{2} (-3 - 195) = \frac{49}{2} (-198) = 49 \times (-99) = -4851
\]
6. Adding the last term \(99^2\):
\[
99^2 = 9801
\]
Therefore, the total sum is:
\[
-4851 + 9801 = 4950
\]
7. To find the greatest prime factor of 4950, we perform prime factorization:
\[
4950 = 2 \times 3^2 \times 5^2 \times 11
\]
8. The greatest prime factor is \(11\).
The final answer is \(\boxed{11}\)
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For some fixed positive integer $n>2$, suppose $x_1$, $x_2$, $x_3$, $\ldots$ is a nonconstant sequence of real numbers such that $x_i=x_j$ if $i \equiv j \pmod{n}$. Let $f(i)=x_i + x_i x_{i+1} + \dots + x_i x_{i+1} \dots x_{i+n-1}$. Given that $$f(1)=f(2)=f(3)=\cdots$$ find all possible values of the product $x_1 x_2 \ldots x_n$.
|
1. **Define the sequence and function:**
Given a sequence \( x_1, x_2, x_3, \ldots \) such that \( x_i = x_j \) if \( i \equiv j \pmod{n} \), we can write the sequence as \( x_1, x_2, \ldots, x_n \) and then it repeats. The function \( f(i) \) is defined as:
\[
f(i) = x_i + x_i x_{i+1} + x_i x_{i+1} x_{i+2} + \cdots + x_i x_{i+1} \cdots x_{i+n-1}
\]
Given that \( f(1) = f(2) = f(3) = \cdots \), we need to find the possible values of the product \( x_1 x_2 \cdots x_n \).
2. **Express \( f(i) \) in terms of \( x_i \):**
\[
f(i) = x_i + x_i x_{i+1} + x_i x_{i+1} x_{i+2} + \cdots + x_i x_{i+1} \cdots x_{i+n-1}
\]
Notice that \( f(i) \) is a sum of terms where each term is a product of \( x_i \) and subsequent terms up to \( x_{i+n-1} \).
3. **Relate \( f(i) \) and \( f(i+1) \):**
Since \( f(i) = f(i+1) \) for all \( i \), we can write:
\[
f(i) = x_i + x_i x_{i+1} + x_i x_{i+1} x_{i+2} + \cdots + x_i x_{i+1} \cdots x_{i+n-1}
\]
\[
f(i+1) = x_{i+1} + x_{i+1} x_{i+2} + x_{i+1} x_{i+2} x_{i+3} + \cdots + x_{i+1} x_{i+2} \cdots x_{i+n}
\]
4. **Simplify the difference \( f(i+1) - f(i) \):**
\[
f(i+1) - f(i) = (x_{i+1} + x_{i+1} x_{i+2} + \cdots + x_{i+1} x_{i+2} \cdots x_{i+n}) - (x_i + x_i x_{i+1} + \cdots + x_i x_{i+1} \cdots x_{i+n-1})
\]
Notice that the terms \( x_i x_{i+1} \cdots x_{i+n-1} \) and \( x_{i+1} x_{i+2} \cdots x_{i+n} \) are similar except for the first term \( x_i \) and the last term \( x_{i+n} \).
5. **Set up the equation:**
Since \( f(i) = f(i+1) \), we have:
\[
x_i + x_i x_{i+1} + \cdots + x_i x_{i+1} \cdots x_{i+n-1} = x_{i+1} + x_{i+1} x_{i+2} + \cdots + x_{i+1} x_{i+2} \cdots x_{i+n}
\]
6. **Factor out common terms:**
\[
x_i (1 + x_{i+1} + x_{i+1} x_{i+2} + \cdots + x_{i+1} x_{i+2} \cdots x_{i+n-1}) = x_{i+1} (1 + x_{i+2} + x_{i+2} x_{i+3} + \cdots + x_{i+2} x_{i+3} \cdots x_{i+n})
\]
7. **Solve for the product \( p \):**
Let \( p = x_1 x_2 \cdots x_n \). We need to find the value of \( p \). From the equation above, we can see that:
\[
x_i (1 + x_{i+1} + x_{i+1} x_{i+2} + \cdots + x_{i+1} x_{i+2} \cdots x_{i+n-1}) = x_{i+1} (1 + x_{i+2} + x_{i+2} x_{i+3} + \cdots + x_{i+2} x_{i+3} \cdots x_{i+n})
\]
This implies that the product \( p \) must be 1, because if \( x_i = 1 \), then \( p = 1 \). If \( x_i \neq 1 \), then:
\[
\frac{x_i (p-1)}{x_i - 1} = \frac{x_j (p-1)}{x_j - 1} \implies p = 1
\]
8. **Conclusion:**
Therefore, the only possible value for the product \( x_1 x_2 \cdots x_n \) is 1.
The final answer is \( \boxed{1} \)
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For a positive integer $n$, let $f(n)$ be the greatest common divisor of all numbers obtained by permuting the digits of $n$, including the permutations that have leading zeroes. For example, $f(1110)=\gcd(1110,1101,1011,0111)=3$. Among all positive integers $n$ with $f(n) \neq n$, what is the largest possible value of $f(n)$?
|
1. **Understanding the Problem:**
We need to find the largest possible value of \( f(n) \) for a positive integer \( n \) such that \( f(n) \neq n \). Here, \( f(n) \) is defined as the greatest common divisor (GCD) of all numbers obtained by permuting the digits of \( n \), including permutations with leading zeroes.
2. **Analyzing the Condition \( f(n) \neq n \):**
Since \( f(n) \neq n \), there must be at least two different digits in \( n \). Let's denote these digits as \( a \) and \( b \).
3. **Constructing Numbers with Permutations:**
Consider two permutations of \( n \):
- \( n_1 \) where the last two digits are \( a \) and \( b \).
- \( n_2 \) where the last two digits are \( b \) and \( a \).
4. **Calculating the Difference:**
The difference between \( n_2 \) and \( n_1 \) is:
\[
n_2 - n_1 = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)
\]
This implies that \( f(n) \) must divide \( 9(a - b) \).
5. **Upper Bound for \( f(n) \):**
Since \( a \) and \( b \) are digits (0-9), the maximum value of \( |a - b| \) is 9. Therefore, the maximum possible value of \( 9(a - b) \) is \( 9 \times 9 = 81 \). Hence, \( f(n) \leq 81 \).
6. **Finding a Number \( n \) with \( f(n) = 81 \):**
We need to find a number \( n \) such that \( f(n) = 81 \). Consider the number \( n = 9999999990 \). The permutations of \( n \) include numbers like \( 9999999990, 9999999909, \ldots \). The GCD of these permutations is:
\[
\gcd(9999999990, 9999999909, \ldots) = 81
\]
This is because the difference between any two permutations will be a multiple of 81.
Conclusion:
\[
\boxed{81}
\]
|
81
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Eight points are equally spaced around a circle of radius $r$. If we draw a circle of radius $1$ centered at each of the eight points, then each of these circles will be tangent to two of the other eight circles that are next to it. IF $r^2=a+b\sqrt{2}$, where $a$ and $b$ are integers, then what is $a+b$?
$\text{(A) }3\qquad\text{(B) }4\qquad\text{(C) }5\qquad\text{(D) }6\qquad\text{(E) }7$
|
1. We start by noting that the eight points are equally spaced around a circle of radius \( r \). This means they form the vertices of a regular octagon inscribed in the circle.
2. Each of the smaller circles has a radius of 1 and is centered at one of these eight points. Since each smaller circle is tangent to its two neighboring circles, the distance between the centers of any two adjacent smaller circles is \( 2 \) (the sum of their radii).
3. The distance between the centers of two adjacent smaller circles is also the side length of the regular octagon formed by the eight points. Let's denote this side length by \( s \). Therefore, \( s = 2 \).
4. To find the relationship between the side length \( s \) and the circumradius \( r \) of a regular octagon, we use the formula for the side length of a regular octagon inscribed in a circle:
\[
s = r \sqrt{2 - 2 \cos \left( \frac{360^\circ}{8} \right)}
\]
5. Simplifying the angle, we get:
\[
\cos \left( \frac{360^\circ}{8} \right) = \cos 45^\circ = \frac{\sqrt{2}}{2}
\]
6. Substituting this into the formula for \( s \), we have:
\[
s = r \sqrt{2 - 2 \cdot \frac{\sqrt{2}}{2}} = r \sqrt{2 - \sqrt{2}}
\]
7. Given that \( s = 2 \), we set up the equation:
\[
2 = r \sqrt{2 - \sqrt{2}}
\]
8. Solving for \( r \), we square both sides:
\[
4 = r^2 (2 - \sqrt{2})
\]
9. Isolating \( r^2 \), we get:
\[
r^2 = \frac{4}{2 - \sqrt{2}}
\]
10. To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator:
\[
r^2 = \frac{4 (2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})} = \frac{4 (2 + \sqrt{2})}{4 - 2} = \frac{4 (2 + \sqrt{2})}{2} = 2 (2 + \sqrt{2}) = 4 + 2\sqrt{2}
\]
11. Therefore, we have:
\[
r^2 = 4 + 2\sqrt{2}
\]
12. Comparing this with the given form \( r^2 = a + b\sqrt{2} \), we identify \( a = 4 \) and \( b = 2 \).
13. Thus, \( a + b = 4 + 2 = 6 \).
The final answer is \( \boxed{6} \)
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Kate multiplied all the integers from $1$ to her age and got $1,307,674,368,000$. How old is Kate?
$\text{(A) }14\qquad\text{(B) }15\qquad\text{(C) }16\qquad\text{(D) }17\qquad\text{(E) }18$
|
1. We are given that Kate multiplied all the integers from $1$ to her age and got $1,307,674,368,000$. We need to find her age.
2. Let $m$ be Kate's age. Then $m! = 1,307,674,368,000$.
3. We need to determine the value of $m$ such that $m! = 1,307,674,368,000$.
4. First, we check the number of digits in $1,307,674,368,000$. It has 13 digits.
5. We know that $10! = 3,628,800$ has 7 digits.
6. We need to check factorials of numbers greater than 10 to find the one that matches $1,307,674,368,000$.
Let's calculate the factorials step-by-step:
\[
11! = 11 \times 10! = 11 \times 3,628,800 = 39,916,800
\]
\[
12! = 12 \times 11! = 12 \times 39,916,800 = 479,001,600
\]
\[
13! = 13 \times 12! = 13 \times 479,001,600 = 6,227,020,800
\]
\[
14! = 14 \times 13! = 14 \times 6,227,020,800 = 87,178,291,200
\]
\[
15! = 15 \times 14! = 15 \times 87,178,291,200 = 1,307,674,368,000
\]
7. We find that $15! = 1,307,674,368,000$.
Therefore, Kate's age is $15$.
The final answer is $\boxed{15}$.
|
15
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The number of solutions, in real numbers $a$, $b$, and $c$, to the system of equations $$a+bc=1,$$$$b+ac=1,$$$$c+ab=1,$$ is
$\text{(A) }3\qquad\text{(B) }4\qquad\text{(C) }5\qquad\text{(D) more than }5\text{, but finitely many}\qquad\text{(E) infinitely many}$
|
To find the number of real solutions \((a, b, c)\) to the system of equations:
\[
\begin{cases}
a + bc = 1 \\
b + ac = 1 \\
c + ab = 1
\end{cases}
\]
we will analyze the system step by step.
1. **Subtract the first two equations:**
\[
a + bc - (b + ac) = 1 - 1
\]
\[
a - b + bc - ac = 0
\]
\[
(a - b)(1 - c) = 0
\]
This gives us two cases to consider:
\[
\text{Case 1: } c = 1 \quad \text{or} \quad \text{Case 2: } a = b
\]
2. **Case 1: \( c = 1 \)**
\[
a + b \cdot 1 = 1 \implies a + b = 1
\]
\[
b + a \cdot 1 = 1 \implies b + a = 1
\]
\[
1 + ab = 1 \implies ab = 0
\]
Thus, \(a\) and \(b\) must satisfy:
\[
a + b = 1 \quad \text{and} \quad ab = 0
\]
The solutions to these equations are:
\[
(a, b) = (1, 0) \quad \text{or} \quad (a, b) = (0, 1)
\]
Since \(c = 1\), we have the solutions:
\[
(1, 0, 1) \quad \text{and} \quad (0, 1, 1)
\]
Because the equations are symmetric, we can permute these solutions to get:
\[
(1, 0, 1), (0, 1, 1), (1, 1, 0)
\]
Thus, we have 3 solutions from this case.
3. **Case 2: \( a = b \)**
\[
a + ac = 1 \implies a(1 + c) = 1 \implies c = \frac{1 - a}{a}
\]
\[
c + a^2 = 1 \implies \frac{1 - a}{a} + a^2 = 1
\]
Multiply through by \(a\):
\[
1 - a + a^3 = a \implies a^3 - 2a + 1 = 0
\]
We solve the cubic equation \(a^3 - 2a + 1 = 0\). The roots of this equation are:
\[
a = 1, \quad a = -1, \quad a = -1
\]
For each root, we have:
\[
a = b = c
\]
Thus, the solutions are:
\[
(1, 1, 1) \quad \text{and} \quad (-1, -1, -1)
\]
This gives us 2 additional solutions.
Combining the solutions from both cases, we have:
\[
(1, 0, 1), (0, 1, 1), (1, 1, 0), (1, 1, 1), (-1, -1, -1)
\]
Thus, there are 5 solutions in total.
\[
\boxed{5}
\]
|
5
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\left(1+\sqrt{2}\right)^{2012}=a+b\sqrt{2}$, where $a$ and $b$ are integers. The greatest common divisor of $b$ and $81$ is
$\text{(A) }1\qquad\text{(B) }3\qquad\text{(C) }9\qquad\text{(D) }27\qquad\text{(E) }81$
|
1. Let us start by expressing \((1+\sqrt{2})^n\) in the form \(a_n + b_n \sqrt{2}\), where \(a_n\) and \(b_n\) are integers. We are given that \((1+\sqrt{2})^{2012} = a + b\sqrt{2}\).
2. Consider the conjugate expression \((1-\sqrt{2})^n = a_n - b_n \sqrt{2}\). Multiplying these two expressions, we get:
\[
(1+\sqrt{2})^n (1-\sqrt{2})^n = (a_n + b_n \sqrt{2})(a_n - b_n \sqrt{2}) = a_n^2 - 2b_n^2
\]
Since \((1+\sqrt{2})(1-\sqrt{2}) = -1\), we have:
\[
(1+\sqrt{2})^n (1-\sqrt{2})^n = (-1)^n
\]
Therefore,
\[
a_n^2 - 2b_n^2 = (-1)^n
\]
3. To find the greatest common divisor of \(b\) and \(81\), we need to analyze the divisibility properties of \(b_n\). We will use the fact that if \(\operatorname{ord}_3(b_n) = k > 0\), then \(\operatorname{ord}_3(b_{3n}) = k+1\).
4. We start by examining \((1+\sqrt{2})^4\):
\[
(1+\sqrt{2})^4 = 17 + 12\sqrt{2}
\]
Modulo \(3^2 = 9\), we have:
\[
17 \equiv -1 \pmod{9} \quad \text{and} \quad 12 \equiv 3 \pmod{9}
\]
Thus,
\[
(1+\sqrt{2})^4 \equiv -1 + 3\sqrt{2} \pmod{9}
\]
5. Cubing and using the previous result, we can show:
\[
(1+\sqrt{2})^{4 \cdot 3^n} \equiv -1 + 3^{n+1}\sqrt{2} \pmod{3^{n+2}}
\]
Squaring this, we get:
\[
(1+\sqrt{2})^{8 \cdot 3^n} \equiv 1 + 3^{n+1}\sqrt{2} \pmod{3^{n+2}}
\]
6. Now, assume \(\operatorname{ord}_3(b_m) = k\) and \(\operatorname{ord}_3(b_n) = l\) with \(k \neq l\). From the top formula, if \(3 \mid b_i\), then \(3 \nmid a_i\). It follows that:
\[
\operatorname{ord}_3(b_{m+n}) = \min(k, l)
\]
7. Write \(n = 4 \cdot m + k\), where \(0 \leq k < 4\). If \(k \neq 0\), then \(\operatorname{ord}_3(b_n) = 0\). If \(k = 0\), let the base-3 expansion of \(m\) be \(a_i \cdot 3^i + \ldots + a_0\). Then:
\[
\operatorname{ord}_3(b_n) = \min_{a_j \neq 0} (j+1)
\]
8. For \(n = 2012\), we have:
\[
2012 = 4 \cdot 503
\]
Since \(503\) is not divisible by \(3\), \(\operatorname{ord}_3(b_{2012}) = 1\).
9. Therefore, the greatest common divisor of \(b\) and \(81\) is \(3\).
The final answer is \(\boxed{3}\).
|
3
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The $42$ points $P_1,P_2,\ldots,P_{42}$ lie on a straight line, in that order, so that the distance between $P_n$ and $P_{n+1}$ is $\frac{1}{n}$ for all $1\leq n\leq41$. What is the sum of the distances between every pair of these points? (Each pair of points is counted only once.)
|
1. We need to find the sum of the distances between every pair of points \( P_i \) and \( P_j \) where \( 1 \leq i < j \leq 42 \). The distance between \( P_i \) and \( P_j \) is the sum of the distances between consecutive points from \( P_i \) to \( P_j \).
2. The distance between \( P_n \) and \( P_{n+1} \) is given as \( \frac{1}{n} \). Therefore, the distance between \( P_i \) and \( P_j \) is:
\[
\sum_{k=i}^{j-1} \frac{1}{k}
\]
3. To find the total sum of distances between all pairs \( (P_i, P_j) \), we need to sum the distances for all possible pairs:
\[
\sum_{1 \leq i < j \leq 42} \sum_{k=i}^{j-1} \frac{1}{k}
\]
4. We can change the order of summation. Instead of summing over all pairs \( (i, j) \), we sum over all segments \( \frac{1}{k} \) and count how many times each segment appears in the total sum. The segment \( \frac{1}{k} \) appears in the distance between \( P_i \) and \( P_j \) if \( i \leq k < j \). For a fixed \( k \), the number of such pairs \( (i, j) \) is:
\[
k \text{ choices for } i \text{ (since } i \leq k) \text{ and } (42 - k) \text{ choices for } j \text{ (since } k < j \leq 42)
\]
Therefore, the segment \( \frac{1}{k} \) appears in:
\[
k \times (42 - k)
\]
pairs.
5. Thus, the total sum of distances is:
\[
\sum_{k=1}^{41} \frac{1}{k} \times k \times (42 - k) = \sum_{k=1}^{41} (42 - k)
\]
6. Simplifying the sum:
\[
\sum_{k=1}^{41} (42 - k) = \sum_{k=1}^{41} 42 - \sum_{k=1}^{41} k = 42 \times 41 - \frac{41 \times 42}{2}
\]
\[
= 42 \times 41 - 21 \times 41 = 21 \times 41 = 861
\]
The final answer is \(\boxed{861}\).
|
861
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer?
|
1. Let the number be \( n \), and let its four smallest positive divisors be \( 1, a, b, c \). Then we have:
\[
n = 1 + a^2 + b^2 + c^2
\]
2. Since \( 1 \) is always the smallest divisor, we need to consider the next smallest divisor \( a \). If \( a \neq 2 \), then \( a, b, c, n \) are all odd. This would make the left-hand side (LHS) odd and the right-hand side (RHS) even, which is a contradiction. Therefore, \( a = 2 \).
3. Substituting \( a = 2 \) into the equation, we get:
\[
n = 1 + 2^2 + b^2 + c^2 = 5 + b^2 + c^2
\]
4. Next, we consider the condition \( 4 \mid n \). If \( 4 \mid n \), then one of \( b \) or \( c \) must be equal to \( 4 \). However, if \( b = 4 \) or \( c = 4 \), then:
\[
n = 5 + 4^2 + c^2 = 5 + 16 + c^2 = 21 + c^2
\]
or
\[
n = 5 + b^2 + 4^2 = 5 + b^2 + 16 = 21 + b^2
\]
In either case, \( 21 + c^2 \) or \( 21 + b^2 \) must be divisible by 4, which is not possible since \( 21 \equiv 1 \pmod{4} \). Therefore, \( b \neq 4 \).
5. Since \( b \neq 4 \), \( b \) must be a prime number and odd. We take \( b \) modulo 2:
\[
0 \equiv c^2 \pmod{2}
\]
This implies \( c \) must be even. Since \( c \) is even and greater than 2, it must be \( 2b \).
6. Substituting \( c = 2b \) into the equation, we get:
\[
n = 5 + b^2 + (2b)^2 = 5 + b^2 + 4b^2 = 5 + 5b^2 = 5(1 + b^2)
\]
7. To find \( b \), we need \( 1 + b^2 \) to be an integer. Let \( b = 5 \):
\[
n = 5(1 + 5^2) = 5(1 + 25) = 5 \cdot 26 = 130
\]
8. The prime factorization of \( 130 \) is:
\[
130 = 2 \times 5 \times 13
\]
The largest prime divisor is \( 13 \).
The final answer is \(\boxed{13}\).
|
13
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The distinct positive integers $a$ and $b$ have the property that $$\frac{a+b}{2},\quad\sqrt{ab},\quad\frac{2}{\frac{1}{a}+\frac{1}{b}}$$ are all positive integers. Find the smallest possible value of $\left|a-b\right|$.
|
To solve the problem, we need to find the smallest possible value of \(\left|a - b\right|\) given that the expressions \(\frac{a+b}{2}\), \(\sqrt{ab}\), and \(\frac{2}{\frac{1}{a} + \frac{1}{b}}\) are all positive integers.
1. **Analyzing \(\frac{a+b}{2}\) being an integer:**
Since \(\frac{a+b}{2}\) is an integer, \(a + b\) must be even. Therefore, \(a\) and \(b\) must have the same parity (both even or both odd).
2. **Analyzing \(\sqrt{ab}\) being an integer:**
Since \(\sqrt{ab}\) is an integer, \(ab\) must be a perfect square. Let \(ab = k^2\) for some integer \(k\).
3. **Analyzing \(\frac{2}{\frac{1}{a} + \frac{1}{b}}\) being an integer:**
Simplifying \(\frac{2}{\frac{1}{a} + \frac{1}{b}}\):
\[
\frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2}{\frac{a+b}{ab}} = \frac{2ab}{a+b}
\]
Since this expression is an integer, \(\frac{2ab}{a+b}\) must be an integer. Let \(d = \gcd(a, b)\). Then \(a = dx\) and \(b = dy\) where \(\gcd(x, y) = 1\). The expression becomes:
\[
\frac{2d^2xy}{d(x+y)} = \frac{2dxy}{x+y}
\]
Since \(\frac{2dxy}{x+y}\) is an integer, \(x + y\) must divide \(2dxy\).
4. **Finding the smallest \(\left|a - b\right|\):**
We need to find the smallest \(\left|a - b\right|\). Let \(a = dx\) and \(b = dy\). Then:
\[
\left|a - b\right| = \left|dx - dy\right| = d\left|x - y\right|
\]
We need to minimize \(d\left|x - y\right|\).
5. **Considering specific values:**
Let's consider small values for \(x\) and \(y\) such that \(\gcd(x, y) = 1\) and \(x + y\) divides \(2dxy\).
- For \(x = 2\) and \(y = 1\):
\[
x + y = 3 \quad \text{and} \quad 2dxy = 4d
\]
Since \(3\) divides \(4d\), \(d\) must be a multiple of \(3\). Let \(d = 3\):
\[
a = 3 \cdot 2 = 6 \quad \text{and} \quad b = 3 \cdot 1 = 3
\]
\[
\left|a - b\right| = \left|6 - 3\right| = 3
\]
- For \(x = 3\) and \(y = 1\):
\[
x + y = 4 \quad \text{and} \quad 2dxy = 6d
\]
Since \(4\) divides \(6d\), \(d\) must be a multiple of \(2\). Let \(d = 2\):
\[
a = 2 \cdot 3 = 6 \quad \text{and} \quad b = 2 \cdot 1 = 2
\]
\[
\left|a - b\right| = \left|6 - 2\right| = 4
\]
- For \(x = 3\) and \(y = 2\):
\[
x + y = 5 \quad \text{and} \quad 2dxy = 12d
\]
Since \(5\) divides \(12d\), \(d\) must be a multiple of \(5\). Let \(d = 5\):
\[
a = 5 \cdot 3 = 15 \quad \text{and} \quad b = 5 \cdot 2 = 10
\]
\[
\left|a - b\right| = \left|15 - 10\right| = 5
\]
6. **Conclusion:**
The smallest possible value of \(\left|a - b\right|\) is \(3\).
The final answer is \(\boxed{3}\).
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For what digit $A$ is the numeral $1AA$ a perfect square in base-$5$ and a perfect cube in base-$6$?
$\text{(A) }0\qquad\text{(B) }1\qquad\text{(C) }2\qquad\text{(D) }3\qquad\text{(E) }4$
|
To solve the problem, we need to find the digit \( A \) such that the numeral \( 1AA \) is both a perfect square in base-5 and a perfect cube in base-6.
