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Define $a_k = (k^2 + 1)k!$ and $b_k = a_1 + a_2 + a_3 + \cdots + a_k$. Let \[\frac{a_{100}}{b_{100}} = \frac{m}{n}\] where $m$ and $n$ are relatively prime natural numbers. Find $n - m$.
|
1. Define the sequences \(a_k\) and \(b_k\) as follows:
\[
a_k = (k^2 + 1)k!
\]
\[
b_k = a_1 + a_2 + a_3 + \cdots + a_k
\]
2. We start by simplifying \(a_k\):
\[
a_k = (k^2 + 1)k! = k^2 k! + k! = k(k \cdot k!) + k! = k(k+1)! - k \cdot k! + k!
\]
\[
= k(k+1)! - k!(k-1)
\]
3. Next, we express \(b_k\) in terms of the sum of \(a_i\):
\[
b_k = a_1 + a_2 + a_3 + \cdots + a_k
\]
Using the expression for \(a_k\) derived above:
\[
b_k = \sum_{i=1}^k \left[ i(i+1)! - i!(i-1) \right]
\]
4. Notice that this sum is telescoping:
\[
b_k = \left[ 1 \cdot 2! - 1! \cdot 0 \right] + \left[ 2 \cdot 3! - 2! \cdot 1 \right] + \left[ 3 \cdot 4! - 3! \cdot 2 \right] + \cdots + \left[ k \cdot (k+1)! - k! \cdot (k-1) \right]
\]
Most terms cancel out, leaving:
\[
b_k = k(k+1)!
\]
5. Now, we need to find \(\frac{a_{100}}{b_{100}}\):
\[
\frac{a_{100}}{b_{100}} = \frac{(100^2 + 1)100!}{100 \cdot 101!}
\]
Simplify the fraction:
\[
\frac{a_{100}}{b_{100}} = \frac{100^2 + 1}{100 \cdot 101} = \frac{10000 + 1}{100 \cdot 101} = \frac{10001}{10100}
\]
6. We need to ensure that the fraction \(\frac{10001}{10100}\) is in its simplest form. Since 10001 and 10100 are relatively prime (they have no common factors other than 1), the fraction is already simplified.
7. Let \(\frac{10001}{10100} = \frac{m}{n}\), where \(m = 10001\) and \(n = 10100\). We are asked to find \(n - m\):
\[
n - m = 10100 - 10001 = 99
\]
The final answer is \(\boxed{99}\)
|
99
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
In the addition problem
\[ \setlength{\tabcolsep}{1mm}\begin{tabular}{cccccc}& W & H & I & T & E\\ + & W & A & T & E & R \\\hline P & I & C & N & I & C\end{tabular} \] each distinct letter represents a different digit. Find the number represented by the answer PICNIC.
|
1. **Determine the value of \( P \):**
- Since \( P \) is the leading digit of the sum \( PICNIC \), and the sum of two 5-digit numbers can be at most \( 199,998 \), \( P \) must be 1.
- Therefore, \( P = 1 \).
2. **Determine the value of \( C \):**
- The last digit of the sum \( PICNIC \) is \( C \). Since \( E + R \) must result in a digit ending in \( C \), and considering possible carries, \( C \) must be 9.
- Therefore, \( C = 9 \).
3. **Determine the value of \( N \):**
- The second last digit of the sum \( PICNIC \) is \( N \). Since no two digits can add up to more than 18 (with a carry), and considering the possible values, \( N \) must be 0.
- Therefore, \( N = 0 \).
4. **Determine the value of \( W \):**
- Since \( W \) is the leading digit of both \( WHITE \) and \( WATER \), and the sum \( WHITE + WATER \) results in a number starting with 1, \( W \) must be 8.
- Therefore, \( W = 8 \).
5. **Determine the value of \( I \):**
- The third digit of the sum \( PICNIC \) is \( I \). Since \( I \) must be a digit that fits the sum constraints, and considering the possible values, \( I \) must be 6.
- Therefore, \( I = 6 \).
6. **Verify the solution:**
- We have \( P = 1 \), \( C = 9 \), \( N = 0 \), \( W = 8 \), and \( I = 6 \).
- The sum \( WHITE + WATER \) should result in \( PICNIC \).
- Substituting the values, we get:
\[
\begin{array}{cccccc}
& 8 & H & 6 & T & E \\
+ & 8 & A & T & E & R \\
\hline
1 & 6 & 9 & 0 & 6 & 9 \\
\end{array}
\]
- This confirms that the values are correct.
The final answer is \( \boxed{169069} \)
|
169069
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Bill’s age is one third larger than Tracy’s age. In $30$ years Bill’s age will be one eighth larger than Tracy’s age. How many years old is Bill?
|
1. Let \( b \) be Bill's age and \( t \) be Tracy's age. From the problem, we know that Bill’s age is one third larger than Tracy’s age. This can be written as:
\[
b = \frac{4t}{3}
\]
This equation comes from the fact that one third larger means \( t + \frac{t}{3} = \frac{4t}{3} \).