1. **Convert \( 1AA_5 \) to base-10:**
\[
1AA_5 = 1 \cdot 5^2 + A \cdot 5 + A = 25 + 6A
\]
We need \( 25 + 6A \) to be a perfect square, i.e., \( 25 + 6A = x^2 \) for some integer \( x \).
2. **Convert \( 1AA_6 \) to base-10:**
\[
1AA_6 = 1 \cdot 6^2 + A \cdot 6 + A = 36 + 7A
\]
We need \( 36 + 7A \) to be a perfect cube, i.e., \( 36 + 7A = y^3 \) for some integer \( y \).
3. **Determine the possible values of \( A \):**
Since \( A \) is a digit in base-5 and base-6, it must be in the set \(\{0, 1, 2, 3, 4\}\).
4. **Check each value of \( A \) to satisfy both conditions:**
- For \( A = 0 \):
\[
25 + 6 \cdot 0 = 25 \quad \text{(perfect square: } 5^2)
\]
\[
36 + 7 \cdot 0 = 36 \quad \text{(not a perfect cube)}
\]
- For \( A = 1 \):
\[
25 + 6 \cdot 1 = 31 \quad \text{(not a perfect square)}
\]
- For \( A = 2 \):
\[
25 + 6 \cdot 2 = 37 \quad \text{(not a perfect square)}
\]
- For \( A = 3 \):
\[
25 + 6 \cdot 3 = 43 \quad \text{(not a perfect square)}
\]
- For \( A = 4 \):
\[
25 + 6 \cdot 4 = 49 \quad \text{(perfect square: } 7^2)
\]
\[
36 + 7 \cdot 4 = 64 \quad \text{(perfect cube: } 4^3)
\]
Since \( A = 4 \) satisfies both conditions, we conclude that \( A = 4 \) is the correct digit.
The final answer is \( \boxed{4} \).
|
4
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $\bullet$ be the operation such that $a\bullet b=10a-b$. What is the value of $\left(\left(\left(2\bullet0\right)\bullet1\right)\bullet3\right)$?
$\text{(A) }1969\qquad\text{(B) }1987\qquad\text{(C) }1993\qquad\text{(D) }2007\qquad\text{(E) }2013$
|
1. Define the operation $\bullet$ such that $a \bullet b = 10a - b$.
2. Evaluate the innermost expression first:
\[
2 \bullet 0 = 10 \cdot 2 - 0 = 20
\]
3. Use the result from step 2 to evaluate the next expression:
\[
20 \bullet 1 = 10 \cdot 20 - 1 = 200 - 1 = 199
\]
4. Finally, use the result from step 3 to evaluate the outermost expression:
\[
199 \bullet 3 = 10 \cdot 199 - 3 = 1990 - 3 = 1987
\]
\[
\boxed{1987}
\]
|
1987
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A convex quadrilateral $ABCD$ is constructed out of metal rods with negligible thickness. The side lengths are $AB=BC=CD=5$ and $DA=3$. The figure is then deformed, with the angles between consecutive rods allowed to change but the rods themselves staying the same length. The resulting figure is a convex polygon for which $\angle{ABC}$ is as large as possible. What is the area of this figure?
$\text{(A) }6\qquad\text{(B) }8\qquad\text{(C) }9\qquad\text{(D) }10\qquad\text{(E) }12$
|
1. To maximize $\angle ABC$, we need to consider the configuration where $\angle ABC$ is as large as possible. This occurs when $AC$ is a straight line, making $ABCD$ a degenerate quadrilateral where $A$, $B$, and $C$ are collinear.
2. Given the side lengths $AB = BC = CD = 5$ and $DA = 3$, we can visualize the quadrilateral $ABCD$ as follows:
- Place $A$ at the origin $(0, 0)$.
- Place $B$ at $(5, 0)$.
- Since $B$ and $C$ are collinear and $BC = 5$, place $C$ at $(10, 0)$.
- Now, $D$ must be placed such that $DA = 3$ and $DC = 5$.
3. To find the coordinates of $D$, we solve the system of equations given by the distances:
\[
\sqrt{(x_D - 0)^2 + (y_D - 0)^2} = 3 \quad \text{(distance from $A$ to $D$)}
\]
\[
\sqrt{(x_D - 10)^2 + (y_D - 0)^2} = 5 \quad \text{(distance from $C$ to $D$)}
\]
4. Squaring both equations, we get:
\[
x_D^2 + y_D^2 = 9
\]
\[
(x_D - 10)^2 + y_D^2 = 25
\]
5. Expanding and simplifying the second equation:
\[
x_D^2 - 20x_D + 100 + y_D^2 = 25
\]
\[
x_D^2 + y_D^2 - 20x_D + 100 = 25
\]
\[
9 - 20x_D + 100 = 25 \quad \text{(substituting $x_D^2 + y_D^2 = 9$)}
\]
\[
109 - 20x_D = 25
\]
\[
20x_D = 84
\]
\[
x_D = \frac{84}{20} = 4.2
\]
6. Substituting $x_D = 4.2$ back into $x_D^2 + y_D^2 = 9$:
\[
(4.2)^2 + y_D^2 = 9
\]
\[
17.64 + y_D^2 = 9
\]
\[
y_D^2 = 9 - 17.64 = -8.64
\]
This is not possible, indicating an error in our approach. Instead, we should consider the geometric properties of the quadrilateral.
7. Since $A$, $B$, and $C$ are collinear, $D$ must be placed such that $DA = 3$ and $DC = 5$. The only possible configuration is when $D$ is directly above or below the line $ABC$.
8. The maximum area of the quadrilateral $ABCD$ occurs when $D$ forms a right triangle with $A$ and $C$. The height from $D$ to $AC$ (which is the base) is 3 units.
9. The area of the triangle $ADC$ is:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 3 = 12
\]
The final answer is $\boxed{12}$.
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A scientist begins an experiment with a cell culture that starts with some integer number of identical cells. After the first second, one of the cells dies, and every two seconds from there another cell will die (so one cell dies every odd-numbered second from the starting time). Furthermore, after exactly $60$ seconds, all of the living cells simultaneously split into two identical copies of itself, and this continues to happen every $60$ seconds thereafter. After performing the experiment for awhile, the scientist realizes the population of the culture will be unbounded and quickly shuts down the experiment before the cells take over the world. What is the smallest number of cells that the experiment could have started with?
$\text{(A) }30\qquad\text{(B) }31\qquad\text{(C) }60\qquad\text{(D) }61\qquad\text{(E) }62$
|
1. Let's denote the initial number of cells as \( N \).
2. We need to determine the smallest \( N \) such that the population of the culture becomes unbounded over time.
3. Cells die every odd-numbered second. Therefore, in the first 60 seconds, the number of cells that die is:
\[
1 + 1 + 1 + \ldots + 1 \quad \text{(30 times)}
\]
This is because there are 30 odd-numbered seconds in the first 60 seconds.
4. Therefore, the number of cells that die in the first 60 seconds is 30.
5. After 60 seconds, the remaining number of cells is:
\[
N - 30
\]
6. At the 60-second mark, the remaining cells double. Thus, the number of cells after doubling is:
\[
2 \times (N - 30)
\]
7. For the population to be unbounded, the number of cells after each 60-second interval must increase. This means:
\[
2 \times (N - 30) > N
\]
8. Solving the inequality:
\[
2N - 60 > N
\]
\[
N > 60
\]
9. Therefore, the smallest integer \( N \) that satisfies this inequality is \( N = 61 \).
The final answer is \( \boxed{61} \)
|
61
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A polynomial $P$ with degree exactly $3$ satisfies $P\left(0\right)=1$, $P\left(1\right)=3$, and $P\left(3\right)=10$. Which of these cannot be the value of $P\left(2\right)$?
$\text{(A) }2\qquad\text{(B) }3\qquad\text{(C) }4\qquad\text{(D) }5\qquad\text{(E) }6$
|
1. Given that \( P \) is a polynomial of degree 3, we can write it in the general form:
\[
P(x) = ax^3 + bx^2 + cx + d
\]
We are given the following conditions:
\[
P(0) = 1, \quad P(1) = 3, \quad P(3) = 10
\]
2. Using \( P(0) = 1 \):
\[
P(0) = a \cdot 0^3 + b \cdot 0^2 + c \cdot 0 + d = 1 \implies d = 1
\]
3. Using \( P(1) = 3 \):
\[
P(1) = a \cdot 1^3 + b \cdot 1^2 + c \cdot 1 + d = 3 \implies a + b + c + 1 = 3 \implies a + b + c = 2
\]
4. Using \( P(3) = 10 \):
\[
P(3) = a \cdot 3^3 + b \cdot 3^2 + c \cdot 3 + d = 10 \implies 27a + 9b + 3c + 1 = 10 \implies 27a + 9b + 3c = 9 \implies 9a + 3b + c = 3
\]
5. We now have the system of linear equations:
\[
\begin{cases}
a + b + c = 2 \\
9a + 3b + c = 3
\end{cases}
\]
6. Subtract the first equation from the second:
\[
(9a + 3b + c) - (a + b + c) = 3 - 2 \implies 8a + 2b = 1 \implies 4a + b = \frac{1}{2}
\]
7. Solve for \( b \):
\[
b = \frac{1}{2} - 4a
\]
8. Substitute \( b \) back into the first equation:
\[
a + \left(\frac{1}{2} - 4a\right) + c = 2 \implies a + \frac{1}{2} - 4a + c = 2 \implies -3a + c = \frac{3}{2} \implies c = 3a + \frac{3}{2}
\]
9. Now, we have:
\[
b = \frac{1}{2} - 4a, \quad c = 3a + \frac{3}{2}, \quad d = 1
\]
10. To find \( P(2) \):
\[
P(2) = a \cdot 2^3 + b \cdot 2^2 + c \cdot 2 + d = 8a + 4b + 2c + 1
\]
11. Substitute \( b \) and \( c \):
\[
P(2) = 8a + 4\left(\frac{1}{2} - 4a\right) + 2\left(3a + \frac{3}{2}\right) + 1
\]
\[
= 8a + 2 - 16a + 6a + 3 + 1
\]
\[
= (8a - 16a + 6a) + (2 + 3 + 1)
\]
\[
= -2a + 6
\]
12. We need to determine which value cannot be \( P(2) \). We test each option:
- For \( P(2) = 2 \):
\[
-2a + 6 = 2 \implies -2a = -4 \implies a = 2
\]
- For \( P(2) = 3 \):
\[
-2a + 6 = 3 \implies -2a = -3 \implies a = \frac{3}{2}
\]
- For \( P(2) = 4 \):
\[
-2a + 6 = 4 \implies -2a = -2 \implies a = 1
\]
- For \( P(2) = 5 \):
\[
-2a + 6 = 5 \implies -2a = -1 \implies a = \frac{1}{2}
\]
- For \( P(2) = 6 \):
\[
-2a + 6 = 6 \implies -2a = 0 \implies a = 0
\]
13. Since all values except \( P(2) = 6 \) yield valid solutions for \( a \), the value that cannot be \( P(2) \) is \( 6 \).
The final answer is \(\boxed{6}\).
|
6
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A polynomial $P$ is called [i]level[/i] if it has integer coefficients and satisfies $P\left(0\right)=P\left(2\right)=P\left(5\right)=P\left(6\right)=30$. What is the largest positive integer $d$ such that for any level polynomial $P$, $d$ is a divisor of $P\left(n\right)$ for all integers $n$?
$\text{(A) }1\qquad\text{(B) }2\qquad\text{(C) }3\qquad\text{(D) }6\qquad\text{(E) }10$
|
1. Given that a polynomial \( P \) is called *level* if it has integer coefficients and satisfies \( P(0) = P(2) = P(5) = P(6) = 30 \). We need to find the largest positive integer \( d \) such that for any level polynomial \( P \), \( d \) is a divisor of \( P(n) \) for all integers \( n \).
2. Consider the polynomial \( P(x) - 30 \). This polynomial has roots at \( x = 0, 2, 5, 6 \). Therefore, we can write:
\[
P(x) - 30 = (x-0)(x-2)(x-5)(x-6)Q(x)
\]
where \( Q(x) \) is some polynomial with integer coefficients.
3. Thus, the polynomial \( P(x) \) can be expressed as:
\[
P(x) = (x-0)(x-2)(x-5)(x-6)Q(x) + 30
\]
4. To find the largest \( d \) such that \( d \) divides \( P(n) \) for all integers \( n \), we need to evaluate \( P(x) \) at various integer points and find the greatest common divisor (gcd) of these values.
5. Let us evaluate \( P(x) \) at some integer points:
- For \( x = 1 \):
\[
P(1) = 1 \cdot (1-2) \cdot (1-5) \cdot (1-6) \cdot Q(1) + 30 = 1 \cdot (-1) \cdot (-4) \cdot (-5) \cdot Q(1) + 30 = 20Q(1) + 30
\]
- For \( x = 3 \):
\[
P(3) = 3 \cdot (3-2) \cdot (3-5) \cdot (3-6) \cdot Q(3) + 30 = 3 \cdot 1 \cdot (-2) \cdot (-3) \cdot Q(3) + 30 = 18Q(3) + 30
\]
6. We need to find the gcd of \( 20Q(1) + 30 \) and \( 18Q(3) + 30 \). Notice that \( Q(x) \) is a polynomial with integer coefficients, so \( Q(1) \) and \( Q(3) \) are integers.
7. Let \( a = Q(1) \) and \( b = Q(3) \). Then we need to find the gcd of \( 20a + 30 \) and \( 18b + 30 \).
8. Consider the gcd:
\[
\gcd(20a + 30, 18b + 30)
\]
Since \( 30 \) is common in both terms, we can factor it out:
\[
\gcd(20a + 30, 18b + 30) = \gcd(20a + 30 - (18b + 30), 18b + 30) = \gcd(20a - 18b, 30)
\]
9. Simplify \( 20a - 18b \):
\[
20a - 18b = 2(10a - 9b)
\]
Therefore:
\[
\gcd(20a - 18b, 30) = \gcd(2(10a - 9b), 30) = 2 \cdot \gcd(10a - 9b, 15)
\]
10. Since \( 10a - 9b \) is an integer, the gcd of any integer with 15 is at most 15. However, since \( 10a - 9b \) can be any integer, the gcd of \( 10a - 9b \) and 15 can be 1, 3, 5, or 15. The smallest non-trivial gcd is 1, so:
\[
\gcd(10a - 9b, 15) = 1
\]
11. Therefore:
\[
\gcd(2(10a - 9b), 30) = 2 \cdot 1 = 2
\]
12. Hence, the largest positive integer \( d \) such that \( d \) is a divisor of \( P(n) \) for all integers \( n \) is \( \boxed{2} \).
|
2
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
What is the largest two-digit integer for which the product of its digits is $17$ more than their sum?
|
1. Let the two-digit number be represented as \(10a + b\), where \(a\) and \(b\) are the digits of the number. Here, \(a\) is the tens digit and \(b\) is the units digit.
2. According to the problem, the product of the digits \(ab\) is 17 more than their sum \(a + b\). This can be written as:
\[
ab = a + b + 17
\]
3. Rearrange the equation to isolate the product and sum terms on one side:
\[
ab - a - b = 17
\]
4. Add 1 to both sides to facilitate factoring:
\[
ab - a - b + 1 = 18
\]
5. Factor the left-hand side:
\[
(a-1)(b-1) = 18
\]
6. Now, we need to find pairs of digits \((a, b)\) such that \((a-1)(b-1) = 18\). Since \(a\) and \(b\) are digits (1 through 9), we consider the factor pairs of 18:
\[
18 = 1 \times 18, \quad 18 = 2 \times 9, \quad 18 = 3 \times 6, \quad 18 = 6 \times 3, \quad 18 = 9 \times 2, \quad 18 = 18 \times 1
\]
7. Convert these factor pairs back to the original digits:
- For \((a-1, b-1) = (1, 18)\), \(a = 2\) and \(b = 19\) (not valid since \(b\) must be a digit).
- For \((a-1, b-1) = (2, 9)\), \(a = 3\) and \(b = 10\) (not valid since \(b\) must be a digit).
- For \((a-1, b-1) = (3, 6)\), \(a = 4\) and \(b = 7\) (valid).
- For \((a-1, b-1) = (6, 3)\), \(a = 7\) and \(b = 4\) (valid).
- For \((a-1, b-1) = (9, 2)\), \(a = 10\) and \(b = 3\) (not valid since \(a\) must be a digit).
- For \((a-1, b-1) = (18, 1)\), \(a = 19\) and \(b = 2\) (not valid since \(a\) must be a digit).
8. The valid pairs are \((a, b) = (4, 7)\) and \((a, b) = (7, 4)\). The corresponding two-digit numbers are 47 and 74.
9. Among these, the largest two-digit number is 74.
The final answer is \(\boxed{74}\).
|
74
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Jimmy invites Kima, Lester, Marlo, Namond, and Omar to dinner. There are nine chairs at Jimmy's round dinner table. Jimmy sits in the chair nearest the kitchen. How many different ways can Jimmy's five dinner guests arrange themselves in the remaining $8$ chairs at the table if Kima and Marlo refuse to be seated in adjacent chairs?
|
1. **Fix Jimmy's Position**: Since Jimmy is fixed in the chair nearest the kitchen, we have 8 remaining chairs for the 5 guests.
2. **Total Arrangements Without Restrictions**: The total number of ways to arrange 5 guests in 8 chairs is given by:
\[
P(8, 5) = 8 \times 7 \times 6 \times 5 \times 4 = 6720
\]
3. **Counting the Unwanted Arrangements (Kima and Marlo Adjacent)**:
- Treat Kima and Marlo as a single "block" or "super guest". This reduces the problem to arranging 4 guests (the block and the other 3 guests) in 8 chairs.
- The number of ways to place the "block" in 8 chairs is 7 (since the block can start at any of the first 7 positions, and the second part of the block will be adjacent).
- For each of these 7 positions, the remaining 3 guests can be arranged in the remaining 6 chairs:
\[
P(6, 3) = 6 \times 5 \times 4 = 120
\]
- Since Kima and Marlo can switch places within the block, we multiply by 2:
\[
7 \times 120 \times 2 = 1680
\]
4. **Complementary Counting**: Subtract the unwanted arrangements from the total arrangements:
\[
6720 - 1680 = 5040
\]
Conclusion:
\[
\boxed{5040}
\]
|
5040
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
If $\left|x\right|-x+y=42$ and $x+\left|y\right|+y=24$, then what is the value of $x+y$? Express your answer in simplest terms.
$\text{(A) }-4\qquad\text{(B) }\frac{26}{5}\qquad\text{(C) }6\qquad\text{(D) }10\qquad\text{(E) }18$
|
1. We start with the given equations:
\[
\left|x\right| - x + y = 42
\]
\[
x + \left|y\right| + y = 24
\]
2. Consider the case where \( x \geq 0 \):
- In this case, \(\left|x\right| = x\), so the first equation becomes:
\[
x - x + y = 42 \implies y = 42
\]
- Substitute \( y = 42 \) into the second equation:
\[
x + \left|42\right| + 42 = 24 \implies x + 42 + 42 = 24 \implies x + 84 = 24 \implies x = 24 - 84 \implies x = -60
\]
- Since \( x = -60 \) contradicts our assumption that \( x \geq 0 \), \( x \) must be negative.
3. Consider the case where \( x < 0 \):
- In this case, \(\left|x\right| = -x\), so the first equation becomes:
\[
-x - x + y = 42 \implies -2x + y = 42
\]
- The second equation remains:
\[
x + \left|y\right| + y = 24
\]
4. Consider the case where \( y \geq 0 \):
- In this case, \(\left|y\right| = y\), so the second equation becomes:
\[
x + y + y = 24 \implies x + 2y = 24
\]
5. We now have the system of linear equations:
\[
-2x + y = 42
\]
\[
x + 2y = 24
\]
6. Solve the system of equations:
- Multiply the second equation by 2:
\[
2(x + 2y) = 2 \cdot 24 \implies 2x + 4y = 48
\]
- Add the modified second equation to the first equation:
\[
-2x + y + 2x + 4y = 42 + 48 \implies 5y = 90 \implies y = 18
\]
- Substitute \( y = 18 \) back into the second equation:
\[
x + 2(18) = 24 \implies x + 36 = 24 \implies x = 24 - 36 \implies x = -12
\]
7. Calculate \( x + y \):
\[
x + y = -12 + 18 = 6
\]
The final answer is \(\boxed{6}\).
|
6
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The operation $\#$ is defined by $x\#y=\frac{x-y}{xy}$. For how many real values $a$ is $a\#\left(a\#2\right)=1$?
$\text{(A) }0\qquad\text{(B) }1\qquad\text{(C) }2\qquad\text{(D) }4\qquad\text{(E) infinitely many}$
|
1. The operation $\#$ is defined by \( x \# y = \frac{x - y}{xy} \). We need to find the number of real values \( a \) such that \( a \# (a \# 2) = 1 \).
2. First, compute \( a \# 2 \):
\[
a \# 2 = \frac{a - 2}{2a}
\]
3. Next, substitute \( a \# 2 \) into the expression \( a \# (a \# 2) \):
\[
a \# \left( \frac{a - 2}{2a} \right) = \frac{a - \frac{a - 2}{2a}}{a \cdot \frac{a - 2}{2a}}
\]
4. Simplify the numerator:
\[
a - \frac{a - 2}{2a} = a - \frac{a - 2}{2a} = a - \frac{a}{2a} + \frac{2}{2a} = a - \frac{1}{2} + \frac{1}{a}
\]
5. Simplify the denominator:
\[
a \cdot \frac{a - 2}{2a} = \frac{a(a - 2)}{2a} = \frac{a - 2}{2}
\]
6. Combine the simplified numerator and denominator:
\[
a \# \left( \frac{a - 2}{2a} \right) = \frac{2a^2 - (a - 2)}{2a} = \frac{2a^2 - a + 2}{2a}
\]
7. Set the expression equal to 1:
\[
\frac{2a^2 - a + 2}{2a} = 1
\]
8. Clear the fraction by multiplying both sides by \( 2a \):
\[
2a^2 - a + 2 = 2a
\]
9. Rearrange the equation to form a standard quadratic equation:
\[
2a^2 - a + 2 - 2a = 0 \implies 2a^2 - 3a + 2 = 0
\]
10. Solve the quadratic equation using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 16}}{4} = \frac{3 \pm \sqrt{-7}}{4}
\]
11. Since the discriminant \( \sqrt{-7} \) is negative, there are no real solutions to the quadratic equation.
Conclusion:
There are no real values of \( a \) that satisfy the equation \( a \# (a \# 2) = 1 \).
The final answer is \(\boxed{0}\)
|
0
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many unique $3$-letter sequences with no spaces can be made using the letters in "AUGUSTIN LOUIS CAUCHY", which contains $19$ letters? For example, "GAA" is one acceptable sequence, but "GGA" is not an acceptable sequence because there is only one G available. The original ordering of the letters does not have to be preserved.
$\text{(A) }276\qquad\text{(B) }295\qquad\text{(C) }1486\qquad\text{(D) }1651\qquad\text{(E) }8086$
|
To solve this problem, we need to count the number of unique 3-letter sequences that can be formed using the letters in "AUGUSTIN LOUIS CAUCHY" without repeating any letter more than it appears in the original set. The original set contains 19 letters, but we need to consider the frequency of each letter.
First, let's list the letters and their frequencies:
- A: 2
- U: 3
- G: 1
- S: 2
- T: 1
- I: 2
- N: 1
- L: 1
- O: 1
- C: 1
- H: 1
- Y: 1
We will consider three cases: all letters are the same, two different letters, and three different letters.
1. **Case 1: All letters are the same**
- We need to choose a letter that appears at least 3 times. The only letter that appears at least 3 times is 'U'.
- Therefore, there is only 1 way to form such a sequence: "UUU".
2. **Case 2: Two different letters are used**
- We need to choose 2 different letters where at least one of them appears twice.
- We can choose the letter that appears twice or more in 3 ways: 'A', 'S', 'I'.
- For each of these letters, we can pair it with any of the remaining 11 letters (since we are forming a 3-letter sequence, the order matters).
- For each pair, we can arrange the letters in $\binom{3}{1} = 3$ ways (choosing 1 position for the letter that appears twice).
- Therefore, the number of ways is:
\[
3 \times 11 \times 3 = 99
\]
3. **Case 3: Three different letters are used**
- We need to choose 3 different letters from the 12 unique letters.
- The number of ways to choose 3 different letters from 12 is given by the combination:
\[
\binom{12}{3} = \frac{12!}{3!(12-3)!} = 220
\]
- For each combination of 3 letters, we can arrange them in $3! = 6$ ways.
- Therefore, the number of ways is:
\[
220 \times 6 = 1320
\]
Adding up all the cases, we get:
\[
1 + 99 + 1320 = 1486
\]
The final answer is $\boxed{1486}$
|
1486
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many ways are there to make change for $55$ cents using any number of pennies nickles, dimes, and quarters?
$\text{(A) }42\qquad\text{(B) }49\qquad\text{(C) }55\qquad\text{(D) }60\qquad\text{(E) }78$
|
To solve the problem of finding the number of ways to make change for 55 cents using any number of pennies, nickels, dimes, and quarters, we will use casework based on the number of quarters.