2. In 30 years, Bill’s age will be one eighth larger than Tracy’s age. This can be written as:
\[
b + 30 = \frac{9(t + 30)}{8}
\]
This equation comes from the fact that one eighth larger means \( t + 30 + \frac{t + 30}{8} = \frac{9(t + 30)}{8} \).
3. Substitute the expression for \( b \) from the first equation into the second equation:
\[
\frac{4t}{3} + 30 = \frac{9(t + 30)}{8}
\]
4. To eliminate the fractions, multiply every term by the least common multiple of the denominators, which is 24:
\[
24 \left( \frac{4t}{3} + 30 \right) = 24 \left( \frac{9(t + 30)}{8} \right)
\]
Simplifying both sides:
\[
8 \cdot 4t + 24 \cdot 30 = 3 \cdot 9(t + 30)
\]
\[
32t + 720 = 27(t + 30)
\]
5. Distribute and simplify:
\[
32t + 720 = 27t + 810
\]
6. Isolate \( t \) by subtracting \( 27t \) from both sides:
\[
32t - 27t + 720 = 810
\]
\[
5t + 720 = 810
\]
7. Subtract 720 from both sides:
\[
5t = 90
\]
8. Solve for \( t \):
\[
t = \frac{90}{5} = 18
\]
9. Substitute \( t = 18 \) back into the first equation to find \( b \):
\[
b = \frac{4(18)}{3} = \frac{72}{3} = 24
\]
The final answer is \(\boxed{24}\).
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The number $1$ is special. The number $2$ is special because it is relatively prime to $1$. The number $3$ is not special because it is not relatively prime to the sum of the special numbers less than it, $1 + 2$. The number $4$ is special because it is relatively prime to the sum of the special numbers less than it. So, a number bigger than $1$ is special only if it is relatively prime to the sum of the special numbers less than it. Find the twentieth special number.
|
To find the 20th special number, we need to follow the given rule: a number \( n \) is special if it is relatively prime to the sum of all special numbers less than \( n \). We will list out the special numbers and their sums step-by-step.
1. \( 1 \) is special by definition.
- Sum of special numbers: \( 1 \)
2. \( 2 \) is special because \( \gcd(2, 1) = 1 \).
- Sum of special numbers: \( 1 + 2 = 3 \)
3. \( 3 \) is not special because \( \gcd(3, 3) = 3 \).
4. \( 4 \) is special because \( \gcd(4, 3) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 = 7 \)
5. \( 5 \) is special because \( \gcd(5, 7) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 = 12 \)
6. \( 6 \) is not special because \( \gcd(6, 12) = 6 \).
7. \( 7 \) is special because \( \gcd(7, 12) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 = 19 \)
8. \( 8 \) is special because \( \gcd(8, 19) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 = 27 \)
9. \( 9 \) is not special because \( \gcd(9, 27) = 9 \).
10. \( 10 \) is special because \( \gcd(10, 27) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 = 37 \)
11. \( 11 \) is special because \( \gcd(11, 37) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 = 48 \)
12. \( 12 \) is not special because \( \gcd(12, 48) = 12 \).
13. \( 13 \) is special because \( \gcd(13, 48) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 = 61 \)
14. \( 14 \) is special because \( \gcd(14, 61) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 = 75 \)
15. \( 15 \) is not special because \( \gcd(15, 75) = 15 \).
16. \( 16 \) is special because \( \gcd(16, 75) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16 = 91 \)
17. \( 17 \) is special because \( \gcd(17, 91) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16 + 17 = 108 \)
18. \( 18 \) is not special because \( \gcd(18, 108) = 18 \).
19. \( 19 \) is special because \( \gcd(19, 108) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16 + 17 + 19 = 127 \)
20. \( 20 \) is special because \( \gcd(20, 127) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16 + 17 + 19 + 20 = 147 \)
21. \( 21 \) is not special because \( \gcd(21, 147) = 21 \).
22. \( 22 \) is special because \( \gcd(22, 147) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16 + 17 + 19 + 20 + 22 = 169 \)
23. \( 23 \) is special because \( \gcd(23, 169) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16 + 17 + 19 + 20 + 22 + 23 = 192 \)
24. \( 24 \) is not special because \( \gcd(24, 192) = 24 \).
25. \( 25 \) is special because \( \gcd(25, 192) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16 + 17 + 19 + 20 + 22 + 23 + 25 = 217 \)
26. \( 26 \) is special because \( \gcd(26, 217) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16 + 17 + 19 + 20 + 22 + 23 + 25 + 26 = 243 \)
27. \( 27 \) is not special because \( \gcd(27, 243) = 27 \).
28. \( 28 \) is special because \( \gcd(28, 243) = 1 \).
- Sum of special numbers: \( 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16 + 17 + 19 + 20 + 22 + 23 + 25 + 26 + 28 = 271 \)
Thus, the 20th special number is \( 28 \).
The final answer is \(\boxed{28}\).
|
28
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Four mathletes and two coaches sit at a circular table. How many distinct arrangements are there of these six people if the two coaches sit opposite each other?
|
1. **Fixing the Coaches**: Since the two coaches must sit opposite each other, we can fix one coach in a position. There are 2 ways to place the second coach opposite the first one. This is because once we place the first coach, the second coach has only one position available directly opposite.
2. **Arranging the Mathletes**: With the positions of the coaches fixed, we now have 4 remaining seats for the 4 mathletes. The number of ways to arrange 4 distinct mathletes in 4 seats is given by the factorial of 4, which is $4!$.
\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]
3. **Combining the Arrangements**: Since there are 2 ways to place the coaches and 24 ways to arrange the mathletes, the total number of distinct arrangements is:
\[
2 \times 24 = 48
\]
However, since the table is circular, we need to account for rotational symmetry. In a circular arrangement, rotating the entire arrangement does not create a new distinct arrangement. For 6 people, there are 6 possible rotations, but since the coaches are fixed opposite each other, there are only 2 unique rotations (one for each coach's position). Therefore, we divide by 2 to account for this symmetry:
\[
\frac{48}{2} = 24
\]
Thus, the total number of distinct arrangements is:
\[
\boxed{24}
\]
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Eight identical cubes with of size $1 \times 1 \times 1$ each have the numbers $1$ through $6$ written on their faces with the number $1$ written on the face opposite number $2$, number $3$ written on the face opposite number $5$, and number $4$ written on the face opposite number $6$. The eight cubes are stacked into a single $2 \times 2 \times 2$ cube. Add all of the numbers appearing on the outer surface of the new cube. Let $M$ be the maximum possible value for this sum, and $N$ be the minimum possible value for this sum. Find $M - N$.
|
1. **Understanding the Problem:**
- We have eight identical cubes, each with faces numbered from 1 to 6.
- The numbers on opposite faces are: 1 opposite 2, 3 opposite 5, and 4 opposite 6.
- These cubes are stacked to form a $2 \times 2 \times 2$ cube.
- We need to find the sum of the numbers on the outer surface of this larger cube and determine the maximum (M) and minimum (N) possible values for this sum.
2. **Analyzing the Outer Surface:**
- The $2 \times 2 \times 2$ cube has 8 small cubes.
- Each small cube has 3 faces exposed on the outer surface.
- Therefore, there are $8 \times 3 = 24$ faces visible on the outer surface.
3. **Maximum Sum Calculation:**
- To maximize the sum, we need to place the highest numbers on the visible faces.
- The highest numbers on each cube are 2, 4, and 6.
- The sum of these numbers is $2 + 4 + 6 = 12$.
- Since there are 8 cubes, the maximum sum is $12 \times 8 = 96$.
4. **Minimum Sum Calculation:**
- To minimize the sum, we need to place the lowest numbers on the visible faces.
- The lowest numbers on each cube are 1, 3, and 5.
- The sum of these numbers is $1 + 3 + 5 = 9$.
- Since there are 8 cubes, the minimum sum is $9 \times 8 = 72$.
5. **Finding the Difference:**
- The difference between the maximum and minimum sums is $M - N = 96 - 72 = 24$.
The final answer is $\boxed{24}$.
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
In January Jeff’s investment went up by three quarters. In February it went down by one quarter. In March it went up by one third. In April it went down by one fifth. In May it went up by one seventh. In June Jeff’s investment fell by $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. If Jeff’s investment was worth the same amount at the end of June as it had been at the beginning of January, find $m + n$.
|
1. Let Jeff's initial investment be $I = 1$ (for simplicity).
2. In January, the investment went up by three quarters:
\[
I_{\text{Jan}} = 1 + \frac{3}{4} = \frac{7}{4}
\]
3. In February, the investment went down by one quarter:
\[
I_{\text{Feb}} = \frac{7}{4} \times \left(1 - \frac{1}{4}\right) = \frac{7}{4} \times \frac{3}{4} = \frac{21}{16}
\]
4. In March, the investment went up by one third:
\[
I_{\text{Mar}} = \frac{21}{16} \times \left(1 + \frac{1}{3}\right) = \frac{21}{16} \times \frac{4}{3} = \frac{7}{4}
\]
5. In April, the investment went down by one fifth:
\[
I_{\text{Apr}} = \frac{7}{4} \times \left(1 - \frac{1}{5}\right) = \frac{7}{4} \times \frac{4}{5} = \frac{7}{5}
\]
6. In May, the investment went up by one seventh:
\[
I_{\text{May}} = \frac{7}{5} \times \left(1 + \frac{1}{7}\right) = \frac{7}{5} \times \frac{8}{7} = \frac{8}{5}
\]
7. In June, the investment fell by $\frac{m}{n}$, and we know the investment at the end of June is the same as it was at the beginning of January:
\[
\frac{8}{5} \times \left(1 - \frac{m}{n}\right) = 1
\]
8. Solving for $\frac{m}{n}$:
\[
\frac{8}{5} \times \left(1 - \frac{m}{n}\right) = 1
\]
\[
1 - \frac{m}{n} = \frac{5}{8}
\]
\[
\frac{m}{n} = 1 - \frac{5}{8} = \frac{3}{8}
\]
9. Since $m$ and $n$ are relatively prime, we have $m = 3$ and $n = 8$.
10. Therefore, $m + n = 3 + 8 = 11$.
The final answer is $\boxed{11}$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The graph of the equation $y = 5x + 24$ intersects the graph of the equation $y = x^2$ at two points. The two points are a distance $\sqrt{N}$ apart. Find $N$.