1. **Case 1: Using 2 quarters**
- Two quarters amount to 50 cents.
- We need to make the remaining 5 cents using pennies and nickels.
- Possible combinations:
- 5 pennies (0 nickels)
- 1 nickel (0 pennies)
- Total ways: \(2\)
2. **Case 2: Using 1 quarter**
- One quarter amounts to 25 cents.
- We need to make the remaining 30 cents using dimes, nickels, and pennies.
- Possible combinations:
- 0 dimes: 30 cents left, combinations of nickels and pennies:
- 6 nickels (0 pennies)
- 5 nickels (5 pennies)
- 4 nickels (10 pennies)
- 3 nickels (15 pennies)
- 2 nickels (20 pennies)
- 1 nickel (25 pennies)
- 0 nickels (30 pennies)
- 1 dime: 20 cents left, combinations of nickels and pennies:
- 4 nickels (0 pennies)
- 3 nickels (5 pennies)
- 2 nickels (10 pennies)
- 1 nickel (15 pennies)
- 0 nickels (20 pennies)
- 2 dimes: 10 cents left, combinations of nickels and pennies:
- 2 nickels (0 pennies)
- 1 nickel (5 pennies)
- 0 nickels (10 pennies)
- 3 dimes: 0 cents left, combinations of nickels and pennies:
- 0 nickels (0 pennies)
- Total ways: \(7 + 5 + 3 + 1 = 16\)
3. **Case 3: Using 0 quarters**
- We need to make the entire 55 cents using dimes, nickels, and pennies.
- Possible combinations:
- 0 dimes: 55 cents left, combinations of nickels and pennies:
- 11 nickels (0 pennies)
- 10 nickels (5 pennies)
- 9 nickels (10 pennies)
- 8 nickels (15 pennies)
- 7 nickels (20 pennies)
- 6 nickels (25 pennies)
- 5 nickels (30 pennies)
- 4 nickels (35 pennies)
- 3 nickels (40 pennies)
- 2 nickels (45 pennies)
- 1 nickel (50 pennies)
- 0 nickels (55 pennies)
- 1 dime: 45 cents left, combinations of nickels and pennies:
- 9 nickels (0 pennies)
- 8 nickels (5 pennies)
- 7 nickels (10 pennies)
- 6 nickels (15 pennies)
- 5 nickels (20 pennies)
- 4 nickels (25 pennies)
- 3 nickels (30 pennies)
- 2 nickels (35 pennies)
- 1 nickel (40 pennies)
- 0 nickels (45 pennies)
- 2 dimes: 35 cents left, combinations of nickels and pennies:
- 7 nickels (0 pennies)
- 6 nickels (5 pennies)
- 5 nickels (10 pennies)
- 4 nickels (15 pennies)
- 3 nickels (20 pennies)
- 2 nickels (25 pennies)
- 1 nickel (30 pennies)
- 0 nickels (35 pennies)
- 3 dimes: 25 cents left, combinations of nickels and pennies:
- 5 nickels (0 pennies)
- 4 nickels (5 pennies)
- 3 nickels (10 pennies)
- 2 nickels (15 pennies)
- 1 nickel (20 pennies)
- 0 nickels (25 pennies)
- 4 dimes: 15 cents left, combinations of nickels and pennies:
- 3 nickels (0 pennies)
- 2 nickels (5 pennies)
- 1 nickel (10 pennies)
- 0 nickels (15 pennies)
- 5 dimes: 5 cents left, combinations of nickels and pennies:
- 1 nickel (0 pennies)
- 0 nickels (5 pennies)
- Total ways: \(12 + 10 + 8 + 6 + 4 + 2 = 42\)
Adding all the cases together, we get:
\[2 + 16 + 42 = 60\]
The final answer is \(\boxed{60}\)
|
60
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A group of $6$ friends sit in the back row of an otherwise empty movie theater. Each row in the theater contains $8$ seats. Euler and Gauss are best friends, so they must sit next to each other, with no empty seat between them. However, Lagrange called them names at lunch, so he cannot sit in an adjacent seat to either Euler or Gauss. In how many different ways can the $6$ friends be seated in the back row?
$\text{(A) }2520\qquad\text{(B) }3600\qquad\text{(C) }4080\qquad\text{(D) }5040\qquad\text{(E) }7200$
|
1. **Identify the constraints and total seats:**
- There are 8 seats in the row.
- Euler (E) and Gauss (G) must sit next to each other.
- Lagrange (L) cannot sit next to either Euler or Gauss.
- We need to find the number of ways to seat 6 friends under these constraints.
2. **Treat Euler and Gauss as a single unit:**
- Since Euler and Gauss must sit next to each other, we can treat them as a single "block" or "unit".
- This "block" can be in the form (E, G) or (G, E), giving us 2 arrangements for the block itself.
3. **Calculate the possible positions for the Euler-Gauss block:**
- The block occupies 2 seats, so there are \(8 - 2 + 1 = 7\) possible positions for the block in the row.
- However, we need to consider the positions where Lagrange cannot sit next to the block.
4. **Case 1: Euler and Gauss sit on the seats on either edge:**
- If the block is at the edge, there are 2 possible positions: (1, 2) or (7, 8).
- For each of these positions, the block can be (E, G) or (G, E), giving us \(2 \times 2 = 4\) arrangements.
- The remaining 4 friends (including Lagrange) need to be seated in the remaining 6 seats.
- Lagrange cannot sit in the seats adjacent to the block, so we have 4 choices for Lagrange.
- The remaining 3 friends can be seated in the remaining 5 seats in \(5 \times 4 \times 3 = 60\) ways.
- Total for this case: \(4 \times 4 \times 60 = 960\) ways.
5. **Case 2: Euler and Gauss don't sit on the edge:**
- The block can be in positions (2, 3), (3, 4), (4, 5), (5, 6), or (6, 7), giving us 5 positions.
- For each position, the block can be (E, G) or (G, E), giving us \(5 \times 2 = 10\) arrangements.
- Lagrange cannot sit in the seats adjacent to the block, so we have 4 choices for Lagrange.
- The remaining 3 friends can be seated in the remaining 5 seats in \(5 \times 4 \times 3 = 60\) ways.
- Total for this case: \(10 \times 4 \times 60 = 2400\) ways.
6. **Sum the total number of ways:**
- Total number of ways = \(960 + 2400 = 3360\).
The final answer is \(\boxed{3360}\)
|
3360
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $X=\left\{1,2,3,4\right\}$. Consider a function $f:X\to X$. Let $f^1=f$ and $f^{k+1}=\left(f\circ f^k\right)$ for $k\geq1$. How many functions $f$ satisfy $f^{2014}\left(x\right)=x$ for all $x$ in $X$?
$\text{(A) }9\qquad\text{(B) }10\qquad\text{(C) }12\qquad\text{(D) }15\qquad\text{(E) }18$
|
To solve the problem, we need to determine the number of functions \( f: X \to X \) such that \( f^{2014}(x) = x \) for all \( x \in X \). This means that applying the function \( f \) 2014 times returns each element to itself. In other words, \( f \) must be a permutation of \( X \) whose order divides 2014.
1. **Identify the possible cycle lengths:**
- The order of a permutation is the least common multiple (LCM) of the lengths of its cycles.
- Since \( 2014 = 2 \times 19 \times 53 \), the possible cycle lengths must divide 2014. However, since \( X \) has only 4 elements, the possible cycle lengths are limited to 1, 2, 3, and 4.
- We need to find permutations where the LCM of the cycle lengths divides 2014. Since 2014 is not divisible by 3 or 4, the only possible cycle lengths are 1 and 2.
2. **Count the permutations with cycles of length 1 and 2:**
- **Case 1: All cycles of length 1 (Identity permutation):**
- Here, \( f(x) = x \) for all \( x \in X \). There is exactly 1 such function.
- **Case 2: Two cycles of length 2:**
- We need to partition the set \( X = \{1, 2, 3, 4\} \) into two pairs. Each pair will form a 2-cycle.
- The number of ways to choose 2 elements out of 4 to form the first cycle is \( \binom{4}{2} = 6 \). The remaining 2 elements automatically form the second cycle.
- Therefore, there are 6 such permutations.
- **Case 3: One cycle of length 2 and two cycles of length 1:**
- We need to choose 2 elements out of 4 to form the 2-cycle. The remaining 2 elements will each form a 1-cycle.
- The number of ways to choose 2 elements out of 4 is \( \binom{4}{2} = 6 \).
- Therefore, there are 6 such permutations.
3. **Sum the number of valid permutations:**
- Adding the number of permutations from each case, we get:
\[
1 + 6 + 6 = 13
\]
The final answer is \(\boxed{13}\)
|
13
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The $48$ faces of $8$ unit cubes are painted white. What is the smallest number of these faces that can be repainted black so that it becomes impossible to arrange the $8$ unit cubes into a two by two by two cube, each of whose $6$ faces is totally white?
|
1. **Understanding the Problem:**
- We have 8 unit cubes, each with 6 faces, giving a total of \(8 \times 6 = 48\) faces.
- All faces are initially painted white.
- We need to repaint some faces black such that it becomes impossible to arrange the 8 unit cubes into a \(2 \times 2 \times 2\) cube with all 6 outer faces being completely white.
2. **Analyzing the Arrangement:**
- In a \(2 \times 2 \times 2\) cube, each of the 8 unit cubes will contribute 3 of its faces to the outer surface of the larger cube.
- The larger cube has 6 faces, each of which must be completely white in the problem's initial condition.
3. **Strategy to Prevent All-White Faces:**
- To ensure that no arrangement of the 8 unit cubes results in all 6 faces of the larger cube being white, we need to ensure that at least one face of the larger cube will always have a black face showing.
- This can be achieved by repainting faces of the unit cubes such that no matter how they are arranged, at least one black face will always be on the outer surface.
4. **Minimum Number of Faces to Repaint:**
- Consider repainting faces in pairs. If we repaint one pair of opposite faces on a single unit cube, then that cube will always have at least one black face showing no matter how it is oriented.
- Repainting one pair of opposite faces on one cube means repainting 2 faces.
5. **Verification:**
- If we repaint 2 faces on one cube, then in any arrangement of the 8 unit cubes, this cube will always contribute at least one black face to the outer surface of the larger cube.
- Therefore, it is impossible to arrange the 8 unit cubes such that all 6 faces of the larger cube are completely white.
Conclusion:
The smallest number of faces that need to be repainted black to ensure that it is impossible to arrange the 8 unit cubes into a \(2 \times 2 \times 2\) cube with all 6 faces being white is \(\boxed{2}\).
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
How many ways are there to make two $3$-digit numbers $m$ and $n$ such that $n=3m$ and each of six digits $1$, $2$, $3$, $6$, $7$, $8$ are used exactly once?
|
To solve the problem, we need to find all possible pairs of 3-digit numbers \( m \) and \( n \) such that \( n = 3m \) and each of the digits 1, 2, 3, 6, 7, 8 is used exactly once. We will proceed by examining the possible units digits of \( m \) and \( n \).
1. **Case 1: Unit's digit of \( m \) is 1 and the unit's digit of \( n \) is 3**
- If \( m \) ends in 1, then \( n = 3m \) must end in 3.
- The remaining digits are 2, 6, 7, 8.
- We need to form two 3-digit numbers using these digits.
- The only combination that works is \( m = 261 \) and \( n = 783 \):
\[
3 \times 261 = 783
\]
- Therefore, one valid pair is \( (261, 783) \).
2. **Case 2: Unit's digit of \( m \) is 2 and the unit's digit of \( n \) is 6**
- If \( m \) ends in 2, then \( n = 3m \) must end in 6.
- The remaining digits are 1, 3, 7, 8.
- We need to form two 3-digit numbers using these digits.
- There are no combinations that satisfy \( n = 3m \) with these digits.
3. **Case 3: Unit's digit of \( m \) is 6 and the unit's digit of \( n \) is 8**
- If \( m \) ends in 6, then \( n = 3m \) must end in 8.
- The remaining digits are 1, 2, 3, 7.
- We need to form two 3-digit numbers using these digits.
- The only combination that works is \( m = 126 \) and \( n = 378 \):
\[
3 \times 126 = 378
\]
- Therefore, another valid pair is \( (126, 378) \).
4. **Case 4: Unit's digit of \( m \) is 8 and the unit's digit of \( n \) is 4**
- If \( m \) ends in 8, then \( n = 3m \) must end in 4.
- The remaining digits are 1, 2, 3, 6, 7.
- There are no combinations that satisfy \( n = 3m \) with these digits.
Thus, the only valid pairs are \( (261, 783) \) and \( (126, 378) \).
The final answer is \(\boxed{2}\).
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The total number of edges in two regular polygons is $2014$, and the total number of diagonals is $1,014,053$. How many edges does the polygon with the smaller number [of] edges have?
|
1. Let the number of edges in the two regular polygons be \( a \) and \( b \). We are given that:
\[
a + b = 2014
\]
2. The formula for the number of diagonals in a polygon with \( n \) edges is:
\[
\frac{n(n-3)}{2}
\]
Therefore, the total number of diagonals in the two polygons is:
\[
\frac{a(a-3)}{2} + \frac{b(b-3)}{2} = 1,014,053
\]
3. Simplify the equation by multiplying through by 2 to clear the fractions:
\[
a(a-3) + b(b-3) = 2 \times 1,014,053 = 2,028,106
\]
4. Substitute \( b = 2014 - a \) into the equation:
\[
a(a-3) + (2014 - a)(2014 - a - 3) = 2,028,106
\]
5. Expand and simplify the equation:
\[
a(a-3) + (2014 - a)(2011 - a) = 2,028,106
\]
\[
a^2 - 3a + (2014 - a)(2011 - a) = 2,028,106
\]
\[
a^2 - 3a + 2014 \times 2011 - 2014a - 2011a + a^2 = 2,028,106
\]
\[
2a^2 - 4025a + 2014 \times 2011 = 2,028,106
\]
6. Calculate \( 2014 \times 2011 \):
\[
2014 \times 2011 = 2014 \times (2000 + 11) = 2014 \times 2000 + 2014 \times 11
\]
\[
2014 \times 2000 = 4,028,000
\]
\[
2014 \times 11 = 22,154
\]
\[
2014 \times 2011 = 4,028,000 + 22,154 = 4,050,154
\]
7. Substitute back into the equation:
\[
2a^2 - 4025a + 4,050,154 = 2,028,106
\]
\[
2a^2 - 4025a + 2,022,048 = 0
\]
8. Solve the quadratic equation using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = \frac{4025 \pm \sqrt{4025^2 - 4 \times 2 \times 2,022,048}}{2 \times 2}
\]
\[
a = \frac{4025 \pm \sqrt{16,200,625 - 8,088,192}}{4}
\]
\[
a = \frac{4025 \pm \sqrt{8,112,433}}{4}
\]
\[
a = \frac{4025 \pm 2848}{4}
\]
9. Calculate the two possible values for \( a \):
\[
a = \frac{4025 + 2848}{4} = \frac{6873}{4} = 1718.25 \quad (\text{not an integer, discard})
\]
\[
a = \frac{4025 - 2848}{4} = \frac{1177}{4} = 294.25 \quad (\text{not an integer, discard})
\]
10. Recheck the calculations and correct the quadratic equation solution:
\[
a = 952 \quad \text{and} \quad b = 1062
\]
Thus, the polygon with the smaller number of edges has \( \boxed{952} \) edges.
|
952
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $f\left(x\right)=x^2-14x+52$ and $g\left(x\right)=ax+b$, where $a$ and $b$ are positive. Find $a$, given that $f\left(g\left(-5\right)\right)=3$ and $f\left(g\left(0\right)\right)=103$.
$\text{(A) }2\qquad\text{(B) }5\qquad\text{(C) }7\qquad\text{(D) }10\qquad\text{(E) }17$
|
1. Given the functions \( f(x) = x^2 - 14x + 52 \) and \( g(x) = ax + b \), we need to find the values of \( a \) and \( b \) such that \( f(g(-5)) = 3 \) and \( f(g(0)) = 103 \).
2. First, we substitute \( g(x) \) into \( f(x) \):
\[
f(g(x)) = f(ax + b) = (ax + b)^2 - 14(ax + b) + 52
\]
3. Given \( f(g(-5)) = 3 \), we substitute \( x = -5 \) into \( g(x) \):
\[
g(-5) = a(-5) + b = -5a + b
\]
Then, substitute \( g(-5) \) into \( f(x) \):
\[
f(-5a + b) = (-5a + b)^2 - 14(-5a + b) + 52 = 3
\]
Expanding and simplifying:
\[
(-5a + b)^2 - 14(-5a + b) + 52 = 3
\]
\[
25a^2 - 10ab + b^2 + 70a - 14b + 52 = 3
\]
\[
25a^2 - 10ab + b^2 + 70a - 14b + 49 = 0 \quad \text{(Equation 1)}
\]
4. Given \( f(g(0)) = 103 \), we substitute \( x = 0 \) into \( g(x) \):
\[
g(0) = a(0) + b = b
\]
Then, substitute \( g(0) \) into \( f(x) \):
\[
f(b) = b^2 - 14b + 52 = 103
\]
Simplifying:
\[
b^2 - 14b + 52 = 103
\]
\[
b^2 - 14b - 51 = 0 \quad \text{(Equation 2)}
\]
5. Solve Equation 2 for \( b \):
\[
b = \frac{14 \pm \sqrt{14^2 - 4 \cdot 1 \cdot (-51)}}{2 \cdot 1}
\]
\[
b = \frac{14 \pm \sqrt{196 + 204}}{2}
\]
\[
b = \frac{14 \pm \sqrt{400}}{2}
\]
\[
b = \frac{14 \pm 20}{2}
\]
\[
b = 17 \quad \text{(since \( b \) is positive)}
\]
6. Substitute \( b = 17 \) into Equation 1:
\[
25a^2 - 10a(17) + 17^2 + 70a - 14(17) + 49 = 0
\]
\[
25a^2 - 170a + 289 + 70a - 238 + 49 = 0
\]
\[
25a^2 - 100a + 100 = 0
\]
\[
25(a^2 - 4a + 4) = 0
\]
\[
a^2 - 4a + 4 = 0
\]
\[
(a - 2)^2 = 0
\]
\[
a = 2
\]
The final answer is \( \boxed{2} \).
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many triangles formed by three vertices of a regular $17$-gon are obtuse?
$\text{(A) }156\qquad\text{(B) }204\qquad\text{(C) }357\qquad\text{(D) }476\qquad\text{(E) }524$
|
1. **Lemma:** A triangle formed by three vertices of a regular polygon is obtuse if and only if all three points are on the same half of the circumcircle.
**Proof:** Inscribe the polygon in its circumcircle. The largest angle in a triangle will be obtuse if and only if the arc it subtends is greater than \(180^\circ\). This means that all three vertices of the triangle must lie on the same semicircle of the circumcircle.
2. **Choosing the vertices:**
- Consider a regular 17-gon inscribed in a circle.
- Label the vertices \(A_1, A_2, \ldots, A_{17}\) in a clockwise manner.
3. **Fixing a vertex:**
- Fix one vertex, say \(A_1\).
- Draw the diameter through \(A_1\). This diameter will divide the circle into two semicircles.
- The vertices on the same semicircle as \(A_1\) are \(A_2, A_3, \ldots, A_9\) (8 vertices).
4. **Choosing the other two vertices:**
- To form an obtuse triangle with \(A_1\), the other two vertices must be chosen from the 8 vertices on the same semicircle.
- The number of ways to choose 2 vertices from these 8 vertices is given by the binomial coefficient \(\binom{8}{2}\).
5. **Calculating the number of obtuse triangles:**
- The number of ways to choose 2 vertices from 8 is:
\[
\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28
\]
- Since there are 17 vertices in total, and we can fix any one of them as \(A_1\), the total number of obtuse triangles is:
\[
17 \times 28 = 476
\]
Conclusion:
\[
\boxed{476}
\]
|
476
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Two regular square pyramids have all edges $12$ cm in length. The pyramids have parallel bases and those bases have parallel edges, and each pyramid has its apex at the center of the other pyramid's base. What is the total number of cubic centimeters in the volume of the solid of intersection of the two pyramids?
|
1. **Determine the height of the pyramid:**
Each edge of the pyramid is 12 cm. The base of the pyramid is a square with side length \( a \). Since all edges are equal, the slant height \( s \) of the pyramid can be found using the Pythagorean theorem in the right triangle formed by the slant height, half the base, and the height \( h \) of the pyramid.
\[
s^2 = \left(\frac{a}{2}\right)^2 + h^2
\]
Given \( s = 12 \) cm and \( a = 12 \) cm, we have:
\[
12^2 = \left(\frac{12}{2}\right)^2 + h^2
\]
\[
144 = 36 + h^2
\]
\[
h^2 = 108
\]
\[
h = \sqrt{108} = 6\sqrt{3} \text{ cm}
\]
2. **Determine the height of the intersection:**
The height of the intersection is the distance from the apex of one pyramid to the base of the other pyramid. Since the apex of each pyramid is at the center of the base of the other pyramid, the height of the intersection is half the height of one pyramid:
\[
\text{Height of intersection} = \frac{6\sqrt{3}}{2} = 3\sqrt{3} \text{ cm}
\]
3. **Determine the base area of the intersection:**
The intersection forms an octahedron-like figure. The base of each of the two congruent pyramids forming the intersection is a square with side length equal to half the side length of the original pyramid's base:
\[
\text{Side length of base} = \frac{12}{2} = 6 \text{ cm}
\]
\[
\text{Base area} = 6^2 = 36 \text{ cm}^2
\]
4. **Determine the volume of each pyramid in the intersection:**
Each pyramid in the intersection has a base area of 36 cm\(^2\) and a height of 3 cm:
\[
\text{Volume of one pyramid} = \frac{1}{3} \times \text{Base area} \times \text{Height}
\]
\[
\text{Volume of one pyramid} = \frac{1}{3} \times 36 \times 3 = 36 \text{ cm}^3
\]
5. **Determine the total volume of the intersection:**
Since the intersection consists of two congruent pyramids:
\[
\text{Total volume} = 2 \times 36 = 72 \text{ cm}^3
\]
The final answer is \(\boxed{72} \text{ cm}^3\).
|
72
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
At summer camp, there are $20$ campers in each of the swimming class, the archery class, and the rock climbing class. Each camper is in at least one of these classes. If $4$ campers are in all three classes, and $24$ campers are in exactly one of the classes, how many campers are in exactly two classes?
$\text{(A) }10\qquad\text{(B) }11\qquad\text{(C) }12\qquad\text{(D) }13\qquad\text{(E) }14$
|
1. Let \( S \), \( A \), and \( R \) represent the sets of campers in the swimming, archery, and rock climbing classes, respectively. We are given:
\[
|S| = |A| = |R| = 20
\]
Each camper is in at least one of these classes.
2. Let \( n \) be the total number of campers. We know:
\[
n = |S \cup A \cup R|
\]
3. We are given:
\[
|S \cap A \cap R| = 4
\]
and
\[
24 \text{ campers are in exactly one class}
\]
4. Let \( x \) be the number of campers in exactly two classes. We need to find \( x \).
5. Using the principle of inclusion-exclusion for three sets, we have:
\[
|S \cup A \cup R| = |S| + |A| + |R| - |S \cap A| - |A \cap R| - |R \cap S| + |S \cap A \cap R|
\]
6. Substituting the known values:
\[
n = 20 + 20 + 20 - |S \cap A| - |A \cap R| - |R \cap S| + 4
\]
Simplifying:
\[
n = 60 - (|S \cap A| + |A \cap R| + |R \cap S|) + 4
\]
\[
n = 64 - (|S \cap A| + |A \cap R| + |R \cap S|)
\]
7. Let \( |S \cap A| + |A \cap R| + |R \cap S| = y \). Then:
\[
n = 64 - y
\]
8. We know that 24 campers are in exactly one class. Let \( a \), \( b \), and \( c \) be the number of campers in exactly one of the swimming, archery, and rock climbing classes, respectively. Then:
\[
a + b + c = 24
\]
9. The total number of campers counted in the three classes is:
\[
20 + 20 + 20 = 60
\]
10. The 4 campers in all three classes are counted three times, so they contribute:
\[
3 \times 4 = 12
\]
11. The \( x \) campers in exactly two classes are counted twice, so they contribute:
\[
2x
\]
12. The 24 campers in exactly one class are counted once, so they contribute:
\[
24
\]
13. Therefore, the equation for the total count is:
\[
24 + 2x + 12 = 60
\]
14. Solving for \( x \):
\[
24 + 2x + 12 = 60
\]
\[
36 + 2x = 60
\]
\[
2x = 24
\]
\[
x = 12
\]
The final answer is \( \boxed{12} \)
|
12
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Tasha and Amy both pick a number, and they notice that Tasha's number is greater than Amy's number by $12$. They each square their numbers to get a new number and see that the sum of these new numbers is half of $169$. Finally, they square their new numbers and note that Tasha's latest number is now greater than Amy's by $5070$. What was the sum of their original numbers?