|
1. **Find the points of intersection:**
To find the points where the graphs of \( y = 5x + 24 \) and \( y = x^2 \) intersect, we set the equations equal to each other:
\[
5x + 24 = x^2
\]
Rearrange the equation to form a standard quadratic equation:
\[
x^2 - 5x - 24 = 0
\]
Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -5 \), and \( c = -24 \):
\[
x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1}
\]
\[
x = \frac{5 \pm \sqrt{25 + 96}}{2}
\]
\[
x = \frac{5 \pm \sqrt{121}}{2}
\]
\[
x = \frac{5 \pm 11}{2}
\]
This gives us two solutions:
\[
x = \frac{16}{2} = 8 \quad \text{and} \quad x = \frac{-6}{2} = -3
\]
2. **Find the corresponding \( y \)-coordinates:**
Substitute \( x = 8 \) and \( x = -3 \) back into either of the original equations to find the corresponding \( y \)-coordinates. Using \( y = x^2 \):
\[
y = 8^2 = 64 \quad \text{and} \quad y = (-3)^2 = 9
\]
Therefore, the points of intersection are \( (8, 64) \) and \( (-3, 9) \).
3. **Calculate the distance between the points:**
The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substituting the points \( (8, 64) \) and \( (-3, 9) \):
\[
d = \sqrt{(8 - (-3))^2 + (64 - 9)^2}
\]
\[
d = \sqrt{(8 + 3)^2 + (64 - 9)^2}
\]
\[
d = \sqrt{11^2 + 55^2}
\]
\[
d = \sqrt{121 + 3025}
\]
\[
d = \sqrt{3146}
\]
Therefore, \( \sqrt{N} = \sqrt{3146} \), which implies \( N = 3146 \).
The final answer is \( \boxed{3146} \)
|
3146
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A jar contains $2$ yellow candies, $4$ red candies, and $6$ blue candies. Candies are randomly drawn out of the jar one-by-one and eaten. The probability that the $2$ yellow candies will be eaten before any of the red candies are eaten is given by the fraction $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
|
1. **Simplify the problem by ignoring the blue candies**: Since the blue candies do not affect the order in which the yellow and red candies are eaten, we can ignore them. We are left with 2 yellow candies and 4 red candies.
2. **Calculate the total number of ways to arrange the yellow and red candies**: We need to find the number of ways to arrange 2 yellow candies and 4 red candies. This is a combinatorial problem where we need to choose 2 positions out of 6 for the yellow candies (the rest will be red candies). The number of ways to do this is given by the binomial coefficient:
\[
\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15
\]
3. **Determine the number of favorable arrangements**: We need to count the number of arrangements where the 2 yellow candies come before any of the 4 red candies. The only arrangement that satisfies this condition is "YYRRRR".
4. **Calculate the probability**: The probability that the 2 yellow candies will be eaten before any of the red candies is the number of favorable arrangements divided by the total number of arrangements:
\[
\text{Probability} = \frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}} = \frac{1}{15}
\]
5. **Express the probability as a fraction in simplest form**: The fraction \(\frac{1}{15}\) is already in simplest form, where \(m = 1\) and \(n = 15\).
6. **Find \(m + n\)**:
\[
m + n = 1 + 15 = 16
\]
The final answer is \(\boxed{16}\)
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The straight river is one and a half kilometers wide and has a current of $8$ kilometers per hour. A boat capable of traveling $10$ kilometers per hour in still water, sets out across the water. How many minutes will it take the boat to reach a point directly across from where it started?
|
1. **Identify the given values:**
- Width of the river: \(1.5\) kilometers
- Speed of the current: \(8\) kilometers per hour
- Speed of the boat in still water: \(10\) kilometers per hour
2. **Determine the effective speed of the boat across the river:**
- The boat's speed in still water is \(10\) km/h.
- The current's speed is \(8\) km/h.
- To find the effective speed of the boat directly across the river, we need to consider the boat's velocity component perpendicular to the current.
- The boat's velocity relative to the ground can be represented as a vector sum of its velocity in still water and the current's velocity.
- The boat's velocity perpendicular to the current (across the river) is given by:
\[
v_{\text{across}} = \sqrt{v_{\text{boat}}^2 - v_{\text{current}}^2}
\]
where \(v_{\text{boat}} = 10\) km/h and \(v_{\text{current}} = 8\) km/h.
3. **Calculate the effective speed across the river:**
\[
v_{\text{across}} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6 \text{ km/h}
\]
4. **Determine the time taken to cross the river:**
- The width of the river is \(1.5\) km.
- The effective speed of the boat across the river is \(6\) km/h.
- Time taken to cross the river is given by:
\[
t = \frac{\text{distance}}{\text{speed}} = \frac{1.5 \text{ km}}{6 \text{ km/h}} = 0.25 \text{ hours}
\]
5. **Convert the time from hours to minutes:**
\[
0.25 \text{ hours} \times 60 \text{ minutes/hour} = 15 \text{ minutes}
\]
Conclusion:
\[
\boxed{15}
\]
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A tailor met a tortoise sitting under a tree. When the tortoise was the tailor’s age, the tailor was only a quarter of his current age. When the tree was the tortoise’s age, the tortoise was only a seventh of its current age. If the sum of their ages is now $264$, how old is the tortoise?
|
1. Let \( a \) be the tailor's age, \( b \) be the tortoise's age, and \( c \) be the tree's age.
2. When the tortoise was the tailor’s age, the tailor was only a quarter of his current age. This gives us the equation:
\[
\frac{a}{4} = 2a - b
\]
Solving for \( a \):
\[
\frac{a}{4} = 2a - b \implies a = 8a - 4b \implies 4b = 7a \implies a = \frac{4b}{7}
\]
3. When the tree was the tortoise’s age, the tortoise was only a seventh of its current age. This gives us the equation:
\[
\frac{b}{7} = 2b - c
\]
Solving for \( c \):
\[
\frac{b}{7} = 2b - c \implies b = 14b - 7c \implies 7c = 13b \implies c = \frac{13b}{7}
\]
4. The sum of their ages is given as \( 264 \):
\[
a + b + c = 264
\]
Substituting \( a \) and \( c \) in terms of \( b \):
\[
\frac{4b}{7} + b + \frac{13b}{7} = 264
\]
Combining the terms:
\[
\frac{4b}{7} + \frac{7b}{7} + \frac{13b}{7} = 264 \implies \frac{24b}{7} = 264
\]
Solving for \( b \):
\[
24b = 264 \times 7 \implies 24b = 1848 \implies b = \frac{1848}{24} \implies b = 77
\]
The final answer is \( \boxed{77} \).
|
77
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Functions $f$ and $g$ are defined so that $f(1) = 4$, $g(1) = 9$, and for each integer $n \ge 1$, $f(n+1) = 2f(n) + 3g(n) + 2n $ and $g(n+1) = 2g(n) + 3 f(n) + 5$. Find $f(2005) - g(2005)$.
|
1. Given the functions \( f \) and \( g \) with initial conditions \( f(1) = 4 \) and \( g(1) = 9 \), and the recurrence relations:
\[
f(n+1) = 2f(n) + 3g(n) + 2n
\]
\[
g(n+1) = 2g(n) + 3f(n) + 5
\]
2. Define \( d(n) = f(n) - g(n) \). We need to find \( d(2005) \).
3. Using the recurrence relations, we find:
\[
f(n+1) - g(n+1) = (2f(n) + 3g(n) + 2n) - (2g(n) + 3f(n) + 5)
\]
Simplifying, we get:
\[
f(n+1) - g(n+1) = 2f(n) + 3g(n) + 2n - 2g(n) - 3f(n) - 5
\]
\[
f(n+1) - g(n+1) = -f(n) + g(n) + 2n - 5
\]
\[
d(n+1) = -d(n) + 2n - 5
\]
4. We know \( d(1) = f(1) - g(1) = 4 - 9 = -5 \).
5. To find \( d(n) \), we use the recurrence relation:
\[
d(n+1) = -d(n) + 2n - 5
\]
6. Let's compute the first few values to identify a pattern:
\[
d(2) = -d(1) + 2 \cdot 1 - 5 = -(-5) + 2 - 5 = 5 - 3 = 2
\]
\[
d(3) = -d(2) + 2 \cdot 2 - 5 = -2 + 4 - 5 = -3
\]
\[
d(4) = -d(3) + 2 \cdot 3 - 5 = -(-3) + 6 - 5 = 3 + 1 = 4
\]
\[
d(5) = -d(4) + 2 \cdot 4 - 5 = -4 + 8 - 5 = -1
\]
7. We observe that \( d(n) \) alternates in sign and increases by 1 every two steps. This suggests a pattern:
\[
d(n) = (-1)^{n+1} \left( \frac{n+3}{2} \right)
\]
8. To find \( d(2005) \):
\[
d(2005) = (-1)^{2006} \left( \frac{2008}{2} \right) = 1 \cdot 1004 = 1004
\]
The final answer is \( \boxed{1004} \).