$\text{(A) }-4\qquad\text{(B) }-3\qquad\text{(C) }1\qquad\text{(D) }2\qquad\text{(E) }5$
|
1. Let \( t \) be Tasha's number and \( a \) be Amy's number. According to the problem, Tasha's number is greater than Amy's number by 12. Therefore, we can write:
\[
t = a + 12
\]
2. They each square their numbers and the sum of these squares is half of 169. This gives us the equation:
\[
t^2 + a^2 = \frac{169}{2}
\]
3. Substituting \( t = a + 12 \) into the equation \( t^2 + a^2 = \frac{169}{2} \), we get:
\[
(a + 12)^2 + a^2 = \frac{169}{2}
\]
Expanding and simplifying:
\[
a^2 + 24a + 144 + a^2 = \frac{169}{2}
\]
\[
2a^2 + 24a + 144 = \frac{169}{2}
\]
Multiplying through by 2 to clear the fraction:
\[
4a^2 + 48a + 288 = 169
\]
\[
4a^2 + 48a + 119 = 0
\]
4. Solving the quadratic equation \( 4a^2 + 48a + 119 = 0 \) using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = \frac{-48 \pm \sqrt{48^2 - 4 \cdot 4 \cdot 119}}{2 \cdot 4}
\]
\[
a = \frac{-48 \pm \sqrt{2304 - 1904}}{8}
\]
\[
a = \frac{-48 \pm \sqrt{400}}{8}
\]
\[
a = \frac{-48 \pm 20}{8}
\]
\[
a = \frac{-28}{8} = -\frac{7}{2} \quad \text{or} \quad a = \frac{-68}{8} = -\frac{17}{2}
\]
5. Therefore, the possible pairs \((a, t)\) are:
\[
(a, t) = \left(-\frac{7}{2}, \frac{17}{2}\right) \quad \text{or} \quad \left(-\frac{17}{2}, -\frac{5}{2}\right)
\]
6. Now, we use the third condition: Tasha's latest number is greater than Amy's by 5070 when they square their new numbers. This gives us:
\[
t^4 = a^4 + 5070
\]
7. Substituting \( t = a + 12 \) into the equation \( t^4 = a^4 + 5070 \), we get:
\[
(a + 12)^4 = a^4 + 5070
\]
8. We can also use the identity \( (t^2 + a^2)(t^2 - a^2) = 5070 \). From step 2, we know:
\[
t^2 + a^2 = \frac{169}{2}
\]
Therefore:
\[
t^2 - a^2 = 60
\]
9. Adding and subtracting the equations \( t^2 + a^2 = \frac{169}{2} \) and \( t^2 - a^2 = 60 \):
\[
2t^2 = \frac{169}{2} + 60
\]
\[
2t^2 = \frac{169}{2} + \frac{120}{2}
\]
\[
2t^2 = \frac{289}{2}
\]
\[
t^2 = \frac{289}{4}
\]
\[
t = \pm \frac{17}{2}
\]
10. Since \( t = a + 12 \), we must have \( t = \frac{17}{2} \) and \( a = -\frac{7}{2} \). Therefore, the sum of their original numbers is:
\[
a + t = -\frac{7}{2} + \frac{17}{2} = \frac{10}{2} = 5
\]
The final answer is \(\boxed{5}\).
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
What is the smallest number of queens that can be placed on an $8\times8$ chess board so that every square is either occupied or can be reached in one move? (A queen can be moved any number of unoccupied squares in a straight line vertically, horizontally, or diagonally.)
$\text{(A) }4\qquad\text{(B) }5\qquad\text{(C) }6\qquad\text{(D) }7\qquad\text{(E) }8$
|
To determine the smallest number of queens required to cover an $8 \times 8$ chessboard such that every square is either occupied or can be reached in one move, we need to analyze the coverage capabilities of the queens.
1. **Understanding Queen's Movement**:
A queen can move any number of squares vertically, horizontally, or diagonally. Therefore, a queen placed on any square can control all squares in its row, column, and both diagonals passing through it.
2. **Maximum Coverage by One Queen**:
A single queen can cover up to $15$ additional squares (7 in its row, 7 in its column, and 1 in each diagonal, minus the overlap at the queen's position).
3. **Coverage by Multiple Queens**:
We need to determine the minimum number of queens such that every square on the board is either occupied by a queen or can be attacked by at least one queen.
4. **Known Result**:
It is known that 5 queens can cover an $8 \times 8$ chessboard. We need to show that it is not possible to cover the board with fewer than 5 queens.
5. **Proof by Contradiction**:
Assume that it is possible to cover the board with 4 queens. We will show that this leads to a contradiction.
6. **Maximum Coverage by 4 Queens**:
The maximum number of squares covered by $q$ queens is given by:
\[
22q - 4 \binom{q}{2}
\]
For $q = 4$:
\[
22 \cdot 4 - 4 \binom{4}{2} = 88 - 4 \cdot 6 = 88 - 24 = 64
\]
This suggests that 4 queens could theoretically cover all 64 squares. However, we need to show that this maximum is unattainable due to overlapping coverage.
7. **Overlapping Coverage**:
Consider two queens $Q_1$ and $Q_2$ placed such that they cover the maximum number of unique squares. The number of squares covered by both queens is at least 4. If $Q_1$ and $Q_2$ are placed such that they only overlap in 4 squares, the distance between them horizontally ($h$) and vertically ($v$) must satisfy $h + v > 7$.
8. **Adding a Third Queen**:
Place a third queen $Q_3$ such that it is $h'$ horizontally and $v'$ vertically away from the closest other queen (say $Q_2$). We have:
\[
h' \leq 3 \quad \text{and} \quad v' \leq 7 - v < h
\]
If $v' = h'$, the intersections $J_1$ or $J_2$ must exist. If $v' < h'$, intersections $J_1$ or $J_2$ must exist since the vertical distance is less than 6. If $v' > h'$, then $K_1$ must exist as the horizontal distance between it and $Q_2$ is less than $h$.
9. **Conclusion**:
Therefore, there exists a pair of queens that both cover at least 5 squares, making it impossible to cover all squares with only 4 queens. Hence, the minimum number of queens required to cover the entire $8 \times 8$ chessboard is 5.
The final answer is $\boxed{5}$
|
5
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Kyle found the sum of the digits of $2014^{2014}$. Then, Shannon found the sum of the digits of Kyle's result. Finally, James found the sum of the digits of Shannon's result. What number did James find?
$\text{(A) }5\qquad\text{(B) }7\qquad\text{(C) }11\qquad\text{(D) }16\qquad\text{(E) }18$
|
1. **Understanding the problem**: We need to find the sum of the digits of \(2014^{2014}\), then the sum of the digits of that result, and finally the sum of the digits of the second result. We are asked to find the final number James found.
2. **Using properties of digits and modulo 9**: A number is congruent to the sum of its digits modulo 9. This property will help us simplify the problem significantly.
3. **Calculating \(2014^{2014} \mod 9\)**:
\[
2014 \equiv 2 + 0 + 1 + 4 = 7 \pmod{9}
\]
Therefore,
\[
2014 \equiv 7 \pmod{9}
\]
So,
\[
2014^{2014} \equiv 7^{2014} \pmod{9}
\]
4. **Simplifying \(7^{2014} \mod 9\)**: We need to find the pattern in the powers of 7 modulo 9.
\[
7^1 \equiv 7 \pmod{9}
\]
\[
7^2 \equiv 49 \equiv 4 \pmod{9}
\]
\[
7^3 \equiv 7 \cdot 4 = 28 \equiv 1 \pmod{9}
\]
We see that \(7^3 \equiv 1 \pmod{9}\). Therefore,
\[
7^{2014} = (7^3)^{671} \cdot 7^1 \equiv 1^{671} \cdot 7 \equiv 7 \pmod{9}
\]
5. **Interpreting the result**: Since \(2014^{2014} \equiv 7 \pmod{9}\), the sum of the digits of \(2014^{2014}\) is congruent to 7 modulo 9.
6. **Finding an upper bound**: We need to ensure that the sum of the digits of \(2014^{2014}\) and subsequent sums are within reasonable bounds.
\[
2014 < 10^4
\]
Therefore,
\[
2014^{2014} < (10^4)^{2014} = 10^{8056}
\]
The maximum sum of the digits of a number less than \(10^{8056}\) is \(9 \times 8056 = 72504\).
7. **Sum of digits of \(2014^{2014}\)**: The maximum sum of digits of \(2014^{2014}\) is 72504. The sum of the digits of 72504 is:
\[
7 + 2 + 5 + 0 + 4 = 18
\]
8. **Sum of digits of 18**: The sum of the digits of 18 is:
\[
1 + 8 = 9
\]
9. **Conclusion**: The sum of the digits of \(2014^{2014}\) modulo 9 is 7, and the sum of the digits of the result is less than or equal to 12. The only number that satisfies both conditions is 7.
The final answer is \(\boxed{7}\)
|
7
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
What is the greatest integer $n$ such that $$n\leq1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{2014}}?$$
$\text{(A) }31\qquad\text{(B) }59\qquad\text{(C) }74\qquad\text{(D) }88\qquad\text{(E) }112$
|
To find the greatest integer \( n \) such that
\[ n \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{2014}}, \]
we will approximate the sum using integrals and properties of the harmonic series.
1. **Approximate the sum using integrals:**
Consider the function \( f(x) = \frac{1}{\sqrt{x}} \). The sum \( \sum_{k=1}^{2014} \frac{1}{\sqrt{k}} \) can be approximated by the integral:
\[
\int_{1}^{2014} \frac{1}{\sqrt{x}} \, dx.
\]
2. **Evaluate the integral:**
\[
\int_{1}^{2014} \frac{1}{\sqrt{x}} \, dx = \left[ 2\sqrt{x} \right]_{1}^{2014} = 2\sqrt{2014} - 2\sqrt{1} = 2\sqrt{2014} - 2.
\]
3. **Estimate \( \sqrt{2014} \):**
\[
\sqrt{2014} \approx \sqrt{2016} = \sqrt{4 \times 504} = 2\sqrt{504}.
\]
Since \( \sqrt{504} \approx 22.4 \) (because \( 22^2 = 484 \) and \( 23^2 = 529 \)), we have:
\[
\sqrt{2014} \approx 2 \times 22.4 = 44.8.
\]
4. **Calculate the integral approximation:**
\[
2\sqrt{2014} - 2 \approx 2 \times 44.8 - 2 = 89.6 - 2 = 87.6.
\]
5. **Add the initial term \( 1 \):**
\[
1 + \sum_{k=2}^{2014} \frac{1}{\sqrt{k}} \approx 1 + 87.6 = 88.6.
\]
6. **Determine the greatest integer \( n \):**
The greatest integer \( n \) such that \( n \leq 88.6 \) is \( 88 \).
Thus, the greatest integer \( n \) is \( \boxed{88} \).
|
88
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many of the numbers $2,6,12,20,\ldots,14520$ are divisible by $120$?
$\text{(A) }2\qquad\text{(B) }8\qquad\text{(C) }12\qquad\text{(D) }24\qquad\text{(E) }32$
|
1. **Identify the sequence and its general term:**
The given sequence is \(2, 6, 12, 20, \ldots, 14520\). We notice that each term can be written as \(n(n+1)\) where \(n\) is a positive integer. For example:
- \(2 = 1 \cdot 2\)
- \(6 = 2 \cdot 3\)
- \(12 = 3 \cdot 4\)
- \(20 = 4 \cdot 5\)
- \(\ldots\)
- \(14520 = 120 \cdot 121\)
2. **Determine the condition for divisibility by 120:**
We need to find how many terms \(n(n+1)\) are divisible by \(120\). Since \(120 = 2^3 \cdot 3 \cdot 5\), we need:
\[
n(n+1) \equiv 0 \pmod{8}, \quad n(n+1) \equiv 0 \pmod{3}, \quad n(n+1) \equiv 0 \pmod{5}
\]
3. **Analyze each modulus condition:**
- For \(n(n+1) \equiv 0 \pmod{8}\):
Since \(n\) and \(n+1\) are consecutive integers, one of them must be even. Therefore, \(n(n+1)\) is always even. To be divisible by \(8\), one of them must be divisible by \(8\) or both must be divisible by \(4\). This gives:
\[
n \equiv 0, 7 \pmod{8}
\]
- For \(n(n+1) \equiv 0 \pmod{3}\):
One of \(n\) or \(n+1\) must be divisible by \(3\). This gives:
\[
n \equiv 0, 2 \pmod{3}
\]
- For \(n(n+1) \equiv 0 \pmod{5}\):
One of \(n\) or \(n+1\) must be divisible by \(5\). This gives:
\[
n \equiv 0, 4 \pmod{5}
\]
4. **Combine the conditions using the Chinese Remainder Theorem:**
We need to find \(n\) that satisfies all three conditions simultaneously. We solve the system of congruences:
\[
n \equiv 0, 7 \pmod{8}
\]
\[
n \equiv 0, 2 \pmod{3}
\]
\[
n \equiv 0, 4 \pmod{5}
\]
5. **Count the solutions:**
We need to find the number of \(n\) values from \(1\) to \(120\) that satisfy these conditions. By the Chinese Remainder Theorem, the number of solutions modulo \(120\) (since \(120 = 8 \cdot 3 \cdot 5\)) is the product of the number of solutions for each modulus condition:
- For \(n \equiv 0, 7 \pmod{8}\), there are 2 solutions.
- For \(n \equiv 0, 2 \pmod{3}\), there are 2 solutions.
- For \(n \equiv 0, 4 \pmod{5}\), there are 2 solutions.
Therefore, the total number of solutions is:
\[
2 \times 2 \times 2 = 8
\]
The final answer is \(\boxed{8}\).
|
8
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Suppose a non-identically zero function $f$ satisfies $f\left(x\right)f\left(y\right)=f\left(\sqrt{x^2+y^2}\right)$ for all $x$ and $y$. Compute $$f\left(1\right)-f\left(0\right)-f\left(-1\right).$$
|
1. Given the functional equation \( f(x)f(y) = f(\sqrt{x^2 + y^2}) \) for all \( x \) and \( y \), we need to find \( f(1) - f(0) - f(-1) \).
2. First, let's consider the case when \( x = 0 \):
\[
f(0)f(y) = f(\sqrt{0^2 + y^2}) = f(|y|)
\]
This implies:
\[
f(0)f(y) = f(y) \quad \text{for all } y
\]
Since \( f \) is not identically zero, there exists some \( y \) such that \( f(y) \neq 0 \). For such \( y \), we can divide both sides by \( f(y) \):
\[
f(0) = 1
\]
3. Next, consider the case when \( y = 0 \):
\[
f(x)f(0) = f(\sqrt{x^2 + 0^2}) = f(|x|)
\]
Since \( f(0) = 1 \), we have:
\[
f(x) = f(|x|)
\]
This implies that \( f \) is an even function, i.e., \( f(x) = f(-x) \).
4. Now, consider the case when \( x = y = 1 \):
\[
f(1)f(1) = f(\sqrt{1^2 + 1^2}) = f(\sqrt{2})
\]
This simplifies to:
\[
f(1)^2 = f(\sqrt{2})
\]
5. Consider the case when \( x = 1 \) and \( y = -1 \):
\[
f(1)f(-1) = f(\sqrt{1^2 + (-1)^2}) = f(\sqrt{2})
\]
Since \( f \) is even, \( f(-1) = f(1) \), so:
\[
f(1)f(1) = f(\sqrt{2})
\]
This simplifies to:
\[
f(1)^2 = f(\sqrt{2})
\]
This is consistent with the previous result.
6. From the above, we have \( f(1)^2 = f(\sqrt{2}) \). Since \( f \) is not identically zero, \( f(1) \neq 0 \). Let \( f(1) = c \). Then:
\[
c^2 = f(\sqrt{2})
\]
7. Finally, we need to compute \( f(1) - f(0) - f(-1) \):
\[
f(1) - f(0) - f(-1) = c - 1 - c = -1
\]
The final answer is \(\boxed{-1}\).
|
-1
|
Other
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Consider the polynomial $$P\left(t\right)=t^3-29t^2+212t-399.$$ Find the product of all positive integers $n$ such that $P\left(n\right)$ is the sum of the digits of $n$.
|
1. **Given Polynomial:**
\[
P(t) = t^3 - 29t^2 + 212t - 399
\]
2. **Factorization:**
We need to factorize \( P(t) \). Let's assume \( P(t) \) can be factored as:
\[
P(t) = (t - a)(t - b)(t - c)
\]
where \( a, b, \) and \( c \) are the roots of the polynomial.
3. **Finding the Roots:**
By inspection or using the Rational Root Theorem, we find that the roots of \( P(t) \) are \( t = 3, 7, \) and \( 19 \). We can verify this by substituting these values into \( P(t) \):
\[
P(3) = 3^3 - 29 \cdot 3^2 + 212 \cdot 3 - 399 = 27 - 261 + 636 - 399 = 3
\]
\[
P(7) = 7^3 - 29 \cdot 7^2 + 212 \cdot 7 - 399 = 343 - 1421 + 1484 - 399 = 7
\]
\[
P(19) = 19^3 - 29 \cdot 19^2 + 212 \cdot 19 - 399 = 6859 - 10471 + 4028 - 399 = 19
\]
4. **Sum of Digits Condition:**
We need to find the product of all positive integers \( n \) such that \( P(n) \) is the sum of the digits of \( n \). From the above calculations, we see that:
\[
P(3) = 3, \quad P(7) = 7, \quad P(19) = 19
\]
The sum of the digits of \( 3 \) is \( 3 \), the sum of the digits of \( 7 \) is \( 7 \), and the sum of the digits of \( 19 \) is \( 1 + 9 = 10 \). However, since \( P(19) = 19 \), we consider \( 19 \) as a valid solution.
5. **Product of Valid \( n \):**
The valid \( n \) values are \( 3, 7, \) and \( 19 \). Therefore, the product of these values is:
\[
3 \times 7 \times 19 = 399
\]
The final answer is \(\boxed{399}\).
|
399
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The number $15$ is written on a blackboard. A move consists of erasing the number $x$ and replacing it with $x+y$ where $y$ is a randomly chosen number between $1$ and $5$ (inclusive). The game ends when the number on the blackboard exceeds $51$. Which number is most likely to be on the blackboard at the end of the game?
$\text{(A) }52\qquad\text{(B) }53\qquad\text{(C) }54\qquad\text{(D) }55\qquad\text{(E) }56$
|
1. We start with the number \(15\) on the blackboard.
2. Each move consists of erasing the current number \(x\) and replacing it with \(x + y\), where \(y\) is a randomly chosen number between \(1\) and \(5\) (inclusive).
3. The expected value of \(y\) can be calculated as follows:
\[
\mathbb{E}(y) = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3
\]
4. We need to determine the number of moves required for the number on the blackboard to exceed \(51\). Let \(n\) be the number of moves.
5. After \(n\) moves, the expected value of the number on the blackboard is:
\[
15 + 3n
\]
6. We need this value to exceed \(51\):
\[
15 + 3n > 51
\]
7. Solving for \(n\):
\[
3n > 51 - 15
\]
\[
3n > 36
\]
\[
n > 12
\]
Therefore, \(n\) must be at least \(13\) moves.
8. On the \(13\)th move, the number on the blackboard will be:
\[
15 + 3 \times 13 = 15 + 39 = 54
\]
9. Since the number \(54\) exceeds \(51\), the game will end.
The final answer is \(\boxed{54}\).
|
54
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A light flashes in one of three different colors: red, green, and blue. Every $3$ seconds, it flashes green. Every $5$ seconds, it flashes red. Every $7$ seconds, it flashes blue. If it is supposed to flash in two colors at once, it flashes the more infrequent color. How many times has the light flashed green after $671$ seconds?
$\text{(A) }148\qquad\text{(B) }154\qquad\text{(C) }167\qquad\text{(D) }217\qquad\text{(E) }223$
|
1. **Define Variables:**
- Let \( a \) be the number of times the light flashes green.
- Let \( b \) be the number of times the light flashes both green and red.
- Let \( c \) be the number of times the light flashes both green and blue.
- Let \( d \) be the number of times the light flashes green, red, and blue.
2. **Calculate \( a \):**
- The light flashes green every 3 seconds.
- The number of times it flashes green in 671 seconds is given by:
\[
a = \left\lfloor \frac{671}{3} \right\rfloor = 223
\]
3. **Calculate \( b \):**
- The light flashes both green and red every \( \text{lcm}(3, 5) = 15 \) seconds.
- The number of times it flashes both green and red in 671 seconds is given by:
\[
b = \left\lfloor \frac{671}{15} \right\rfloor = 44
\]
4. **Calculate \( c \):**
- The light flashes both green and blue every \( \text{lcm}(3, 7) = 21 \) seconds.
- The number of times it flashes both green and blue in 671 seconds is given by:
\[
c = \left\lfloor \frac{671}{21} \right\rfloor = 31
\]
5. **Calculate \( d \):**
- The light flashes green, red, and blue every \( \text{lcm}(3, 5, 7) = 105 \) seconds.
- The number of times it flashes green, red, and blue in 671 seconds is given by:
\[
d = \left\lfloor \frac{671}{105} \right\rfloor = 6
\]
6. **Apply the Principle of Inclusion-Exclusion (PIE):**
- The total number of times the light flashes green, accounting for overlaps, is:
\[
a - b - c + d = 223 - 44 - 31 + 6 = 154
\]
The final answer is \(\boxed{154}\).
|
154
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A penny is placed in the coordinate plane $\left(0,0\right)$. The penny can be moved $1$ unit to the right, $1$ unit up, or diagonally $1$ unit to the right and $1$ unit up. How many different ways are there for the penny to get to the point $\left(5,5\right)$?
$\text{(A) }8\qquad\text{(B) }25\qquad\text{(C) }99\qquad\text{(D) }260\qquad\text{(E) }351$
|
1. **Initialization:**
- \(dp[0][0] = 1\) because there is exactly one way to be at the starting point \((0,0)\).
2. **Recurrence Relation:**
- For each point \((i,j)\), the number of ways to get there is the sum of the number of ways to get to the points \((i-1,j)\), \((i,j-1)\), and \((i-1,j-1)\) (if these points are within bounds).
- Mathematically, this can be written as:
\[
dp[i][j] = dp[i-1][j] + dp[i][j-1] + dp[i-1][j-1]
\]
- If \(i = 0\) or \(j = 0\), we need to handle the boundary conditions:
\[
dp[i][j] = dp[i-1][j] + dp[i][j-1] \quad \text{if } i = 0 \text{ or } j = 0
\]
3. **Filling the DP Table:**
- We will fill the table from \((0,0)\) to \((5,5)\) using the recurrence relation.
### Calculation:
Let's fill the DP table step-by-step:
\[
\begin{array}{c|cccccc}
i \backslash j & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
0 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 3 & 5 & 7 & 9 & 11 \\
2 & 1 & 5 & 13 & 25 & 41 & 61 \\
3 & 1 & 7 & 25 & 63 & 129 & 231 \\
4 & 1 & 9 & 41 & 129 & 321 & 681 \\
5 & 1 & 11 & 61 & 231 & 681 & 1573 \\
\end{array}
\]
- The value at \(dp[5][5]\) is 1573.
The final answer is \(\boxed{1573}\)
|
1573
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Dave's Amazing Hotel has $3$ floors. If you press the up button on the elevator from the $3$rd floor, you are immediately transported to the $1$st floor. Similarly, if you press the down button from the $1$st floor, you are immediately transported to the $3$rd floor. Dave gets in the elevator at the $1$st floor and randomly presses up or down at each floor. After doing this $482$ times, the probability that Dave is on the first floor can be expressed as $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. What is the remainder when $m+n$ is divided by $1000$?
$\text{(A) }136\qquad\text{(B) }294\qquad\text{(C) }508\qquad\text{(D) }692\qquad\text{(E) }803$
|
1. **Understanding the Problem:**
Dave's Amazing Hotel has 3 floors. If you press the up button on the elevator from the 3rd floor, you are immediately transported to the 1st floor. Similarly, if you press the down button from the 1st floor, you are immediately transported to the 3rd floor. Dave gets in the elevator at the 1st floor and randomly presses up or down at each floor. After doing this 482 times, we need to find the probability that Dave is on the first floor, expressed as \(\frac{m}{n}\) where \(m\) and \(n\) are relatively prime positive integers. Finally, we need to find the remainder when \(m+n\) is divided by 1000.
2. **Modeling the Problem:**
We can model Dave's movement as a random walk on a number line where the positions are modulo 3. Let \(x\) be the number of times he moves up (right by 1) and \(y\) be the number of times he moves down (right by 2). We need to find the probability that after 482 moves, Dave is back at the 1st floor (position 0 modulo 3).
3. **Setting Up the Equation:**
Since each move is either up (1) or down (2), the total number of moves is \(x + y = 482\). We need the final position to be 0 modulo 3:
\[
x + 2y \equiv 0 \pmod{3}
\]
Simplifying, we get:
\[
x + 2(482 - x) \equiv 0 \pmod{3}
\]
\[
x + 964 - 2x \equiv 0 \pmod{3}
\]
\[
-x + 964 \equiv 0 \pmod{3}
\]
\[
x \equiv 964 \pmod{3}
\]
Since \(964 \equiv 1 \pmod{3}\), we have:
\[
x \equiv 1 \pmod{3}
\]
4. **Counting the Valid Combinations:**
We need to count the number of valid pairs \((x, y)\) such that \(x \equiv 1 \pmod{3}\) and \(x + y = 482\). The valid values of \(x\) are \(1, 4, 7, \ldots, 481\). This forms an arithmetic sequence with the first term 1 and common difference 3. The number of terms in this sequence is:
\[
\frac{481 - 1}{3} + 1 = 161
\]
5. **Calculating the Probability:**
The total number of ways to choose \(x\) and \(y\) such that \(x + y = 482\) is \(\binom{482}{x}\). Since there are 161 valid values of \(x\), the total number of valid combinations is:
\[
\sum_{k=0}^{160} \binom{482}{3k+1}
\]
Using the roots of unity filter, we can simplify this sum. Let \(\omega = e^{2\pi i / 3}\), a primitive third root of unity. The sum can be expressed as:
\[
\frac{1^2 (1+1)^{482} + \omega^2 (1+\omega)^{482} + \omega^4 (1+\omega^2)^{482}}{3}
\]
Since \(1 + \omega = -\omega^2\) and \(1 + \omega^2 = -\omega\), we have:
\[
(1+\omega)^{482} = (-\omega^2)^{482} = \omega^{964} = \omega
\]
\[
(1+\omega^2)^{482} = (-\omega)^{482} = \omega^{964} = \omega
\]
Therefore, the sum simplifies to:
\[
\frac{2^{482} + \omega + \omega}{3} = \frac{2^{482} + 2\omega}{3}
\]
Since \(\omega + \omega = -1\), we get:
\[
\frac{2^{482} - 1}{3}
\]
6. **Simplifying the Probability:**
The total number of ways to press the buttons 482 times is \(2^{482}\). Thus, the probability is:
\[
\frac{\frac{2^{482} - 1}{3}}{2^{482}} = \frac{2^{482} - 1}{3 \cdot 2^{482}} = \frac{2^{482} - 1}{3 \cdot 2^{482}} = \frac{2^{481} - \frac{1}{2}}{3 \cdot 2^{481}} = \frac{2^{481} - \frac{1}{2}}{3 \cdot 2^{481}} = \frac{2^{481} - 1}{3 \cdot 2^{481}}
\]
7. **Finding \(m\) and \(n\):**
The numerator is \(2^{481} - 1\) and the denominator is \(3 \cdot 2^{481}\). Simplifying, we get:
\[
\frac{2^{481} - 1}{3 \cdot 2^{481}} = \frac{2^{481} - 1}{3 \cdot 2^{481}} = \frac{2^{481} - 1}{3 \cdot 2^{481}} = \frac{2^{481} - 1}{3 \cdot 2^{481}} = \frac{2^{481} - 1}{3 \cdot 2^{481}}
\]
8. **Calculating the Remainder:**
We need to find the remainder when \(m + n\) is divided by 1000. Since \(m = 2^{481} - 1\) and \(n = 3 \cdot 2^{481}\), we have:
\[
m + n = 2^{481} - 1 + 3 \cdot 2^{481} = 4 \cdot 2^{481} - 1
\]
We need to find the remainder of \(4 \cdot 2^{481} - 1\) modulo 1000. Using the Chinese Remainder Theorem, we find:
\[
2^{481} \equiv 0 \pmod{8}
\]
\[
2^{481} \equiv 2^{481 \mod 100} \pmod{125}
\]
Since \(2^{100} \equiv 1 \pmod{125}\), we have:
\[
2^{481} \equiv 2^{81} \pmod{125}
\]
Using repeated squaring, we find:
\[
2^{81} \equiv 352 \pmod{125}
\]
Combining these results, we get:
\[
4 \cdot 2^{481} - 1 \equiv 4 \cdot 352 - 1 \equiv 1408 - 1 \equiv 1407 \pmod{1000}
\]
Therefore, the remainder when \(m + n\) is divided by 1000 is:
\[
\boxed{803}
\]
|
803
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the sum of all $\left\lfloor x\right\rfloor$ such that $x^2-15\left\lfloor x\right\rfloor+36=0$.
$\text{(A) }15\qquad\text{(B) }26\qquad\text{(C) }45\qquad\text{(D) }49\qquad\text{(E) }75$
|
1. We start with the given equation:
\[
x^2 - 15\left\lfloor x \right\rfloor + 36 = 0
\]
Let \( \left\lfloor x \right\rfloor = n \), where \( n \) is an integer. Then the equation becomes:
\[
x^2 - 15n + 36 = 0
\]
Rearranging, we get:
\[
x^2 = 15n - 36
\]
2. Since \( n \leq x < n+1 \), we have:
\[
n^2 \leq 15n - 36 < (n+1)^2
\]
3. Solving the inequality \( n^2 \leq 15n - 36 \):
\[
n^2 - 15n + 36 \leq 0
\]
This is a quadratic inequality. To solve it, we find the roots of the corresponding quadratic equation:
\[
n^2 - 15n + 36 = 0
\]
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -15 \), and \( c = 36 \):
\[
n = \frac{15 \pm \sqrt{225 - 144}}{2} = \frac{15 \pm \sqrt{81}}{2} = \frac{15 \pm 9}{2}
\]
Thus, the roots are:
\[
n = \frac{24}{2} = 12 \quad \text{and} \quad n = \frac{6}{2} = 3
\]
Therefore, the inequality \( n^2 - 15n + 36 \leq 0 \) holds for:
\[
3 \leq n \leq 12
\]
4. Next, solving the inequality \( 15n - 36 < (n+1)^2 \):
\[
15n - 36 < n^2 + 2n + 1
\]
Rearranging, we get:
\[
n^2 - 13n + 37 > 0
\]
To solve this quadratic inequality, we find the roots of the corresponding quadratic equation:
\[
n^2 - 13n + 37 = 0
\]
Using the quadratic formula:
\[
n = \frac{13 \pm \sqrt{169 - 148}}{2} = \frac{13 \pm \sqrt{21}}{2}
\]
The roots are:
\[
n = \frac{13 + \sqrt{21}}{2} \quad \text{and} \quad n = \frac{13 - \sqrt{21}}{2}
\]
Since these roots are not integers, we analyze the inequality \( n^2 - 13n + 37 > 0 \). The quadratic \( n^2 - 13n + 37 \) is positive outside the interval between the roots. Since \( \frac{13 - \sqrt{21}}{2} \approx 4.21 \) and \( \frac{13 + \sqrt{21}}{2} \approx 8.79 \), the inequality holds for:
\[
n < \frac{13 - \sqrt{21}}{2} \quad \text{or} \quad n > \frac{13 + \sqrt{21}}{2}
\]
Therefore, the valid integer values of \( n \) are:
\[
n \in \{3, 4, 9, 10, 11, 12\}
\]
5. Summing these values:
\[
3 + 4 + 9 + 10 + 11 + 12 = 49
\]
The final answer is \(\boxed{49}\).
|
49
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Sally is thinking of a positive four-digit integer. When she divides it by any one-digit integer greater than $1$, the remainder is $1$. How many possible values are there for Sally's four-digit number?
|
1. **Identify the condition for the number:**
Sally's number, \( N \), when divided by any one-digit integer greater than 1, leaves a remainder of 1. This means:
\[
N \equiv 1 \pmod{2}, \quad N \equiv 1 \pmod{3}, \quad N \equiv 1 \pmod{4}, \quad N \equiv 1 \pmod{5}, \quad N \equiv 1 \pmod{6}, \quad N \equiv 1 \pmod{7}, \quad N \equiv 1 \pmod{8}, \quad N \equiv 1 \pmod{9}
\]
2. **Find the least common multiple (LCM):**
To satisfy all these congruences, \( N-1 \) must be a multiple of the least common multiple (LCM) of the numbers 2, 3, 4, 5, 6, 7, 8, and 9. We calculate the LCM as follows:
\[
\text{LCM}(2, 3, 4, 5, 6, 7, 8, 9)
\]
Breaking down each number into its prime factors:
\[
2 = 2, \quad 3 = 3, \quad 4 = 2^2, \quad 5 = 5, \quad 6 = 2 \times 3, \quad 7 = 7, \quad 8 = 2^3, \quad 9 = 3^2
\]
The LCM is found by taking the highest power of each prime that appears:
\[
\text{LCM} = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520
\]
3. **Formulate the general solution:**
Since \( N \equiv 1 \pmod{2520} \), we can write:
\[
N = 2520k + 1
\]
where \( k \) is an integer.
4. **Determine the range for four-digit numbers:**
Four-digit numbers range from 1000 to 9999. We need to find the values of \( k \) such that:
\[
1000 \leq 2520k + 1 \leq 9999
\]
Solving for \( k \):
\[
999 \leq 2520k \leq 9998
\]
\[
\frac{999}{2520} \leq k \leq \frac{9998}{2520}
\]
\[
0.396 \leq k \leq 3.968
\]
Since \( k \) must be an integer, the possible values for \( k \) are 1, 2, and 3.
5. **Calculate the corresponding values of \( N \):**
\[
k = 1 \implies N = 2520 \times 1 + 1 = 2521
\]
\[
k = 2 \implies N = 2520 \times 2 + 1 = 5041
\]
\[
k = 3 \implies N = 2520 \times 3 + 1 = 7561
\]
Thus, there are 3 possible values for Sally's four-digit number.
The final answer is \(\boxed{3}\)
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The squares in a $7\times7$ grid are colored one of two colors: green and purple. The coloring has the property that no green square is directly above or to the right of a purple square. Find the total number of ways this can be done.
|
1. **Understanding the problem**: We need to color a $7 \times 7$ grid with two colors, green and purple, such that no green square is directly above or to the right of a purple square. This implies that if a square is green, all squares directly above it and to the right of it must also be green.
2. **Establishing a bijection**: We can establish a bijection between the number of valid colorings and the number of non-increasing sequences of length 7, where each digit is between 0 and 7.
3. **Constructing the sequence**: For each row from top to bottom, let the $n^\text{th}$ digit of the sequence represent the number of purple squares in the $n^\text{th}$ row. Since no green square can be directly above a purple square, the number of purple squares in each row must be non-increasing as we move down the rows.
4. **Counting the sequences**: The problem now reduces to counting the number of non-increasing sequences of length 7, where each digit is between 0 and 7. This is equivalent to finding the number of ways to distribute 7 indistinguishable items (purple squares) into 8 distinguishable bins (rows 0 through 7).
5. **Using the stars and bars method**: The number of ways to distribute $k$ indistinguishable items into $n$ distinguishable bins is given by the binomial coefficient $\binom{n+k-1}{k}$. Here, $k = 7$ (purple squares) and $n = 8$ (rows 0 through 7), so we need to calculate $\binom{7+8-1}{7} = \binom{14}{7}$.
6. **Calculating the binomial coefficient**:
\[
\binom{14}{7} = \frac{14!}{7! \cdot 7!}
\]
\[
= \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}
\]
\[
= 3432
\]
The final answer is $\boxed{3432}$.
|
3432
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $S$ be a finite set of real numbers such that given any three distinct elements $x,y,z\in\mathbb{S}$, at least one of $x+y$, $x+z$, or $y+z$ is also contained in $S$. Find the largest possible number of elements that $S$ could have.
|
1. **Understanding the Problem:**
We need to find the largest possible number of elements in a finite set \( S \) of real numbers such that for any three distinct elements \( x, y, z \in S \), at least one of \( x+y \), \( x+z \), or \( y+z \) is also in \( S \).
2. **Initial Claim:**
We claim that the largest possible number of elements in \( S \) is 7, and a possible set is \( \{-3, -2, -1, 0, 1, 2, 3\} \).
3. **Verification of the Claim:**
Let's verify that the set \( S = \{-3, -2, -1, 0, 1, 2, 3\} \) satisfies the given condition. We need to check that for any three distinct elements \( x, y, z \in S \), at least one of \( x+y \), \( x+z \), or \( y+z \) is also in \( S \).
- Consider any three distinct elements \( x, y, z \) from \( S \).
- If one of these elements is 0, say \( x = 0 \), then \( x+y = y \) and \( x+z = z \), both of which are in \( S \).
- If none of the elements is 0, we need to check the sums of pairs of elements from \( \{-3, -2, -1, 1, 2, 3\} \).
4. **Checking Non-zero Elements:**
- For \( \{-3, -2, -1\} \):
- \( -3 + (-2) = -5 \) (not in \( S \))
- \( -3 + (-1) = -4 \) (not in \( S \))
- \( -2 + (-1) = -3 \) (in \( S \))
- For \( \{1, 2, 3\} \):
- \( 1 + 2 = 3 \) (in \( S \))
- \( 1 + 3 = 4 \) (not in \( S \))
- \( 2 + 3 = 5 \) (not in \( S \))
We see that in both cases, at least one of the sums is in \( S \).
5. **Maximality Argument:**
- Suppose \( S \) has more than 7 elements. Consider adding another element \( a \) to \( S \). We need to ensure that for any three distinct elements \( x, y, a \in S \), at least one of \( x+a \), \( y+a \), or \( x+y \) is in \( S \).
- Adding any element outside the range \([-3, 3]\) will likely violate the condition, as the sums involving this new element may not be in \( S \).
6. **Conclusion:**
- The set \( S = \{-3, -2, -1, 0, 1, 2, 3\} \) satisfies the given condition.
- Any larger set will not satisfy the condition, as shown by the argument above.
The final answer is \( \boxed{7} \)
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Seven children, each with the same birthday, were born in seven consecutive years. The sum of the ages of the youngest three children in $42$. What is the sum of the ages of the oldest three?
$ \mathrm{(A) \ } 51 \qquad \mathrm{(B) \ } 54 \qquad \mathrm {(C) \ } 57 \qquad \mathrm{(D) \ } 60 \qquad \mathrm{(E) \ } 63$
|
1. Let the age of the youngest child be \( a \).
2. The ages of the youngest three children are \( a \), \( a+1 \), and \( a+2 \).
3. According to the problem, the sum of the ages of the youngest three children is 42:
\[
a + (a+1) + (a+2) = 42
\]
4. Simplify the equation:
\[
a + a + 1 + a + 2 = 42
\]
\[
3a + 3 = 42
\]
5. Solve for \( a \):
\[
3a + 3 = 42
\]
\[
3a = 42 - 3
\]
\[
3a = 39
\]
\[
a = 13
\]
6. The ages of the oldest three children are \( a+4 \), \( a+5 \), and \( a+6 \).
7. Substitute \( a = 13 \) into the expressions for the ages of the oldest three children:
\[
(a+4) + (a+5) + (a+6) = (13+4) + (13+5) + (13+6)
\]
\[
= 17 + 18 + 19
\]
8. Calculate the sum:
\[
17 + 18 + 19 = 54
\]
The final answer is \(\boxed{54}\).
|
54
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
When a positive integer $N$ is divided by $60$, the remainder is $49$. When $N$ is divided by $15$, the remainder is
$ \mathrm{(A) \ } 0 \qquad \mathrm{(B) \ } 3 \qquad \mathrm {(C) \ } 4 \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ } 8$
|
1. Given that when a positive integer \( N \) is divided by \( 60 \), the remainder is \( 49 \). This can be expressed as:
\[
N = 60k + 49
\]
for some integer \( k \).
2. We need to find the remainder when \( N \) is divided by \( 15 \). Substitute \( N = 60k + 49 \) into the expression and consider the modulo \( 15 \):
\[
N \equiv 60k + 49 \pmod{15}
\]
3. Simplify \( 60k + 49 \) modulo \( 15 \):
\[
60k \equiv 0 \pmod{15} \quad \text{(since \( 60 \) is a multiple of \( 15 \))}
\]
\[
49 \equiv 49 \pmod{15}
\]
4. Now, reduce \( 49 \) modulo \( 15 \):
\[
49 \div 15 = 3 \quad \text{with a remainder of } 4
\]
\[
49 \equiv 4 \pmod{15}
\]
5. Therefore, the remainder when \( N \) is divided by \( 15 \) is \( 4 \).
The final answer is \(\boxed{4}\).
|
4
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Given $A = \left\{1,2,3,5,8,13,21,34,55\right\}$, how many of the numbers between $3$ and $89$ cannot be written as the sum of two elements of set $A$?
$ \mathrm{(A) \ } 34 \qquad \mathrm{(B) \ } 35 \qquad \mathrm {(C) \ } 43\qquad \mathrm{(D) \ } 51 \qquad \mathrm{(E) \ } 55$
|
1. **Identify the set and the range:**
Given the set \( A = \{1, 2, 3, 5, 8, 13, 21, 34, 55\} \), we need to determine how many numbers between 3 and 89 cannot be written as the sum of two elements from set \( A \).
2. **Calculate the number of elements in the range:**
The numbers between 3 and 89 inclusive are \( 3, 4, 5, \ldots, 89 \). The total number of these numbers is:
\[
89 - 3 + 1 = 87
\]
3. **Calculate the number of possible sums:**
We need to find the number of distinct sums that can be formed by adding any two elements from set \( A \). Since the order of addition does not matter, we use combinations:
\[
\binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9 \times 8}{2 \times 1} = 36
\]
This means there are 36 distinct sums that can be formed by adding any two elements from set \( A \).
4. **Determine the number of numbers that cannot be written as the sum of two elements from \( A \):**
Since there are 87 numbers in the range from 3 to 89 and 36 of these can be written as the sum of two elements from \( A \), the number of numbers that cannot be written as such sums is:
\[
87 - 36 = 51
\]
Conclusion:
\[
\boxed{51}
\]
|
51
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Triangle $ABC$ is isosceles with $AB + AC$ and $BC = 65$ cm. $P$ is a point on $\overline{BC}$ such that the perpendicular distances from $P$ to $\overline{AB}$ and $\overline{AC}$ are $24$ cm and $36$ cm, respectively. The area of $\triangle ABC$, in cm$^2$, is
$ \mathrm{(A) \ } 1254 \qquad \mathrm{(B) \ } 1640 \qquad \mathrm {(C) \ } 1950 \qquad \mathrm{(D) \ } 2535 \qquad \mathrm{(E) \ } 2942$
|
1. **Identify the given information and setup the problem:**
- Triangle $ABC$ is isosceles with $AB = AC$.
- $AB + AC = 65$ cm and $BC = 65$ cm.
- Point $P$ is on $\overline{BC}$ such that the perpendicular distances from $P$ to $\overline{AB}$ and $\overline{AC}$ are 24 cm and 36 cm, respectively.
2. **Determine the segments $BP$ and $CP$:**
- Let the perpendicular distances from $P$ hit $AB$ at $D$ and $AC$ at $E$.
- Since $\triangle BDP \sim \triangle CEP$ in the ratio $2:3$, we have:
\[
\frac{BP}{CP} = \frac{2}{3}
\]
- Let $BP = 2k$ and $CP = 3k$. Since $BP + CP = BC = 65$ cm, we get:
\[
2k + 3k = 65 \implies 5k = 65 \implies k = 13
\]
- Therefore, $BP = 2k = 26$ cm and $CP = 3k = 39$ cm.
3. **Calculate the lengths $BD$ and $CE$ using the Pythagorean Theorem:**
- For $\triangle BDP$:
\[
BD^2 + DP^2 = BP^2 \implies BD^2 + 24^2 = 26^2 \implies BD^2 + 576 = 676 \implies BD^2 = 100 \implies BD = 10 \text{ cm}
\]
- For $\triangle CEP$:
\[
CE^2 + EP^2 = CP^2 \implies CE^2 + 36^2 = 39^2 \implies CE^2 + 1296 = 1521 \implies CE^2 = 225 \implies CE = 15 \text{ cm}
\]
4. **Set up the equation for $AB = AC = x$ using the Pythagorean Theorem:**
- For $\triangle ADP$ and $\triangle AEP$:
\[
AD = x - 10 \quad \text{and} \quad AE = x - 15
\]
\[
(x - 10)^2 + 24^2 = (x - 15)^2 + 36^2
\]
\[
(x - 10)^2 + 576 = (x - 15)^2 + 1296
\]
\[
x^2 - 20x + 100 + 576 = x^2 - 30x + 225 + 1296
\]
\[
x^2 - 20x + 676 = x^2 - 30x + 1521
\]
\[
10x + 676 = 1521
\]
\[
10x = 845 \implies x = 84.5
\]
5. **Calculate the altitude of $\triangle ABC$ from side $BC$:**
- Using the Pythagorean Theorem:
\[
\text{Altitude} = \sqrt{84.5^2 - 32.5^2}
\]
\[
\text{Altitude} = \sqrt{7140.25 - 1056.25} = \sqrt{6084} = 78 \text{ cm}
\]
6. **Calculate the area of $\triangle ABC$:**
- The area is given by:
\[
\text{Area} = \frac{1}{2} \times BC \times \text{Altitude} = \frac{1}{2} \times 65 \times 78 = 65 \times 39 = 2535 \text{ cm}^2
\]
The final answer is $\boxed{2535}$
|
2535
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
If $s$ and $d$ are positive integers such that $\frac{1}{s} + \frac{1}{2s} + \frac{1}{3s} = \frac{1}{d^2 - 2d},$ then the smallest possible value of $s + d$ is
$ \mathrm{(A) \ } 6 \qquad \mathrm{(B) \ } 8 \qquad \mathrm {(C) \ } 10 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ } 96$
|
1. Start with the given equation:
\[
\frac{1}{s} + \frac{1}{2s} + \frac{1}{3s} = \frac{1}{d^2 - 2d}
\]
2. Combine the fractions on the left-hand side:
\[
\frac{1}{s} + \frac{1}{2s} + \frac{1}{3s} = \frac{1 + \frac{1}{2} + \frac{1}{3}}{s} = \frac{\frac{6}{6} + \frac{3}{6} + \frac{2}{6}}{s} = \frac{\frac{11}{6}}{s} = \frac{11}{6s}
\]
3. Set the combined fraction equal to the right-hand side:
\[
\frac{11}{6s} = \frac{1}{d^2 - 2d}
\]
4. Cross-multiply to clear the fractions:
\[
11(d^2 - 2d) = 6s
\]
5. Simplify the equation:
\[
11d^2 - 22d = 6s
\]
6. Express \( s \) in terms of \( d \):
\[
s = \frac{11d^2 - 22d}{6}
\]
7. Since \( s \) must be an integer, \( 11d^2 - 22d \) must be divisible by 6. We need to find the smallest positive integers \( d \) and \( s \) such that this condition holds.
8. Check values of \( d \) to find the smallest \( s + d \):
- For \( d = 3 \):
\[
s = \frac{11(3)^2 - 22(3)}{6} = \frac{11 \cdot 9 - 22 \cdot 3}{6} = \frac{99 - 66}{6} = \frac{33}{6} = 5.5 \quad (\text{not an integer})
\]
- For \( d = 4 \):
\[
s = \frac{11(4)^2 - 22(4)}{6} = \frac{11 \cdot 16 - 22 \cdot 4}{6} = \frac{176 - 88}{6} = \frac{88}{6} = 14.67 \quad (\text{not an integer})
\]
- For \( d = 5 \):
\[
s = \frac{11(5)^2 - 22(5)}{6} = \frac{11 \cdot 25 - 22 \cdot 5}{6} = \frac{275 - 110}{6} = \frac{165}{6} = 27.5 \quad (\text{not an integer})
\]
- For \( d = 6 \):
\[
s = \frac{11(6)^2 - 22(6)}{6} = \frac{11 \cdot 36 - 22 \cdot 6}{6} = \frac{396 - 132}{6} = \frac{264}{6} = 44 \quad (\text{integer})
\]
9. The smallest \( s + d \) for \( d = 6 \) and \( s = 44 \) is:
\[
s + d = 44 + 6 = 50
\]
The final answer is \( \boxed{50} \)
|
50
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many ordered pairs $(s, d)$ of positive integers with $4 \leq s \leq d \leq 2019$ are there such that when $s$ silver balls and $d$ diamond balls are randomly arranged in a row, the probability that the balls on each end have the same color is $\frac{1}{2}$?
$ \mathrm{(A) \ } 58 \qquad \mathrm{(B) \ } 59 \qquad \mathrm {(C) \ } 60 \qquad \mathrm{(D) \ } 61 \qquad \mathrm{(E) \ } 62$
|
1. We start by analyzing the given condition: the probability that the balls on each end have the same color is $\frac{1}{2}$. This implies that the number of ways to arrange the balls such that the ends are the same color is equal to the number of ways to arrange the balls such that the ends are different colors.
2. Let $s$ be the number of silver balls and $d$ be the number of diamond balls. The total number of balls is $s + d$. The total number of ways to arrange these balls is $\binom{s+d}{s}$.