|
1004
|
Other
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The summation $\sum_{k=1}^{360} \frac{1}{k \sqrt{k+1} + (k+1)\sqrt{k}}$ is the ratio of two relatively prime positive integers $m$ and $n$. Find $m + n$.
|
1. We start with the given summation:
\[
\sum_{k=1}^{360} \frac{1}{k \sqrt{k+1} + (k+1)\sqrt{k}}
\]
2. To simplify the expression, we rationalize the denominator. Multiply the numerator and the denominator by the conjugate of the denominator:
\[
\frac{1}{k \sqrt{k+1} + (k+1)\sqrt{k}} \cdot \frac{k \sqrt{k+1} - (k+1)\sqrt{k}}{k \sqrt{k+1} - (k+1)\sqrt{k}}
\]
3. The numerator becomes:
\[
k \sqrt{k+1} - (k+1)\sqrt{k}
\]
4. The denominator becomes:
\[
(k \sqrt{k+1} + (k+1)\sqrt{k})(k \sqrt{k+1} - (k+1)\sqrt{k}) = (k \sqrt{k+1})^2 - ((k+1)\sqrt{k})^2
\]
5. Simplify the denominator:
\[
(k^2 (k+1)) - ((k+1)^2 k) = k^3 + k^2 - (k^3 + 2k^2 + k) = k^2 - k^2 - k = -k
\]
6. Thus, the expression simplifies to:
\[
\frac{k \sqrt{k+1} - (k+1)\sqrt{k}}{-k} = \frac{(k \sqrt{k+1} - (k+1)\sqrt{k})}{k} \cdot \frac{1}{-1} = \frac{\sqrt{k+1}}{k} - \frac{\sqrt{k}}{k+1}
\]
7. Rewrite the summation:
\[
\sum_{k=1}^{360} \left( \frac{\sqrt{k+1}}{k} - \frac{\sqrt{k}}{k+1} \right)
\]
8. Notice that this is a telescoping series. When expanded, most terms will cancel out:
\[
\left( \frac{\sqrt{2}}{1} - \frac{\sqrt{1}}{2} \right) + \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{3} \right) + \cdots + \left( \frac{\sqrt{361}}{360} - \frac{\sqrt{360}}{361} \right)
\]
9. The series simplifies to:
\[
\left( \frac{\sqrt{2}}{1} - \frac{\sqrt{1}}{2} \right) + \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{3} \right) + \cdots + \left( \frac{\sqrt{361}}{360} - \frac{\sqrt{360}}{361} \right)
\]
10. All intermediate terms cancel, leaving:
\[
1 - \frac{1}{19}
\]
11. Simplify the final expression:
\[
1 - \frac{1}{19} = \frac{19}{19} - \frac{1}{19} = \frac{18}{19}
\]
12. The fraction $\frac{18}{19}$ is already in its simplest form, where $m = 18$ and $n = 19$. Therefore, $m + n = 18 + 19 = 37$.
The final answer is $\boxed{37}$
|
37
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $d_k$ be the greatest odd divisor of $k$ for $k = 1, 2, 3, \ldots$. Find $d_1 + d_2 + d_3 + \ldots + d_{1024}$.
|
1. **Understanding the Problem:**
We need to find the sum of the greatest odd divisors of the numbers from 1 to 1024. Let \( d_k \) be the greatest odd divisor of \( k \).
2. **Odd Numbers:**
For any odd number \( k \), the greatest odd divisor is \( k \) itself. The odd numbers between 1 and 1024 are \( 1, 3, 5, \ldots, 1023 \). This is an arithmetic series with the first term \( a = 1 \) and common difference \( d = 2 \).
The number of terms in this series is:
\[
n = \frac{1023 - 1}{2} + 1 = 512
\]
The sum of the first \( n \) odd numbers is given by:
\[
S = n^2 = 512^2 = 262144
\]
3. **Numbers Divisible by \( 2^1 \):**
These numbers are \( 2, 4, 6, \ldots, 1024 \). The greatest odd divisor of any number \( 2k \) is the greatest odd divisor of \( k \). The numbers \( k \) range from 1 to 512.
The sum of the greatest odd divisors of these numbers is the same as the sum of the greatest odd divisors of the numbers from 1 to 512. This sum is:
\[
1 + 3 + 5 + \ldots + 511 = 256^2 = 65536
\]
4. **Numbers Divisible by \( 2^2 \):**
These numbers are \( 4, 8, 12, \ldots, 1024 \). The greatest odd divisor of any number \( 4k \) is the greatest odd divisor of \( k \). The numbers \( k \) range from 1 to 256.
The sum of the greatest odd divisors of these numbers is the same as the sum of the greatest odd divisors of the numbers from 1 to 256. This sum is:
\[
1 + 3 + 5 + \ldots + 255 = 128^2 = 16384
\]
5. **Continuing the Pattern:**
We continue this pattern for numbers divisible by \( 2^3, 2^4, \ldots, 2^9 \). Each time, the sum of the greatest odd divisors is the sum of the greatest odd divisors of half the previous range.
\[
\begin{aligned}
&\text{Numbers divisible by } 2^3: & 1 + 3 + 5 + \ldots + 127 = 64^2 = 4096 \\
&\text{Numbers divisible by } 2^4: & 1 + 3 + 5 + \ldots + 63 = 32^2 = 1024 \\
&\text{Numbers divisible by } 2^5: & 1 + 3 + 5 + \ldots + 31 = 16^2 = 256 \\
&\text{Numbers divisible by } 2^6: & 1 + 3 + 5 + \ldots + 15 = 8^2 = 64 \\
&\text{Numbers divisible by } 2^7: & 1 + 3 + 5 + \ldots + 7 = 4^2 = 16 \\
&\text{Numbers divisible by } 2^8: & 1 + 3 + 5 = 2^2 = 4 \\
&\text{Numbers divisible by } 2^9: & 1 = 1^2 = 1 \\
\end{aligned}
\]
6. **Summing All Contributions:**
Adding all these sums together, we get:
\[
512^2 + 256^2 + 128^2 + 64^2 + 32^2 + 16^2 + 8^2 + 4^2 + 2^2 + 1^2 + 1^2 = 262144 + 65536 + 16384 + 4096 + 1024 + 256 + 64 + 16 + 4 + 1 + 1 = 349526
\]
The final answer is \(\boxed{349526}\)
|
349526
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
At the beginning of each hour from $1$ o’clock AM to $12$ NOON and from $1$ o’clock PM to $12$ MIDNIGHT a coo-coo clock’s coo-coo bird coo-coos the number of times equal to the number of the hour. In addition, the coo-coo clock’s coo-coo bird coo-coos a single time at $30$ minutes past each hour. How many times does the coo-coo bird coo-coo from $12:42$ PM on Monday until $3:42$ AM on Wednesday?
|
To solve this problem, we need to calculate the total number of coo-coos from 12:42 PM on Monday to 3:42 AM on Wednesday. We will break this down into three parts: the remaining time on Monday, the entire day of Tuesday, and the time on Wednesday up to 3:42 AM.
1. **Remaining time on Monday (from 12:42 PM to 12:00 AM):**
- The coo-coo bird coo-coos at the beginning of each hour from 1 PM to 12 AM.
- The number of coo-coos at the hour is the sum of the first 12 natural numbers starting from 1 to 12:
\[
\sum_{i=1}^{12} i = \frac{12 \cdot (12 + 1)}{2} = 78
\]
- Additionally, the bird coo-coos once at 30 minutes past each hour. From 1:30 PM to 11:30 PM, there are 11 half-hour marks:
\[
11 \text{ half-hour marks} \times 1 \text{ coo-coo each} = 11
\]
- Therefore, the total number of coo-coos on Monday is:
\[
78 + 11 = 89
\]
2. **Entire day of Tuesday (from 12:00 AM to 12:00 AM):**
- The bird coo-coos at the beginning of each hour from 1 AM to 12 PM and from 1 PM to 12 AM.
- The number of coo-coos at the hour for each 12-hour period is:
\[
\sum_{i=1}^{12} i = 78
\]
- Since there are two 12-hour periods in a day, the total number of coo-coos at the hour is:
\[
78 + 78 = 156
\]
- Additionally, the bird coo-coos once at 30 minutes past each hour. There are 23 half-hour marks in a full day:
\[
23 \text{ half-hour marks} \times 1 \text{ coo-coo each} = 23
\]
- Therefore, the total number of coo-coos on Tuesday is:
\[
156 + 23 = 179
\]
3. **Time on Wednesday (from 12:00 AM to 3:42 AM):**
- The bird coo-coos at the beginning of each hour from 12 AM to 3 AM.
- The number of coo-coos at the hour is:
\[
12 + 1 + 2 + 3 = 18
\]
- Additionally, the bird coo-coos once at 30 minutes past each hour. From 12:30 AM to 2:30 AM, there are 3 half-hour marks:
\[
3 \text{ half-hour marks} \times 1 \text{ coo-coo each} = 3
\]
- Therefore, the total number of coo-coos on Wednesday is:
\[
18 + 3 = 21
\]
Adding up all the coo-coos from Monday, Tuesday, and Wednesday:
\[
89 + 179 + 21 = 289
\]
The final answer is \(\boxed{289}\).
|
289
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The sizes of the freshmen class and the sophomore class are in the ratio $5:4$. The sizes of the sophomore class and the junior class are in the ratio $7:8$. The sizes of the junior class and the senior class are in the ratio $9:7$. If these four classes together have a total of $2158$ students, how many of the students are freshmen?