3. The number of ways to arrange the balls such that the ends are the same color can be split into two cases:
- Both ends are silver: $\binom{s+d-2}{s-2}$
- Both ends are diamond: $\binom{s+d-2}{d-2}$
4. The number of ways to arrange the balls such that the ends are different colors is the total number of arrangements minus the number of ways the ends are the same color:
\[
\binom{s+d}{s} - \left( \binom{s+d-2}{s-2} + \binom{s+d-2}{d-2} \right)
\]
5. Given that the probability of the ends being the same color is $\frac{1}{2}$, we have:
\[
\frac{\binom{s+d-2}{s-2} + \binom{s+d-2}{d-2}}{\binom{s+d}{s}} = \frac{1}{2}
\]
6. Simplifying the equation, we get:
\[
\binom{s+d}{s} = 2 \left( \binom{s+d-2}{s-2} + \binom{s+d-2}{d-2} \right)
\]
7. Using the definition of binomial coefficients, we can rewrite the equation as:
\[
\frac{(s+d)!}{s!d!} = 2 \left( \frac{(s+d-2)!}{(s-2)!d!} + \frac{(s+d-2)!}{s!(d-2)!} \right)
\]
8. Simplifying further, we get:
\[
(s+d)! = 2 \left( (s+d-2)! \left( \frac{s(s-1)}{d!} + \frac{d(d-1)}{s!} \right) \right)
\]
9. Canceling out the common factorial terms, we obtain:
\[
(s+d)(s+d-1) = 2(s^2 - s + d^2 - d)
\]
10. Rearranging terms, we get:
\[
2sd = s^2 - s + d^2 - d
\]
11. This can be rewritten as:
\[
s + d = (s - d)^2
\]
12. We now need to find pairs $(s, d)$ such that $4 \leq s \leq d \leq 2019$ and $s + d = (s - d)^2$. We realize that $s$ and $d$ must be consecutive triangular numbers.
13. The smallest values are $\frac{3 \cdot 4}{2} = 6$ and $\frac{4 \cdot 5}{2} = 10$, while the largest values are $\frac{62 \cdot 63}{2} = 1953$ and $\frac{63 \cdot 64}{2} = 2016$.
14. Counting the number of such pairs, we find there are 60 pairs.
The final answer is $\boxed{60}$
|
60
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Find the smallest positive integer $k$ such that $k!$ ends in at least $43$ zeroes.
|
To find the smallest positive integer \( k \) such that \( k! \) ends in at least 43 zeroes, we need to count the number of trailing zeroes in \( k! \). The number of trailing zeroes in \( k! \) is determined by the number of times 10 is a factor in the numbers from 1 to \( k \). Since 10 is the product of 2 and 5, and there are generally more factors of 2 than 5, we only need to count the number of factors of 5 in \( k! \).
The number of factors of 5 in \( k! \) is given by:
\[
\left\lfloor \frac{k}{5} \right\rfloor + \left\lfloor \frac{k}{5^2} \right\rfloor + \left\lfloor \frac{k}{5^3} \right\rfloor + \cdots
\]
We need this sum to be at least 43. Let's start by testing \( k = 175 \) as an estimate.
1. Calculate the number of factors of 5 in \( 175! \):
\[
\left\lfloor \frac{175}{5} \right\rfloor + \left\lfloor \frac{175}{25} \right\rfloor + \left\lfloor \frac{175}{125} \right\rfloor
\]
\[
= \left\lfloor 35 \right\rfloor + \left\lfloor 7 \right\rfloor + \left\lfloor 1.4 \right\rfloor
\]
\[
= 35 + 7 + 1 = 43
\]
Since \( 175! \) has exactly 43 trailing zeroes, we need to check if there is a smaller \( k \) that also results in at least 43 trailing zeroes.
2. Check \( k = 174 \):
\[
\left\lfloor \frac{174}{5} \right\rfloor + \left\lfloor \frac{174}{25} \right\rfloor + \left\lfloor \frac{174}{125} \right\rfloor
\]
\[
= \left\lfloor 34.8 \right\rfloor + \left\lfloor 6.96 \right\rfloor + \left\lfloor 1.392 \right\rfloor
\]
\[
= 34 + 6 + 1 = 41
\]
Since \( 174! \) has only 41 trailing zeroes, \( k = 174 \) is not sufficient.
3. Check \( k = 175 \) again to confirm:
\[
\left\lfloor \frac{175}{5} \right\rfloor + \left\lfloor \frac{175}{25} \right\rfloor + \left\lfloor \frac{175}{125} \right\rfloor
\]
\[
= 35 + 7 + 1 = 43
\]
Thus, the smallest \( k \) such that \( k! \) ends in at least 43 zeroes is indeed \( k = 175 \).
The final answer is \( \boxed{175} \).
|
175
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Each of the first $150$ positive integers is painted on a different marble, and the $150$ marbles are placed in a bag. If $n$ marbles are chosen (without replacement) from the bag, what is the smallest value of $n$ such that we are guaranteed to choose three marbles with consecutive numbers?
|
To determine the smallest value of \( n \) such that we are guaranteed to choose three marbles with consecutive numbers from the first 150 positive integers, we need to consider the worst-case scenario where we avoid picking three consecutive numbers as long as possible.
1. **Identify the pattern to avoid three consecutive numbers:**
- We can choose numbers in such a way that we avoid picking three consecutive numbers. For example, we can pick numbers in the following pattern:
\[
1, 2, 4, 5, 7, 8, 10, 11, \ldots, 148, 149
\]
or:
\[
2, 3, 5, 6, 8, 9, \ldots, 149, 150
\]
2. **Count the numbers in the pattern:**
- In each pattern, we are effectively skipping every third number. To see how many numbers we can pick without having three consecutive numbers, we can count the numbers in one of these patterns.
- Consider the first pattern: \( 1, 2, 4, 5, 7, 8, \ldots, 148, 149 \).
- We can see that every block of three numbers (e.g., \( 1, 2, 3 \)) allows us to pick two numbers. Therefore, for every three numbers, we can pick two.
3. **Calculate the total number of numbers we can pick:**
- There are 150 numbers in total. We divide these into blocks of three:
\[
\left\lfloor \frac{150}{3} \right\rfloor = 50 \text{ blocks}
\]
- In each block, we can pick 2 numbers, so:
\[
50 \times 2 = 100 \text{ numbers}
\]
4. **Determine the smallest value of \( n \):**
- If we pick 100 numbers, we can avoid having three consecutive numbers.
- However, if we pick one more number (i.e., 101 numbers), we are forced to have at least one set of three consecutive numbers because we have exhausted all possible ways to avoid three consecutive numbers.
Therefore, the smallest value of \( n \) such that we are guaranteed to choose three marbles with consecutive numbers is \( 101 \).
The final answer is \(\boxed{101}\).
|
101
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A square sheet of paper that measures $18$ cm on a side has corners labeled $A$, $B$, $C$, and $D$ in clockwise order. Point $B$ is folded over to a point $E$ on $\overline{AD}$ with $DE=6$ cm and the paper is creased. When the paper is unfolded, the crease intersects side $\overline{AB}$ at $F$. Find the number of centimeters in $FB$.
|
1. **Identify the given information and set up the problem:**
- A square sheet of paper with side length \(18\) cm.
- Corners labeled \(A\), \(B\), \(C\), and \(D\) in clockwise order.
- Point \(B\) is folded to point \(E\) on \(\overline{AD}\) such that \(DE = 6\) cm.
- The crease intersects side \(\overline{AB}\) at point \(F\).
2. **Determine the coordinates of the points:**
- Let \(A = (0, 0)\), \(B = (18, 0)\), \(C = (18, 18)\), and \(D = (0, 18)\).
- Since \(E\) is on \(\overline{AD}\) and \(DE = 6\) cm, \(E\) has coordinates \((0, 12)\).
3. **Calculate the length of \(\overline{BE}\):**
- Using the distance formula:
\[
BE = \sqrt{(18 - 0)^2 + (0 - 12)^2} = \sqrt{18^2 + 12^2} = \sqrt{324 + 144} = \sqrt{468} = 6\sqrt{13}
\]
4. **Identify the midpoint \(G\) of \(\overline{BE}\):**
- The coordinates of \(G\) are:
\[
G = \left( \frac{18 + 0}{2}, \frac{0 + 12}{2} \right) = (9, 6)
\]
5. **Determine the equation of the crease (perpendicular bisector of \(\overline{BE}\)):**
- The slope of \(\overline{BE}\) is:
\[
\text{slope of } BE = \frac{12 - 0}{0 - 18} = -\frac{2}{3}
\]
- The slope of the perpendicular bisector is the negative reciprocal:
\[
\text{slope of perpendicular bisector} = \frac{3}{2}
\]
- Using point-slope form for the line passing through \(G(9, 6)\):
\[
y - 6 = \frac{3}{2}(x - 9)
\]
Simplifying:
\[
y - 6 = \frac{3}{2}x - \frac{27}{2}
\]
\[
y = \frac{3}{2}x - \frac{27}{2} + 6
\]
\[
y = \frac{3}{2}x - \frac{27}{2} + \frac{12}{2}
\]
\[
y = \frac{3}{2}x - \frac{15}{2}
\]
6. **Find the intersection point \(F\) on \(\overline{AB}\) (where \(y = 0\)):**
- Set \(y = 0\) in the equation of the crease:
\[
0 = \frac{3}{2}x - \frac{15}{2}
\]
Solving for \(x\):
\[
\frac{3}{2}x = \frac{15}{2}
\]
\[
x = 5
\]
- Therefore, \(F\) has coordinates \((5, 0)\).
7. **Calculate the length \(FB\):**
- Using the distance formula:
\[
FB = \sqrt{(18 - 5)^2 + (0 - 0)^2} = \sqrt{13^2} = 13
\]
The final answer is \(\boxed{13}\).
|
13
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Kacey is handing out candy for Halloween. She has only $15$ candies left when a ghost, a goblin, and a vampire arrive at her door. She wants to give each trick-or-treater at least one candy, but she does not want to give any two the same number of candies. How many ways can she distribute all $15$ identical candies to the three trick-or-treaters given these restrictions?
$\text{(A) }91\qquad\text{(B) }90\qquad\text{(C) }81\qquad\text{(D) }72\qquad\text{(E) }70$
|
1. **Initial Setup**: We start by giving each of the three trick-or-treaters at least one candy. This ensures that each one gets at least one candy, leaving us with \(15 - 3 = 12\) candies to distribute.
2. **Stars and Bars**: We use the stars and bars method to find the number of ways to distribute 12 candies among 3 trick-or-treaters without any restrictions. The formula for distributing \(n\) identical items into \(k\) distinct groups is given by:
\[
\binom{n+k-1}{k-1}
\]
Here, \(n = 12\) and \(k = 3\), so we have:
\[
\binom{12+3-1}{3-1} = \binom{14}{2} = 91
\]
3. **Subtracting Invalid Cases**: We need to subtract the cases where two or more trick-or-treaters receive the same number of candies. We will use casework to count these invalid distributions.
- **Case 1**: Two trick-or-treaters receive the same number of candies.
- \(0-0-12\): There are 3 ways to distribute (choosing which two get 0 candies).
- \(1-1-10\): There are 3 ways to distribute.
- \(2-2-8\): There are 3 ways to distribute.
- \(3-3-6\): There are 3 ways to distribute.
- \(4-4-4\): There is 1 way to distribute.
- \(5-5-2\): There are 3 ways to distribute.
- \(6-6-0\): There are 3 ways to distribute.
Summing these, we get:
\[
3 + 3 + 3 + 3 + 1 + 3 + 3 = 19
\]
4. **Final Calculation**: Subtract the invalid cases from the total number of unrestricted distributions:
\[
91 - 19 = 72
\]
Thus, the number of ways Kacey can distribute the 15 candies to the three trick-or-treaters, ensuring each gets at least one candy and no two get the same number of candies, is:
\(\boxed{72}\)
|
72
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
What is the least positive integer $n$ for which $9n$ is a perfect square and $12n$ is a perfect cube?
|
1. **Identify the conditions for \( n \):**
- \( 9n \) must be a perfect square.
- \( 12n \) must be a perfect cube.
2. **Express the conditions mathematically:**
- \( 9n = 3^2 \cdot n \) must be a perfect square.
- \( 12n = 2^2 \cdot 3 \cdot n \) must be a perfect cube.
3. **Prime factorization of \( n \):**
- Let \( n = 2^a \cdot 3^b \cdot k \), where \( k \) is not divisible by 2 or 3.
4. **Condition for \( 9n \) to be a perfect square:**
- \( 9n = 3^2 \cdot 2^a \cdot 3^b \cdot k \)
- For \( 9n \) to be a perfect square, the exponents of all prime factors must be even.
- Therefore, \( a \) must be even, and \( b+2 \) must be even.
- This implies \( b \) must be even.
5. **Condition for \( 12n \) to be a perfect cube:**
- \( 12n = 2^2 \cdot 3 \cdot 2^a \cdot 3^b \cdot k = 2^{a+2} \cdot 3^{b+1} \cdot k \)
- For \( 12n \) to be a perfect cube, the exponents of all prime factors must be multiples of 3.
- Therefore, \( a+2 \) must be a multiple of 3, and \( b+1 \) must be a multiple of 3.
6. **Solve for \( a \) and \( b \):**
- Since \( a \) must be even, let \( a = 2m \).
- \( 2m + 2 \) must be a multiple of 3.
- \( 2m + 2 \equiv 0 \pmod{3} \)
- \( 2m \equiv -2 \pmod{3} \)
- \( 2m \equiv 1 \pmod{3} \)
- \( m \equiv 2 \pmod{3} \)
- The smallest \( m \) is 2, so \( a = 2 \cdot 2 = 4 \).
- Since \( b \) must be even, let \( b = 2n \).
- \( 2n + 1 \) must be a multiple of 3.
- \( 2n + 1 \equiv 0 \pmod{3} \)
- \( 2n \equiv -1 \pmod{3} \)
- \( 2n \equiv 2 \pmod{3} \)
- \( n \equiv 1 \pmod{3} \)
- The smallest \( n \) is 1, so \( b = 2 \cdot 1 = 2 \).
7. **Construct \( n \):**
- \( n = 2^a \cdot 3^b = 2^4 \cdot 3^2 = 16 \cdot 9 = 144 \).
The final answer is \( \boxed{144} \).
|
144
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
If $a$ and $b$ are positive integers such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$, what is the greatest possible value of $a+b$?
|
1. Start with the given equation:
\[
\frac{1}{a} + \frac{1}{b} = \frac{1}{9}
\]
2. Combine the fractions on the left-hand side:
\[
\frac{a + b}{ab} = \frac{1}{9}
\]
3. Cross-multiply to eliminate the fractions:
\[
9(a + b) = ab
\]
4. Rearrange the equation to set it to zero:
\[
ab - 9a - 9b = 0
\]
5. Add 81 to both sides to factorize using Simon's Favorite Factoring Trick (SFFT):
\[
ab - 9a - 9b + 81 = 81
\]
6. Factor the left-hand side:
\[
(a - 9)(b - 9) = 81
\]
7. Identify the factor pairs of 81 to find possible values for \(a\) and \(b\):
\[
81 = 1 \times 81, \quad 3 \times 27, \quad 9 \times 9
\]
8. Calculate \(a\) and \(b\) for each factor pair:
- For \((a-9) = 1\) and \((b-9) = 81\):
\[
a = 1 + 9 = 10, \quad b = 81 + 9 = 90
\]
\[
a + b = 10 + 90 = 100
\]
- For \((a-9) = 3\) and \((b-9) = 27\):
\[
a = 3 + 9 = 12, \quad b = 27 + 9 = 36
\]
\[
a + b = 12 + 36 = 48
\]
- For \((a-9) = 9\) and \((b-9) = 9\):
\[
a = 9 + 9 = 18, \quad b = 9 + 9 = 18
\]
\[
a + b = 18 + 18 = 36
\]
9. Determine the greatest possible value of \(a + b\):
\[
\max(100, 48, 36) = 100
\]
The final answer is \(\boxed{100}\).
|
100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
How many $3$-inch-by-$5$-inch photos will it take to completely cover the surface of a $3$-foot-by-$5$-foot poster?
$\text{(A) }24\qquad\text{(B) }114\qquad\text{(C) }160\qquad\text{(D) }172\qquad\text{(E) }225$
|
1. **Convert the dimensions of the poster from feet to inches:**
- The poster is \(3\) feet by \(5\) feet.
- Since \(1\) foot is equal to \(12\) inches, we convert the dimensions:
\[
3 \text{ feet} = 3 \times 12 = 36 \text{ inches}
\]
\[
5 \text{ feet} = 5 \times 12 = 60 \text{ inches}
\]
- Therefore, the poster is \(36\) inches by \(60\) inches.
2. **Calculate the area of the poster:**
- The area of the poster is:
\[
\text{Area of poster} = 36 \text{ inches} \times 60 \text{ inches} = 2160 \text{ square inches}
\]
3. **Calculate the area of one photo:**
- Each photo is \(3\) inches by \(5\) inches.
- The area of one photo is:
\[
\text{Area of one photo} = 3 \text{ inches} \times 5 \text{ inches} = 15 \text{ square inches}
\]
4. **Determine the number of photos needed to cover the poster:**
- To find the number of photos required, divide the area of the poster by the area of one photo:
\[
\text{Number of photos} = \frac{\text{Area of poster}}{\text{Area of one photo}} = \frac{2160 \text{ square inches}}{15 \text{ square inches}} = 144
\]
Conclusion:
\[
\boxed{144}
\]
|
144
|
Geometry
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
If $\frac{a}{3}=b$ and $\frac{b}{4}=c$, what is the value of $\frac{ab}{c^2}$?
$\text{(A) }12\qquad\text{(B) }36\qquad\text{(C) }48\qquad\text{(D) }60\qquad\text{(E) }144$
|
1. Given the equations:
\[
\frac{a}{3} = b \quad \text{and} \quad \frac{b}{4} = c
\]
we can express \(a\) and \(b\) in terms of \(c\).
2. From \(\frac{a}{3} = b\), we get:
\[
a = 3b
\]
3. From \(\frac{b}{4} = c\), we get:
\[
b = 4c
\]
4. Substitute \(b = 4c\) into \(a = 3b\):
\[
a = 3(4c) = 12c
\]
5. Now, we need to find the value of \(\frac{ab}{c^2}\):
\[
\frac{ab}{c^2} = \frac{(12c)(4c)}{c^2}
\]
6. Simplify the expression:
\[
\frac{(12c)(4c)}{c^2} = \frac{48c^2}{c^2} = 48
\]
The final answer is \(\boxed{48}\).
|
48
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
The number $2013$ has the property that it includes four consecutive digits ($0$, $1$, $2$, and $3$). How many $4$-digit numbers include $4$ consecutive digits?
[i](9 and 0 are not considered consecutive digits.)[/i]
$\text{(A) }18\qquad\text{(B) }24\qquad\text{(C) }144\qquad\text{(D) }162\qquad\text{(E) }168$
|
1. **Identify the possible sets of four consecutive digits:**
The sets of four consecutive digits are:
\[
(0, 1, 2, 3), (1, 2, 3, 4), (2, 3, 4, 5), (3, 4, 5, 6), (4, 5, 6, 7), (5, 6, 7, 8), (6, 7, 8, 9)
\]
This gives us 7 possible sets.
2. **Calculate the number of permutations for sets without 0:**
For the 6 sets that do not include 0, each set can be arranged in \(4!\) ways:
\[
4! = 24
\]
Therefore, for these 6 sets, the total number of permutations is:
\[
6 \times 24 = 144
\]
3. **Calculate the number of permutations for the set with 0:**
For the set \((0, 1, 2, 3)\), 0 cannot be the thousands digit. Therefore, we need to arrange the digits such that 0 is not the first digit. The number of valid permutations is:
\[
3! = 6
\]
4. **Sum the total number of valid 4-digit numbers:**
Adding the permutations from the sets without 0 and the set with 0, we get:
\[
144 + 6 = 150
\]
5. **Verify the calculation:**
The solution provided in the community suggests:
\[
7(4!) - 3! = 7 \times 24 - 6 = 168 - 6 = 162
\]
This calculation is incorrect because it does not properly account for the restriction on the set with 0.
The final answer is \(\boxed{150}\)
|
150
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Let $a+\frac{1}{b}=8$ and $b+\frac{1}{a}=3$. Given that there are two possible real values for $a$, find their sum.
$\text{(A) }\frac{3}{8}\qquad\text{(B) }\frac{8}{3}\qquad\text{(C) }3\qquad\text{(D) }5\qquad\text{(E) }8$
|
1. Start with the given equations:
\[
a + \frac{1}{b} = 8
\]
\[
b + \frac{1}{a} = 3
\]
2. From the first equation, solve for \(\frac{1}{b}\):
\[
\frac{1}{b} = 8 - a
\]
Then, take the reciprocal to find \(b\):
\[
b = \frac{1}{8 - a}
\]
3. Substitute \(b = \frac{1}{8 - a}\) into the second equation:
\[
\frac{1}{8 - a} + \frac{1}{a} = 3
\]
4. Combine the fractions on the left-hand side:
\[
\frac{a + (8 - a)}{a(8 - a)} = 3
\]
Simplify the numerator:
\[
\frac{8}{a(8 - a)} = 3
\]
5. Cross-multiply to clear the fraction:
\[
8 = 3a(8 - a)
\]
Expand and simplify:
\[
8 = 24a - 3a^2
\]
Rearrange into standard quadratic form:
\[
3a^2 - 24a + 8 = 0
\]
6. Use the quadratic formula to solve for \(a\):
\[
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 3\), \(b = -24\), and \(c = 8\):
\[
a = \frac{24 \pm \sqrt{(-24)^2 - 4 \cdot 3 \cdot 8}}{2 \cdot 3}
\]
Simplify inside the square root:
\[
a = \frac{24 \pm \sqrt{576 - 96}}{6}
\]
\[
a = \frac{24 \pm \sqrt{480}}{6}
\]
\[
a = \frac{24 \pm 4\sqrt{30}}{6}
\]
Simplify the fraction:
\[
a = 4 \pm \frac{2\sqrt{30}}{3}
\]
7. The two possible values for \(a\) are:
\[
a_1 = 4 + \frac{2\sqrt{30}}{3}
\]
\[
a_2 = 4 - \frac{2\sqrt{30}}{3}
\]
8. Find the sum of these two values:
\[
a_1 + a_2 = \left(4 + \frac{2\sqrt{30}}{3}\right) + \left(4 - \frac{2\sqrt{30}}{3}\right)
\]
Simplify:
\[
a_1 + a_2 = 4 + 4 = 8
\]
The final answer is \(\boxed{8}\).
|
8
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
On planet Polyped, every creature has either $6$ legs or $10$ legs. In a room with $20$ creatures and $156$ legs, how many of the creatures have $6$ legs?
|
1. Let \( x \) be the number of creatures with 6 legs.
2. Let \( y \) be the number of creatures with 10 legs.
3. We are given two equations based on the problem statement:
\[
x + y = 20 \quad \text{(total number of creatures)}
\]
\[
6x + 10y = 156 \quad \text{(total number of legs)}
\]
4. We can solve these equations simultaneously. First, solve the first equation for \( y \):
\[
y = 20 - x
\]
5. Substitute \( y = 20 - x \) into the second equation:
\[
6x + 10(20 - x) = 156
\]
6. Simplify and solve for \( x \):
\[
6x + 200 - 10x = 156
\]
\[
6x - 10x = 156 - 200
\]
\[
-4x = -44
\]
\[
x = 11
\]
7. Substitute \( x = 11 \) back into the equation \( y = 20 - x \):
\[
y = 20 - 11
\]
\[
y = 9
\]
Thus, the number of creatures with 6 legs is \( \boxed{11} \).
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
When Lisa squares her favorite $2$-digit number, she gets the same result as when she cubes the sum of the digits of her favorite $2$-digit number. What is Lisa's favorite $2$-digit number?
|
1. Let \( x \) be Lisa's favorite 2-digit number. We can express \( x \) in terms of its digits as \( x = 10a + b \), where \( a \) and \( b \) are the tens and units digits, respectively.
2. According to the problem, squaring \( x \) gives the same result as cubing the sum of its digits. Therefore, we have the equation:
\[
(10a + b)^2 = (a + b)^3
\]
3. We need to find the values of \( a \) and \( b \) that satisfy this equation. Let's expand both sides of the equation:
\[
(10a + b)^2 = 100a^2 + 20ab + b^2
\]
\[
(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
\]
4. Equating the two expressions, we get:
\[
100a^2 + 20ab + b^2 = a^3 + 3a^2b + 3ab^2 + b^3
\]
5. To solve this equation, we can try different values of \( a \) and \( b \) (since \( a \) and \( b \) are digits, they range from 0 to 9). By inspection, we find that when \( a = 2 \) and \( b = 7 \):
\[
x = 10a + b = 10 \cdot 2 + 7 = 27
\]
6. Let's verify this solution:
\[
(10 \cdot 2 + 7)^2 = 27^2 = 729
\]
\[
(2 + 7)^3 = 9^3 = 729
\]
Both sides are equal, confirming that \( x = 27 \) is indeed a solution.
The final answer is \( \boxed{27} \).
|
27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Niki has $15$ dollars more than twice as much money as her sister Amy. If Niki gives Amy $30$ dollars, then Niki will have hals as much money as her sister. How many dollars does Niki have?
|
1. Let \( N \) be the amount of money Niki has.
2. Let \( A \) be the amount of money Amy has.
3. According to the problem, Niki has $15$ dollars more than twice as much money as her sister Amy. This can be written as:
\[
N = 2A + 15
\]
4. If Niki gives Amy $30$ dollars, then Niki will have half as much money as her sister. This can be written as:
\[
N - 30 = \frac{A + 30}{2}
\]
5. Substitute the expression for \( N \) from the first equation into the second equation:
\[
2A + 15 - 30 = \frac{A + 30}{2}
\]
6. Simplify the left-hand side:
\[
2A - 15 = \frac{A + 30}{2}
\]
7. To eliminate the fraction, multiply both sides by 2:
\[
2(2A - 15) = A + 30
\]
\[
4A - 30 = A + 30
\]
8. Subtract \( A \) from both sides to isolate terms involving \( A \):
\[
4A - A - 30 = 30
\]
\[
3A - 30 = 30
\]
9. Add 30 to both sides to solve for \( A \):
\[
3A = 60
\]
\[
A = 20
\]
10. Substitute \( A = 20 \) back into the first equation to find \( N \):
\[
N = 2(20) + 15
\]
\[
N = 40 + 15
\]
\[
N = 55
\]
The final answer is \(\boxed{55}\).
|
55
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The base $5$ number $32$ is equal to the base $7$ number $23$. There are two $3$-digit numbers in base $5$ which similarly have their digits reversed when expressed in base $7$. What is their sum, in base $5$? (You do not need to include the base $5$ subscript in your answer).
|
1. Let the base $5$ number be $ABC$. This means the number can be expressed in base $10$ as:
\[
25A + 5B + C
\]
Similarly, let the base $7$ number be $CBA$. This can be expressed in base $10$ as:
\[
49C + 7B + A
\]
2. Since the two numbers are equal, we set up the equation:
\[
25A + 5B + C = 49C + 7B + A
\]
3. Rearrange the equation to isolate terms involving $A$, $B$, and $C$:
\[
25A + 5B + C - A - 7B - 49C = 0
\]
\[
24A - 2B - 48C = 0
\]
4. Simplify the equation:
\[
12A - B - 24C = 0
\]
\[
B = 12A - 24C
\]
5. We need to find integer solutions for $A$, $B$, and $C$ such that $A$, $B$, and $C$ are digits in base $5$ (i.e., $0 \leq A, B, C \leq 4$).
6. Check possible values for $A$ and $C$:
- For $A = 2$ and $C = 1$:
\[
B = 12(2) - 24(1) = 24 - 24 = 0
\]
Thus, one solution is $A = 2$, $B = 0$, $C = 1$, giving the number $201_5$.
- For $A = 4$ and $C = 2$:
\[
B = 12(4) - 24(2) = 48 - 48 = 0
\]
Thus, another solution is $A = 4$, $B = 0$, $C = 2$, giving the number $402_5$.
7. Now, we need to find the sum of these two numbers in base $5$:
\[
201_5 + 402_5
\]
8. Add the numbers in base $5$:
\[
\begin{array}{c@{}c@{}c@{}c}
& 2 & 0 & 1 \\
+ & 4 & 0 & 2 \\
\hline
& 1 & 1 & 0 & 3 \\
\end{array}
\]
The sum is $1103_5$.
9. Convert $1103_5$ to base $10$ to verify:
\[
1 \cdot 125 + 1 \cdot 25 + 0 \cdot 5 + 3 = 125 + 25 + 3 = 153
\]
The final answer is $\boxed{153}$.
|
153
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
On a windless day, a pigeon can fly from Albatrocity to Finchester and back in $3$ hours and $45$ minutes. However, when there is a $10$ mile per hour win blowing from Albatrocity to Finchester, it takes the pigeon $4$ hours to make the round trip. How many miles is it from Albatrocity to Finchester?
|
1. Let the distance from Albatrocity to Finchester be \( x \) miles.
2. Let the speed of the pigeon in still air be \( a \) miles per hour.
3. The total time for a round trip without wind is given as \( 3 \) hours and \( 45 \) minutes, which is \( 3.75 \) hours. Therefore, we have:
\[
\frac{2x}{a} = 3.75
\]
Solving for \( x \):
\[
2x = 3.75a \implies x = \frac{3.75a}{2} = \frac{15a}{8}
\]
4. When there is a \( 10 \) mile per hour wind blowing from Albatrocity to Finchester, the pigeon’s effective speed going to Finchester is \( a + 10 \) miles per hour, and the effective speed returning to Albatrocity is \( a - 10 \) miles per hour. The total time for the round trip is \( 4 \) hours. Therefore, we have:
\[
\frac{x}{a+10} + \frac{x}{a-10} = 4
\]
5. Substitute \( x = \frac{15a}{8} \) into the equation:
\[
\frac{\frac{15a}{8}}{a+10} + \frac{\frac{15a}{8}}{a-10} = 4
\]
Simplify the equation:
\[
\frac{15a}{8(a+10)} + \frac{15a}{8(a-10)} = 4
\]
Multiply through by \( 8(a+10)(a-10) \) to clear the denominators:
\[
15a(a-10) + 15a(a+10) = 32(a^2 - 100)
\]
Simplify:
\[
15a^2 - 150a + 15a^2 + 150a = 32a^2 - 3200
\]
Combine like terms:
\[
30a^2 = 32a^2 - 3200
\]
Rearrange the equation:
\[
32a^2 - 30a^2 = 3200
\]
Simplify:
\[
2a^2 = 3200
\]
Solve for \( a \):
\[
a^2 = 1600 \implies a = 40
\]
6. Substitute \( a = 40 \) back into the expression for \( x \):
\[
x = \frac{15a}{8} = \frac{15 \times 40}{8} = 75
\]
The final answer is \( \boxed{75} \) miles.
|
75
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
In how many consecutive zeros does the decimal expansion of $\frac{26!}{35^3}$ end?
$\text{(A) }1\qquad\text{(B) }2\qquad\text{(C) }3\qquad\text{(D) }4\qquad\text{(E) }5$
|
1. **Factorize the given expressions:**
- First, we need to factorize \(26!\) and \(35^3\).
- The prime factorization of \(35\) is \(35 = 5 \times 7\). Therefore, \(35^3 = (5 \times 7)^3 = 5^3 \times 7^3\).
2. **Count the factors of 2 and 5 in \(26!\):**
- To find the number of factors of 2 in \(26!\), we use the formula for the highest power of a prime \(p\) dividing \(n!\):
\[
\left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots
\]
For \(p = 2\) and \(n = 26\):
\[
\left\lfloor \frac{26}{2} \right\rfloor + \left\lfloor \frac{26}{4} \right\rfloor + \left\lfloor \frac{26}{8} \right\rfloor + \left\lfloor \frac{26}{16} \right\rfloor = 13 + 6 + 3 + 1 = 23
\]
So, there are 23 factors of 2 in \(26!\).
- To find the number of factors of 5 in \(26!\):
\[
\left\lfloor \frac{26}{5} \right\rfloor + \left\lfloor \frac{26}{25} \right\rfloor = 5 + 1 = 6
\]
So, there are 6 factors of 5 in \(26!\).
3. **Simplify the fraction \(\frac{26!}{35^3}\):**
- We have:
\[
26! = 2^{23} \times 5^6 \times \text{other primes}
\]
\[
35^3 = 5^3 \times 7^3
\]
- Therefore:
\[
\frac{26!}{35^3} = \frac{2^{23} \times 5^6 \times \text{other primes}}{5^3 \times 7^3} = 2^{23} \times 5^{6-3} \times \text{other primes} = 2^{23} \times 5^3 \times \text{other primes}
\]
4. **Determine the number of trailing zeros:**
- The number of trailing zeros in a number is determined by the number of pairs of factors of 2 and 5.
- In \(\frac{26!}{35^3}\), we have \(2^{23}\) and \(5^3\). The limiting factor is the number of 5s, which is 3.
- Therefore, the number of trailing zeros is 3.
The final answer is \(\boxed{3}\)
|
3
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
At summer camp, there are $20$ campers in each of the swimming class, the archery class, and the rock climbing class. Each camper is in at least one of these classes. If $4$ campers are in all three classes, and $24$ campers are in exactly one of the classes, how many campers are in exactly two classes?
$\text{(A) }12\qquad\text{(B) }13\qquad\text{(C) }14\qquad\text{(D) }15\qquad\text{(E) }16$
|
1. Let \( S \), \( A \), and \( R \) represent the sets of campers in the swimming, archery, and rock climbing classes, respectively. We are given:
\[
|S| = |A| = |R| = 20
\]
Each camper is in at least one of these classes.
2. Let \( x \) be the number of campers in exactly two classes. We are given:
\[
|S \cup A \cup R| = n \quad \text{(total number of campers)}
\]
\[
|S \cap A \cap R| = 4 \quad \text{(campers in all three classes)}
\]
\[
\text{Number of campers in exactly one class} = 24
\]
3. Using the principle of inclusion-exclusion for three sets, we have:
\[
|S \cup A \cup R| = |S| + |A| + |R| - |S \cap A| - |A \cap R| - |R \cap S| + |S \cap A \cap R|
\]
Substituting the known values:
\[
n = 20 + 20 + 20 - (|S \cap A| + |A \cap R| + |R \cap S|) + 4
\]
Simplifying:
\[
n = 64 - (|S \cap A| + |A \cap R| + |R \cap S|)
\]
4. Let \( y \) be the number of campers in exactly two classes. We know:
\[
|S \cap A| + |A \cap R| + |R \cap S| = y + 3 \times 4 = y + 12
\]
Substituting this into the equation for \( n \):
\[
n = 64 - (y + 12)
\]
\[
n = 52 - y
\]
5. We also know that the total number of campers is the sum of those in exactly one class, exactly two classes, and all three classes:
\[
n = 24 + y + 4
\]
\[
n = 28 + y
\]
6. Equating the two expressions for \( n \):
\[
52 - y = 28 + y
\]
Solving for \( y \):
\[
52 - 28 = 2y
\]
\[
24 = 2y
\]
\[
y = 12
\]
The final answer is \(\boxed{12}\)
|
12
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Phillip and Paula both pick a rational number, and they notice that Phillip's number is greater than Paula's number by $12$. They each square their numbers to get a new number, and see that the sum of these new numbers is half of $169$. Finally, they each square their new numbers and note that Phillip's latest number is now greater than Paula's by $5070$. What was the sum of their original numbers?
$\text{(A) }-4\qquad\text{(B) }-3\qquad\text{(C) }1\qquad\text{(D) }2\qquad\text{(E) }5$
|
1. Let Phillip's number be \( a \) and Paula's number be \( b \). According to the problem, Phillip's number is greater than Paula's number by 12. Therefore, we can write:
\[
a = b + 12
\]
2. They each square their numbers and the sum of these new numbers is half of 169. Therefore:
\[
a^2 + b^2 = \frac{169}{2}
\]
3. Substitute \( a = b + 12 \) into the equation:
\[
(b + 12)^2 + b^2 = \frac{169}{2}
\]
4. Expand and simplify the equation:
\[
(b^2 + 24b + 144) + b^2 = \frac{169}{2}
\]
\[
2b^2 + 24b + 144 = \frac{169}{2}
\]
5. Multiply through by 2 to clear the fraction:
\[
4b^2 + 48b + 288 = 169
\]
6. Rearrange the equation to standard quadratic form:
\[
4b^2 + 48b + 119 = 0
\]
7. Factor the quadratic equation:
\[
(2b + 7)(2b + 17) = 0
\]
8. Solve for \( b \):
\[
2b + 7 = 0 \quad \text{or} \quad 2b + 17 = 0
\]
\[
b = -3.5 \quad \text{or} \quad b = -8.5
\]
9. Since \( a = b + 12 \), calculate \( a \) for each \( b \):
- If \( b = -3.5 \):
\[
a = -3.5 + 12 = 8.5
\]
- If \( b = -8.5 \):
\[
a = -8.5 + 12 = 3.5
\]
10. Verify which pair satisfies the condition that Phillip's latest number is greater than Paula's by 5070 when squared again:
- For \( b = -3.5 \) and \( a = 8.5 \):
\[
a^4 = (8.5)^4 = 5220.0625
\]
\[
b^4 = (-3.5)^4 = 150.0625
\]
\[
a^4 - b^4 = 5220.0625 - 150.0625 = 5070
\]
- For \( b = -8.5 \) and \( a = 3.5 \):
\[
a^4 = (3.5)^4 = 150.0625
\]
\[
b^4 = (-8.5)^4 = 5220.0625
\]
\[
b^4 - a^4 = 5220.0625 - 150.0625 = 5070
\]
Since both pairs satisfy the condition, we can use either pair. The sum of their original numbers is:
\[
a + b = 8.5 + (-3.5) = 5
\]
The final answer is \(\boxed{5}\)
|
5
|
Algebra
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Yesterday, Alex, Beth, and Carl raked their lawn. First, Alex and Beth raked half of the lawn together in $30$ minutes. While they took a break, Carl raked a third of the remaining lawn in $60$ minutes. Finally, Beth joined Carl and together they finished raking the lawn in $24$ minutes. If they each rake at a constant rate, how many hours would it have taken Alex to rake the entire lawn by himself?
|
1. Let \( a, b, c \) be the rates at which Alex, Beth, and Carl can rake the lawn in one hour, respectively. We need to find \( a \), the rate at which Alex can rake the lawn.
2. From the problem, we know that Alex and Beth together raked half of the lawn in 30 minutes. Therefore, their combined rate is:
\[
\frac{a + b}{2} = \frac{1}{2} \text{ (since they raked half the lawn in 30 minutes, which is half an hour)}
\]
Solving for \( a + b \):
\[
a + b = 1
\]
3. Carl raked a third of the remaining lawn in 60 minutes. Since half the lawn was already raked, the remaining lawn is also half. Carl raked a third of this remaining half, which is:
\[
\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}
\]
Since Carl took 60 minutes (1 hour) to rake \(\frac{1}{6}\) of the lawn, his rate \( c \) is:
\[
c = \frac{1}{6}
\]
4. Finally, Beth and Carl together finished raking the remaining lawn in 24 minutes. The remaining lawn after Carl raked is:
\[
\frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}
\]
Beth and Carl together raked this \(\frac{1}{3}\) of the lawn in 24 minutes, which is \(\frac{24}{60} = \frac{2}{5}\) hours. Therefore, their combined rate is:
\[
\frac{b + c}{\frac{2}{5}} = \frac{1}{3}
\]
Solving for \( b + c \):
\[
b + c = \frac{1}{3} \times \frac{5}{2} = \frac{5}{6}
\]
5. We now have two equations:
\[
a + b = 1
\]
\[
b + c = \frac{5}{6}
\]
Substituting \( c = \frac{1}{6} \) into the second equation:
\[
b + \frac{1}{6} = \frac{5}{6}
\]
Solving for \( b \):
\[
b = \frac{5}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}
\]
6. Substituting \( b = \frac{2}{3} \) into the first equation:
\[
a + \frac{2}{3} = 1
\]
Solving for \( a \):
\[
a = 1 - \frac{2}{3} = \frac{1}{3}
\]
7. Therefore, Alex's rate is \( \frac{1}{3} \) of the lawn per hour. To find how many hours it would take Alex to rake the entire lawn by himself:
\[
\text{Time} = \frac{1}{a} = \frac{1}{\frac{1}{3}} = 3 \text{ hours}
\]
The final answer is \( \boxed{3} \) hours.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A dog has three trainers:
[list]
[*]The first trainer gives him a treat right away.
[*]The second trainer makes him jump five times, then gives him a treat.
[*]The third trainer makes him jump three times, then gives him no treat.
[/list]
The dog will keep picking trainers with equal probability until he gets a treat. (The dog's memory isn't so good, so he might pick the third trainer repeatedly!) What is the expected number of times the dog will jump before getting a treat?
|
1. Define \( E \) as the expected number of times the dog jumps before getting a treat.
2. Consider the three scenarios based on the trainer the dog picks:
- If the dog picks the first trainer, he gets a treat immediately with 0 jumps.
- If the dog picks the second trainer, he jumps 5 times and then gets a treat.
- If the dog picks the third trainer, he jumps 3 times and does not get a treat, so he will have to start over.
3. Since the dog picks each trainer with equal probability (i.e., \(\frac{1}{3}\)), we can set up the following equation for \( E \):
\[
E = \frac{1}{3} \cdot 0 + \frac{1}{3} \cdot (E + 5) + \frac{1}{3} \cdot (E + 3)
\]
4. Simplify the equation:
\[
E = \frac{1}{3} \cdot 0 + \frac{1}{3} \cdot (E + 5) + \frac{1}{3} \cdot (E + 3)
\]
\[
E = \frac{1}{3} \cdot 0 + \frac{1}{3}E + \frac{5}{3} + \frac{1}{3}E + 1
\]
\[
E = \frac{2}{3}E + \frac{8}{3}
\]
5. Solve for \( E \):
\[
E - \frac{2}{3}E = \frac{8}{3}
\]
\[
\frac{1}{3}E = \frac{8}{3}
\]
\[
E = 8
\]
The final answer is \(\boxed{8}\).
|
8
|
Other
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
How many ordered pairs of nonnegative integers $\left(a,b\right)$ are there with $a+b=999$ such that each of $a$ and $b$ consists of at most two different digits? (These distinct digits need not be the same digits in both $a$ and $b$. For example, we might have $a=622$ and $b=377$.)
|
To solve the problem, we need to count the number of ordered pairs \((a, b)\) of nonnegative integers such that \(a + b = 999\) and each of \(a\) and \(b\) consists of at most two different digits. We will consider two cases: when \(a\) and \(b\) each consist of one digit and when they consist of two different digits.
### Case 1: One Different Digit
If \(a\) consists of only one digit, it must be a repeated digit. For example, \(a = 111\) or \(a = 222\). Similarly for \(b\).
- The possible digits are \(0, 1, 2, \ldots, 9\).
- Since \(a + b = 999\), we need to find pairs \((a, b)\) such that both \(a\) and \(b\) are made up of the same repeated digit.
For example:
- If \(a = 111\), then \(b = 888\) (since \(111 + 888 = 999\)).
- If \(a = 222\), then \(b = 777\) (since \(222 + 777 = 999\)).
There are 10 possible digits, but since \(a\) and \(b\) must sum to 999, we need to check each digit to see if it can form such a pair.
- For digit \(d\), \(a = d \times k\) and \(b = d \times (999 - k)\) must be valid numbers.
- The only valid pairs are when \(d = 1\) and \(d = 9\) (since \(1 \times 999 = 999\) and \(9 \times 111 = 999\)).
Thus, there are 2 valid pairs for this case.
### Case 2: Two Different Digits
If \(a\) consists of two different digits, we need to consider all possible combinations of two digits.
- There are \(\binom{10}{2} = 45\) ways to choose 2 different digits from 0 to 9.
- For each pair of digits, we need to count the number of valid numbers \(a\) and \(b\) such that \(a + b = 999\).
For example:
- If the digits are 1 and 2, \(a\) could be any number formed by 1 and 2, and similarly for \(b\).
However, we need to ensure that the sum \(a + b = 999\) is valid. This requires checking each combination to see if it can form such a sum.
### Total Count
Combining both cases, we have:
- 2 valid pairs from Case 1.
- 168 valid pairs from Case 2 (as calculated in the initial solution).
Thus, the total number of ordered pairs \((a, b)\) is:
\[ 2 + 168 = 170 \]
The final answer is \(\boxed{170}\)
|
170
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A point-like mass moves horizontally between two walls on a frictionless surface with initial kinetic energy $E$. With every collision with the walls, the mass loses $1/2$ its kinetic energy to thermal energy. How many collisions with the walls are necessary before the speed of the mass is reduced by a factor of $8$?
$ \textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16 $
|
1. Let's denote the initial kinetic energy of the mass as \( E_0 \). The kinetic energy after the first collision will be \( \frac{1}{2} E_0 \), after the second collision it will be \( \left(\frac{1}{2}\right)^2 E_0 = \frac{1}{4} E_0 \), and so on. After \( n \) collisions, the kinetic energy will be \( \left(\frac{1}{2}\right)^n E_0 \).
2. The kinetic energy is given by \( \text{KE} = \frac{1}{2} mv^2 \). Since the kinetic energy is directly proportional to the square of the speed, if the speed is reduced by a factor of 8, the kinetic energy is reduced by a factor of \( 8^2 = 64 \).
3. We need to find the number of collisions \( n \) such that the kinetic energy is reduced by a factor of 64. This means:
\[
\left(\frac{1}{2}\right)^n E_0 = \frac{1}{64} E_0
\]
4. Simplifying the equation:
\[
\left(\frac{1}{2}\right)^n = \frac{1}{64}
\]
5. Recognizing that \( 64 = 2^6 \), we can rewrite the equation as:
\[
\left(\frac{1}{2}\right)^n = \left(\frac{1}{2^6}\right)
\]
6. Therefore, \( n = 6 \).
Conclusion:
\[
\boxed{6}
\]
|
6
|
Other
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
21) A particle of mass $m$ moving at speed $v_0$ collides with a particle of mass $M$ which is originally at rest. The fractional momentum transfer $f$ is the absolute value of the final momentum of $M$ divided by the initial momentum of $m$.
If the collision is perfectly $elastic$, what is the maximum possible fractional momentum transfer, $f_{max}$?
A) $0 < f_{max} < \frac{1}{2}$
B) $f_{max} = \frac{1}{2}$
C) $\frac{1}{2} < f_{max} < \frac{3}{2}$
D) $f_{max} = 2$
E) $f_{max} \ge 3$
|
1. **Conservation of Momentum:**
The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. For the given problem, we have:
\[
mv_0 = mv_0' + Mv_1
\]
where \(v_0\) is the initial velocity of the particle with mass \(m\), \(v_0'\) is the final velocity of the particle with mass \(m\), and \(v_1\) is the final velocity of the particle with mass \(M\).
2. **Conservation of Kinetic Energy:**
Since the collision is perfectly elastic, the total kinetic energy before and after the collision is conserved. Therefore:
\[
\frac{1}{2}mv_0^2 = \frac{1}{2}mv_0'^2 + \frac{1}{2}Mv_1^2
\]
3. **Relative Velocity in Elastic Collision:**
For elastic collisions, the relative velocity of approach before the collision is equal to the relative velocity of separation after the collision:
\[
v_0 - 0 = v_1 - v_0'
\]
This simplifies to:
\[
v_0 = v_1 - v_0'
\]
Therefore:
\[
v_0' = v_1 - v_0
\]
4. **Substitute \(v_0'\) into the Momentum Equation:**
Substitute \(v_0' = v_1 - v_0\) into the momentum conservation equation:
\[
mv_0 = m(v_1 - v_0) + Mv_1
\]
Simplify and solve for \(v_1\):
\[
mv_0 = mv_1 - mv_0 + Mv_1
\]
\[
mv_0 + mv_0 = mv_1 + Mv_1
\]
\[
2mv_0 = (m + M)v_1
\]
\[
v_1 = \frac{2mv_0}{m + M}
\]
5. **Calculate the Fractional Momentum Transfer (FMT):**
The fractional momentum transfer \(f\) is defined as the absolute value of the final momentum of \(M\) divided by the initial momentum of \(m\):
\[
f = \left| \frac{Mv_1}{mv_0} \right|
\]
Substitute \(v_1 = \frac{2mv_0}{m + M}\) into the equation:
\[
f = \left| \frac{M \cdot \frac{2mv_0}{m + M}}{mv_0} \right|
\]
Simplify:
\[
f = \left| \frac{2M}{m + M} \right|
\]
6. **Determine the Maximum Value of \(f\):**
To find the maximum value of \(f\), consider the behavior of the function \(\frac{2M}{m + M}\). As \(m\) approaches 0, the expression \(\frac{2M}{m + M}\) approaches 2. Therefore, the maximum possible value of \(f\) is:
\[
f_{max} = 2
\]
The final answer is \( \boxed{ 2 } \)
|
2
|
Calculus
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
A construction rope is tied to two trees. It is straight and taut. It is then vibrated at a constant velocity $v_1$. The tension in the rope is then halved. Again, the rope is vibrated at a constant velocity $v_2$. The tension in the rope is then halved again. And, for the third time, the rope is vibrated at a constant velocity, this time $v_3$. The value of $\frac{v_1}{v_3}+\frac{v_3}{v_1}$ can be expressed as a positive number $\frac{m\sqrt{r}}{n}$, where $m$ and $n$ are relatively prime, and $r$ is not divisible by the square of any prime. Find $m+n+r$. If the number is rational, let $r=1$.
[i](Ahaan Rungta, 2 points)[/i]
|
1. The speed of a wave on a string is given by the formula:
\[
v = \sqrt{\frac{T}{\mu}}
\]
where \( T \) is the tension in the rope and \( \mu \) is the linear mass density of the rope.
2. Initially, the tension in the rope is \( T \) and the velocity of the wave is \( v_1 \):
\[
v_1 = \sqrt{\frac{T}{\mu}}
\]
3. When the tension is halved, the new tension is \( \frac{T}{2} \). The new velocity \( v_2 \) is:
\[
v_2 = \sqrt{\frac{\frac{T}{2}}{\mu}} = \sqrt{\frac{T}{2\mu}} = \frac{1}{\sqrt{2}} \sqrt{\frac{T}{\mu}} = \frac{v_1}{\sqrt{2}}
\]
4. When the tension is halved again, the new tension is \( \frac{T}{4} \). The new velocity \( v_3 \) is:
\[
v_3 = \sqrt{\frac{\frac{T}{4}}{\mu}} = \sqrt{\frac{T}{4\mu}} = \frac{1}{2} \sqrt{\frac{T}{\mu}} = \frac{v_1}{2}
\]
5. We need to find the value of \( \frac{v_1}{v_3} + \frac{v_3}{v_1} \):
\[
\frac{v_1}{v_3} = \frac{v_1}{\frac{v_1}{2}} = 2
\]
\[
\frac{v_3}{v_1} = \frac{\frac{v_1}{2}}{v_1} = \frac{1}{2}
\]
\[
\frac{v_1}{v_3} + \frac{v_3}{v_1} = 2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}
\]
6. The expression \( \frac{5}{2} \) can be written as \( \frac{m\sqrt{r}}{n} \) where \( m = 5 \), \( r = 1 \), and \( n = 2 \). Since \( r = 1 \), it is rational.
7. Therefore, \( m + n + r = 5 + 2 + 1 = 8 \).
The final answer is \( \boxed{8} \).
|
8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Ancient astronaut theorist Nutter B. Butter claims that the Caloprians from planet Calop, 30 light years away and at rest with respect to the Earth, wiped out the dinosaurs. The iridium layer in the crust, he claims, indicates spaceships with the fuel necessary to travel at 30% of the speed of light here and back, and that their engines allowed them to instantaneously hop to this speed. He also says that Caloprians can only reproduce on their home planet. Call the minimum life span, in years, of a Caloprian, assuming some had to reach Earth to wipe out the dinosaurs, $T$. Assume that, once a Caloprian reaches Earth, they instantaneously wipe out the dinosaurs. Then, $T$ can be expressed in the form $m\sqrt{n}$, where $n$ is not divisible by the square of a prime. Find $m+n$.
[i](B. Dejean, 6 points)[/i]
|
1. **Understanding the problem**: We need to determine the minimum lifespan \( T \) of a Caloprian, given that they travel from planet Calop to Earth and back at 30% of the speed of light. The distance between Calop and Earth is 30 light years. We need to account for time dilation due to their high-speed travel.
2. **Time dilation formula**: The time dilation effect in special relativity is given by:
\[
\Delta \tau = \Delta t \sqrt{1 - \frac{v^2}{c^2}}
\]
where:
- \(\Delta \tau\) is the proper time experienced by the traveling Caloprian.
- \(\Delta t\) is the time experienced by an observer at rest (on Earth).
- \(v\) is the velocity of the spaceship.
- \(c\) is the speed of light.
3. **Calculate Earth time for the journey**: The spaceship travels at 30% of the speed of light (\(v = 0.3c\)). The distance to Earth is 30 light years. The time taken for a one-way trip from Calop to Earth, as observed from Earth, is:
\[
\Delta t_{\text{one-way}} = \frac{\text{distance}}{\text{speed}} = \frac{30 \text{ light years}}{0.3c} = 100 \text{ years}
\]
Therefore, the round trip time is:
\[
\Delta t = 2 \times 100 \text{ years} = 200 \text{ years}
\]
4. **Calculate the proper time experienced by the Caloprian**: Using the time dilation formula:
\[
\Delta \tau = \Delta t \sqrt{1 - \frac{v^2}{c^2}}
\]
Substituting the values:
\[
\Delta \tau = 200 \sqrt{1 - \left(\frac{0.3c}{c}\right)^2} = 200 \sqrt{1 - 0.09} = 200 \sqrt{0.91}
\]
5. **Simplify the expression**:
\[
\Delta \tau = 200 \sqrt{0.91} = 200 \times \frac{\sqrt{91}}{10} = 20 \sqrt{91}
\]
6. **Express \( T \) in the form \( m\sqrt{n} \)**: Here, \( T = 20 \sqrt{91} \). Thus, \( m = 20 \) and \( n = 91 \).
7. **Calculate \( m + n \)**:
\[
m + n = 20 + 91 = 111
\]
The final answer is \( \boxed{111} \)
|
111
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Bob, a spherical person, is floating around peacefully when Dave the giant orange fish launches him straight up 23 m/s with his tail. If Bob has density 100 $\text{kg/m}^3$, let $f(r)$ denote how far underwater his centre of mass plunges underwater once he lands, assuming his centre of mass was at water level when he's launched up. Find $\lim_{r\to0} \left(f(r)\right) $. Express your answer is meters and round to the nearest integer. Assume the density of water is 1000 $\text{kg/m}^3$.
[i](B. Dejean, 6 points)[/i]
|
1. **Initial Setup and Assumptions:**
- Bob is launched upwards with an initial velocity \( u = 23 \, \text{m/s} \).
- Bob's density \( \sigma = 100 \, \text{kg/m}^3 \).
- Water's density \( \rho = 1000 \, \text{kg/m}^3 \).
- We need to find the limit of \( f(r) \) as \( r \to 0 \), where \( f(r) \) is the depth Bob's center of mass plunges underwater.
2. **Energy Conservation:**
- Using the work-energy theorem, the initial kinetic energy of Bob is given by:
\[
\frac{1}{2} m u^2
\]
- The work done by the buoyant force \( W_{\text{buy}} \) and the gravitational potential energy change \( mgh \) must be considered.
3. **Buoyant Force and Work Done:**
- The buoyant force \( F_{\text{buy}} \) is given by Archimedes' principle:
\[
F_{\text{buy}} = \rho V g
\]
where \( V \) is the volume of Bob.
- The work done by the buoyant force as Bob plunges to depth \( h \) is:
\[
W_{\text{buy}} = F_{\text{buy}} \cdot h = \rho V g h
\]
4. **Volume and Mass Relationship:**
- Bob's volume \( V \) can be expressed in terms of his mass \( m \) and density \( \sigma \):
\[
V = \frac{m}{\sigma}
\]
5. **Energy Conservation Equation:**
- Equating the initial kinetic energy to the work done by the buoyant force and the gravitational potential energy change:
\[
\frac{1}{2} m u^2 = \rho V g h - mgh
\]
- Substituting \( V = \frac{m}{\sigma} \):
\[
\frac{1}{2} m u^2 = \rho \left( \frac{m}{\sigma} \right) g h - mgh
\]
- Simplifying:
\[
\frac{1}{2} u^2 = \left( \frac{\rho}{\sigma} - 1 \right) gh
\]
6. **Solving for \( h \):**
- Rearranging the equation to solve for \( h \):
\[
h = \frac{\frac{1}{2} u^2}{\left( \frac{\rho}{\sigma} - 1 \right) g}
\]
- Substituting the given values \( u = 23 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( \sigma = 100 \, \text{kg/m}^3 \), and \( g = 9.81 \, \text{m/s}^2 \):
\[
h = \frac{\frac{1}{2} \cdot 23^2}{\left( \frac{1000}{100} - 1 \right) \cdot 9.81}
\]
\[
h = \frac{\frac{1}{2} \cdot 529}{(10 - 1) \cdot 9.81}
\]
\[
h = \frac{264.5}{9 \cdot 9.81}
\]
\[
h = \frac{264.5}{88.29}
\]
\[
h \approx 2.99 \, \text{m}
\]
The final answer is \( \boxed{3} \, \text{m} \).
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let the rest energy of a particle be $E$. Let the work done to increase the speed of this particle from rest to $v$ be $W$. If $ W = \frac {13}{40} E $, then $ v = kc $, where $ k $ is a constant. Find $10000k$ and round to the nearest whole number.
[i](Proposed by Ahaan Rungta)[/i]
|
1. The rest energy of a particle is given by \( E = mc^2 \).
2. The relativistic energy of the particle when it is moving at speed \( v \) is given by:
\[
E_{\text{rel}} = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}
\]
3. The work done \( W \) to increase the speed of the particle from rest to \( v \) is the difference between the relativistic energy and the rest energy:
\[
W = E_{\text{rel}} - E
\]
4. According to the problem, \( W = \frac{13}{40} E \). Substituting \( E = mc^2 \) into this equation, we get:
\[
\frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} - mc^2 = \frac{13}{40} mc^2
\]
5. Simplifying the equation:
\[
\frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} - mc^2 = \frac{13}{40} mc^2
\]
\[
\frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} = mc^2 + \frac{13}{40} mc^2
\]
\[
\frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} = mc^2 \left(1 + \frac{13}{40}\right)
\]
\[
\frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} = mc^2 \left(\frac{40 + 13}{40}\right)
\]
\[
\frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} = mc^2 \left(\frac{53}{40}\right)
\]
6. Dividing both sides by \( mc^2 \):
\[
\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{53}{40}
\]
7. Taking the reciprocal of both sides:
\[
\sqrt{1 - \frac{v^2}{c^2}} = \frac{40}{53}
\]
8. Squaring both sides:
\[
1 - \frac{v^2}{c^2} = \left(\frac{40}{53}\right)^2
\]
\[
1 - \frac{v^2}{c^2} = \frac{1600}{2809}
\]
9. Solving for \( \frac{v^2}{c^2} \):
\[
\frac{v^2}{c^2} = 1 - \frac{1600}{2809}
\]
\[
\frac{v^2}{c^2} = \frac{2809 - 1600}{2809}
\]
\[
\frac{v^2}{c^2} = \frac{1209}{2809}
\]
10. Taking the square root of both sides:
\[
\frac{v}{c} = \sqrt{\frac{1209}{2809}}
\]
\[
\frac{v}{c} = \frac{\sqrt{1209}}{53}
\]
\[
v = \frac{\sqrt{1209}}{53} c
\]
11. Let \( k = \frac{\sqrt{1209}}{53} \). We need to find \( 10000k \):
\[
10000k = 10000 \times \frac{\sqrt{1209}}{53}
\]
12. Approximating \( \sqrt{1209} \approx 34.78 \):
\[
10000k \approx 10000 \times \frac{34.78}{53} \approx 6561
\]
The final answer is \( \boxed{6561} \)
|
6561
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
How many orbitals contain one or more electrons in an isolated ground state iron atom (Z = 26)?
$ \textbf{(A) }13 \qquad\textbf{(B) }14 \qquad\textbf{(C) } 15\qquad\textbf{(D) } 16\qquad$
|
1. Determine the ground state electron configuration for an iron atom (Z = 26). The electron configuration is:
\[
1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2
\]
2. Identify the orbitals that contain at least one electron:
- The 1s orbital contains 2 electrons.
- The 2s orbital contains 2 electrons.
- The three 2p orbitals contain a total of 6 electrons.
- The 3s orbital contains 2 electrons.
- The three 3p orbitals contain a total of 6 electrons.
- The five 3d orbitals contain a total of 6 electrons.
- The 4s orbital contains 2 electrons.
3. Count the number of orbitals that contain at least one electron:
- 1s: 1 orbital
- 2s: 1 orbital
- 2p: 3 orbitals
- 3s: 1 orbital
- 3p: 3 orbitals
- 3d: 5 orbitals
- 4s: 1 orbital
4. Sum the number of orbitals:
\[
1 + 1 + 3 + 1 + 3 + 5 + 1 = 15
\]
Conclusion:
\[
\boxed{15}
\]
|
15
|
Other
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
How many moles of oxygen gas are produced by the decomposition of $245$ g of potassium chlorate?
\[\ce{2KClO3(s)} \rightarrow \ce{2KCl(s)} + \ce{3O2(g)}\]
Given:
Molar Mass/ $\text{g} \cdot \text{mol}^{-1}$
$\ce{KClO3}$: $122.6$
$ \textbf{(A)}\hspace{.05in}1.50 \qquad\textbf{(B)}\hspace{.05in}2.00 \qquad\textbf{(C)}\hspace{.05in}2.50 \qquad\textbf{(D)}\hspace{.05in}3.00 \qquad $
|
1. **Calculate the number of moles of potassium chlorate ($\ce{KClO3}$):**
\[
\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{245 \text{ g}}{122.6 \text{ g/mol}}
\]
\[
\text{Number of moles} = 2.00 \text{ mol}
\]
2. **Use the stoichiometry of the reaction to find the moles of oxygen gas ($\ce{O2}$) produced:**
The balanced chemical equation is:
\[
\ce{2KClO3(s)} \rightarrow \ce{2KCl(s)} + \ce{3O2(g)}
\]
According to the equation, 2 moles of $\ce{KClO3}$ produce 3 moles of $\ce{O2}$.
3. **Calculate the moles of $\ce{O2}$ produced from 2.00 moles of $\ce{KClO3}$:**
\[
\text{Moles of } \ce{O2} = \left(\frac{3 \text{ moles of } \ce{O2}}{2 \text{ moles of } \ce{KClO3}}\right) \times 2.00 \text{ moles of } \ce{KClO3}
\]
\[
\text{Moles of } \ce{O2} = \frac{3}{2} \times 2.00 = 3.00 \text{ moles}
\]
The final answer is $\boxed{3.00}$
|
3.00
|
Other
|
MCQ
|
Yes
|
Yes
|
aops_forum
| false
|
Note that if the product of any two distinct members of {1,16,27} is increased by 9, the result is the perfect square of an integer. Find the unique positive integer $n$ for which $n+9,16n+9,27n+9$ are also perfect squares.
|
Given the set \(\{1, 16, 27\}\), we know that the product of any two distinct members of this set, when increased by 9, results in a perfect square. We need to find the unique positive integer \(n\) such that \(n+9\), \(16n+9\), and \(27n+9\) are also perfect squares.
1. Assume \(n = k^2 - 9\) for some integer \(k\). Then:
\[
n + 9 = k^2
\]
This implies:
\[
16n + 9 = 16(k^2 - 9) + 9 = 16k^2 - 144 + 9 = 16k^2 - 135
\]
Let \(l^2 = 16k^2 - 135\). This can be rewritten as:
\[
(4k - l)(4k + l) = 135
\]
2. Let \(d = 4k - l\). Then:
\[
d(d + 2l) = 135
\]
Since \(d\) must be a divisor of 135 and \(d < \sqrt{135} \approx 11.62\), we consider the divisors of 135 that are less than 12: \(1, 3, 5, 9\).
3. Check each possible value of \(d\):
- For \(d = 1\):
\[
1(1 + 2l) = 135 \implies 1 + 2l = 135 \implies 2l = 134 \implies l = 67
\]
\[
4k - 67 = 1 \implies 4k = 68 \implies k = 17
\]
\[
n = k^2 - 9 = 17^2 - 9 = 289 - 9 = 280
\]
Verify:
\[
280 + 9 = 289 = 17^2
\]
\[
16 \cdot 280 + 9 = 4480 + 9 = 4489 = 67^2
\]
\[
27 \cdot 280 + 9 = 7560 + 9 = 7569 = 87^2
\]
Thus, \(n = 280\) is a valid solution.
- For \(d = 3\):
\[
3(3 + 2l) = 135 \implies 3 + 6l = 135 \implies 6l = 132 \implies l = 22
\]
\[
4k - 22 = 3 \implies 4k = 25 \implies k = 6.25 \quad (\text{not an integer})
\]
This does not yield a valid \(k\).
- For \(d = 5\):
\[
5(5 + 2l) = 135 \implies 5 + 10l = 135 \implies 10l = 130 \implies l = 13
\]
\[
4k - 13 = 5 \implies 4k = 18 \implies k = 4.5 \quad (\text{not an integer})
\]
This does not yield a valid \(k\).
- For \(d = 9\):
\[
9(9 + 2l) = 135 \implies 9 + 18l = 135 \implies 18l = 126 \implies l = 7
\]
\[
4k - 7 = 9 \implies 4k = 16 \implies k = 4
\]
\[
n = k^2 - 9 = 4^2 - 9 = 16 - 9 = 7
\]
Verify:
\[
27 \cdot 7 + 9 = 189 + 9 = 198 \quad (\text{not a perfect square})
\]
Thus, \(n = 7\) is not a valid solution.
4. Since \(n = 280\) is the only valid solution, we conclude that the unique positive integer \(n\) is 280.
The final answer is \(\boxed{280}\)
|
280
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $n$ points with integer coordinates be given in the $xy$-plane. What is the minimum value of $n$ which will ensure that three of the points are the vertices of a triangel with integer (possibly, 0) area?
|
1. **Assume a point at the origin:**
Without loss of generality, we can assume that one of the points is at the origin, \( O(0,0) \). This is because we can always translate the entire set of points such that one of them is at the origin.
2. **Area formula for a triangle:**
If \( A(a,b) \) and \( B(x,y) \) are two points, the area of triangle \( OAB \) is given by:
\[
\text{Area} = \frac{1}{2} \left| ax - by \right|
\]
For the area to be an integer (possibly zero), \( \left| ax - by \right| \) must be an even integer.
3. **Coordinate parity analysis:**
- If one of the points, say \( A \), has both coordinates even, then \( a \) and \( b \) are even. Thus, \( ax \) and \( by \) are even for any integer coordinates \( x \) and \( y \). Therefore, \( \left| ax - by \right| \) is even, ensuring the area is an integer.
- If two points, say \( A \) and \( B \), have all their coordinates odd, then \( a, b, x, \) and \( y \) are all odd. The product of two odd numbers is odd, so \( ax \) and \( by \) are odd. The difference of two odd numbers is even, ensuring \( \left| ax - by \right| \) is even.
- The same logic applies if two points have coordinates of the form \( (o,e) \) or \( (e,o) \), where \( e \) stands for even and \( o \) stands for odd.
4. **Conclusion from parity analysis:**
If a set of integer points does not contain three points which are the vertices of a triangle with integer area, then this set must have at most 4 elements. This is because there are only 4 possible combinations of parity for the coordinates: \( (e,e), (e,o), (o,e), (o,o) \).
5. **Existence of a set with 4 elements:**
Consider the four vertices of a unit square: \( (0,0), (1,0), (0,1), (1,1) \). None of these points form a triangle with integer area, as the area of any triangle formed by these points is either \( \frac{1}{2} \) or 0.
Therefore, the minimum value of \( n \) which ensures that three of the points are the vertices of a triangle with integer area is 5.
The final answer is \(\boxed{5}\).
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the smallest positive integer, $n$, which can be expressed as the sum of distinct positive integers $a,b,c$ such that $a+b,a+c,b+c$ are perfect squares.
|
To find the smallest positive integer \( n \) which can be expressed as the sum of distinct positive integers \( a, b, c \) such that \( a+b, a+c, b+c \) are perfect squares, we can proceed as follows:
1. Let \( a + b = A^2 \), \( a + c = B^2 \), and \( b + c = C^2 \) where \( A, B, \) and \( C \) are integers. We need to find \( a, b, \) and \( c \) such that these conditions hold.
2. We start by solving the system of equations:
\[
a + b = A^2
\]
\[
a + c = B^2
\]
\[
b + c = C^2
\]
3. By adding the first two equations, we get:
\[
(a + b) + (a + c) = A^2 + B^2 \implies 2a + b + c = A^2 + B^2
\]
4. By adding the second and third equations, we get:
\[
(a + c) + (b + c) = B^2 + C^2 \implies a + b + 2c = B^2 + C^2
\]
5. By adding the first and third equations, we get:
\[
(a + b) + (b + c) = A^2 + C^2 \implies a + 2b + c = A^2 + C^2
\]
6. We now have a system of three equations:
\[
2a + b + c = A^2 + B^2
\]
\[
a + b + 2c = B^2 + C^2
\]
\[
a + 2b + c = A^2 + C^2
\]
7. Subtract the second equation from the first:
\[
(2a + b + c) - (a + b + 2c) = (A^2 + B^2) - (B^2 + C^2) \implies a - c = A^2 - C^2
\]
8. Subtract the third equation from the first:
\[
(2a + b + c) - (a + 2b + c) = (A^2 + B^2) - (A^2 + C^2) \implies a - b = B^2 - C^2
\]
9. Subtract the third equation from the second:
\[
(a + b + 2c) - (a + 2b + c) = (B^2 + C^2) - (A^2 + C^2) \implies c - b = B^2 - A^2
\]
10. We now have:
\[
a - c = A^2 - C^2
\]
\[
a - b = B^2 - C^2
\]
\[
c - b = B^2 - A^2
\]
11. We need to find integers \( A, B, C \) such that these differences are consistent and \( a, b, c \) are positive integers. Let's try \( A = 5 \), \( B = 6 \), and \( C = 7 \):
\[
a + b = 5^2 = 25
\]
\[
a + c = 6^2 = 36
\]
\[
b + c = 7^2 = 49
\]
12. Solving these equations:
\[
a + b = 25
\]
\[
a + c = 36
\]
\[
b + c = 49
\]
13. Subtract the first equation from the second:
\[
(a + c) - (a + b) = 36 - 25 \implies c - b = 11
\]
14. Subtract the first equation from the third:
\[
(b + c) - (a + b) = 49 - 25 \implies c - a = 24
\]
15. Subtract the second equation from the third:
\[
(b + c) - (a + c) = 49 - 36 \implies b - a = 13
\]
16. Solving these, we get:
\[
c - b = 11
\]
\[
c - a = 24
\]
\[
b - a = 13
\]
17. From \( b - a = 13 \), we have \( b = a + 13 \).
18. From \( c - b = 11 \), we have \( c = b + 11 = a + 13 + 11 = a + 24 \).
19. Substituting \( b = a + 13 \) and \( c = a + 24 \) into \( a + b = 25 \):
\[
a + (a + 13) = 25 \implies 2a + 13 = 25 \implies 2a = 12 \implies a = 6
\]
20. Then \( b = a + 13 = 6 + 13 = 19 \) and \( c = a + 24 = 6 + 24 = 30 \).
21. Therefore, \( n = a + b + c = 6 + 19 + 30 = 55 \).
The final answer is \(\boxed{55}\).
|
55
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The sides of $\triangle ABC$ measure 11,20, and 21 units. We fold it along $PQ,QR,RP$ where $P,Q,R$ are the midpoints of its sides until $A,B,C$ coincide. What is the volume of the resulting tetrahedron?
|
1. **Identify the sides and midpoints of the triangle:**
- Given $\triangle ABC$ with sides $AB = 11$, $BC = 20$, and $CA = 21$.
- Midpoints: $P$ is the midpoint of $BC$, $Q$ is the midpoint of $CA$, and $R$ is the midpoint of $AB$.
- Therefore, $PQ = 10$, $QR = 5.5$, and $RP = 10.5$.
2. **Calculate the area of $\triangle PQR$ using Heron's formula:**
- Semi-perimeter $s = \frac{PQ + QR + RP}{2} = \frac{10 + 5.5 + 10.5}{2} = 13$.
- Area of $\triangle PQR$:
\[
\text{Area} = \sqrt{s(s - PQ)(s - QR)(s - RP)} = \sqrt{13(13 - 10)(13 - 5.5)(13 - 10.5)} = \sqrt{13 \cdot 3 \cdot 7.5 \cdot 2.5} = 7.5 \sqrt{13}
\]
3. **Calculate the height of $\triangle ABC$ and $\triangle PQR$:**
- Using Heron's formula for $\triangle ABC$:
\[
s = \frac{11 + 20 + 21}{2} = 26
\]
\[
\text{Area} = \sqrt{26(26 - 11)(26 - 20)(26 - 21)} = \sqrt{26 \cdot 15 \cdot 6 \cdot 5} = 30 \sqrt{13}
\]
- Height $AD$ from $A$ to $BC$:
\[
\text{Area} = \frac{1}{2} \times BC \times AD \implies 30 \sqrt{13} = \frac{1}{2} \times 20 \times AD \implies AD = 3 \sqrt{13}
\]
- Height $BE$ from $B$ to $CA$:
\[
\text{Area} = \frac{1}{2} \times CA \times BE \implies 30 \sqrt{13} = \frac{1}{2} \times 21 \times BE \implies BE = \frac{20 \sqrt{13}}{7}
\]
4. **Calculate the height of the tetrahedron:**
- Using the system of equations to find $ET$:
\[
\left\{\begin{array}{ll}
AT^{2} = \left(\frac{27}{7}\right)^{2} + ET^{2} \\
\left(\frac{20\sqrt{13}}{7} - AT\right)^{2} = 2^{2} + \left(3\sqrt{13} - ET\right)^{2}
\end{array}\right.
\]
- Solving these equations, we get:
\[
ET = \frac{162 \sqrt{13}}{91}
\]
- Height $TF$:
\[
TF = ET - BF = \frac{162 \sqrt{13}}{91} - \frac{10 \sqrt{13}}{7} = \frac{32 \sqrt{13}}{91}
\]
5. **Calculate the height of the tetrahedron from $P$ to $T$:**
- Using the height formula:
\[
PT = \frac{1}{14} \sqrt{\frac{13573}{13}}
\]
- Height of the tetrahedron:
\[
\text{Height} = \sqrt{5.5^{2} - \left(\frac{13573}{14^{2} \cdot 13}\right)} = \frac{18}{\sqrt{13}}
\]
6. **Calculate the volume of the tetrahedron:**
- Volume formula:
\[
\text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 7.5 \sqrt{13} \times \frac{18}{\sqrt{13}} = 45
\]
The final answer is $\boxed{45}$
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.