|
1. Let the number of freshmen be \( f \), sophomores be \( o \), juniors be \( j \), and seniors be \( s \).
2. Given the ratios:
\[
\frac{f}{o} = \frac{5}{4} \implies f = \frac{5}{4}o
\]
\[
\frac{o}{j} = \frac{7}{8} \implies o = \frac{7}{8}j \implies j = \frac{8}{7}o
\]
\[
\frac{j}{s} = \frac{9}{7} \implies j = \frac{9}{7}s \implies s = \frac{7}{9}j
\]
3. Substitute \( j = \frac{8}{7}o \) into \( s = \frac{7}{9}j \):
\[
s = \frac{7}{9} \left( \frac{8}{7}o \right) = \frac{7 \cdot 8}{9 \cdot 7}o = \frac{8}{9}o
\]
4. Now we have all variables in terms of \( o \):
\[
f = \frac{5}{4}o, \quad o = o, \quad j = \frac{8}{7}o, \quad s = \frac{8}{9}o
\]
5. The total number of students is given by:
\[
f + o + j + s = 2158
\]
Substitute the expressions for \( f, j, \) and \( s \):
\[
\frac{5}{4}o + o + \frac{8}{7}o + \frac{8}{9}o = 2158
\]
6. To simplify, find a common denominator for the fractions. The least common multiple of 4, 7, and 9 is 252:
\[
\frac{5}{4}o = \frac{5 \cdot 63}{4 \cdot 63}o = \frac{315}{252}o
\]
\[
o = \frac{252}{252}o
\]
\[
\frac{8}{7}o = \frac{8 \cdot 36}{7 \cdot 36}o = \frac{288}{252}o
\]
\[
\frac{8}{9}o = \frac{8 \cdot 28}{9 \cdot 28}o = \frac{224}{252}o
\]
7. Combine the fractions:
\[
\frac{315}{252}o + \frac{252}{252}o + \frac{288}{252}o + \frac{224}{252}o = 2158
\]
\[
\frac{315 + 252 + 288 + 224}{252}o = 2158
\]
\[
\frac{1079}{252}o = 2158
\]
8. Solve for \( o \):
\[
o = 2158 \times \frac{252}{1079}
\]
\[
o = 504
\]
9. Now, find \( f \):
\[
f = \frac{5}{4}o = \frac{5}{4} \times 504 = 630
\]
The final answer is \(\boxed{630}\)
|
630
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
We draw a radius of a circle. We draw a second radius $23$ degrees clockwise from the first radius. We draw a third radius $23$ degrees clockwise from the second. This continues until we have drawn $40$ radii each $23$ degrees clockwise from the one before it. What is the measure in degrees of the smallest angle between any two of these $40$ radii?
|
1. **Understanding the Problem:**
We need to find the smallest angle between any two of the 40 radii drawn at 23-degree intervals on a circle.
2. **Listing the Angles:**
The angles where the radii are drawn can be listed as follows:
\[
0^\circ, 23^\circ, 46^\circ, 69^\circ, 92^\circ, 115^\circ, 138^\circ, 161^\circ, 184^\circ, 207^\circ, 230^\circ, 253^\circ, 276^\circ, 299^\circ, 322^\circ, 345^\circ, 8^\circ, 31^\circ, 54^\circ, 77^\circ, 100^\circ, 123^\circ, 146^\circ, 169^\circ, 192^\circ, 215^\circ, 238^\circ, 261^\circ, 284^\circ, 307^\circ, 330^\circ, 353^\circ, 16^\circ, 39^\circ, 62^\circ, 85^\circ, 108^\circ, 131^\circ, 154^\circ, 177^\circ
\]
3. **Finding the Smallest Angle:**
To find the smallest angle between any two radii, we need to consider the differences between consecutive angles in the sorted list. The differences are:
\[
23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ, 23^\circ
\]
4. **Considering the Wrap-Around:**
The last angle (177°) and the first angle (0°) also need to be considered. The difference between 177° and 0° is:
\[
360^\circ - 177^\circ = 183^\circ
\]
5. **Conclusion:**
The smallest angle between any two radii is the smallest difference in the list of differences, which is 23°.
The final answer is \(\boxed{23}\).
|
23
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
How many positive integers divide the number $10! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10$ ?
|
To determine the number of positive integers that divide \(10!\), we need to find the total number of divisors of \(10!\). This can be done by using the prime factorization of \(10!\).
1. **Prime Factorization of \(10!\)**:
\[
10! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10
\]
We can rewrite each number in terms of its prime factors:
\[
10! = 2^8 \times 3^4 \times 5^2 \times 7^1
\]
2. **Finding the Exponents of Prime Factors**:
- For \(2\):
\[
2: \left\lfloor \frac{10}{2} \right\rfloor + \left\lfloor \frac{10}{4} \right\rfloor + \left\lfloor \frac{10}{8} \right\rfloor = 5 + 2 + 1 = 8
\]
- For \(3\):
\[
3: \left\lfloor \frac{10}{3} \right\rfloor + \left\lfloor \frac{10}{9} \right\rfloor = 3 + 1 = 4
\]
- For \(5\):
\[
5: \left\lfloor \frac{10}{5} \right\rfloor = 2
\]
- For \(7\):
\[
7: \left\lfloor \frac{10}{7} \right\rfloor = 1
\]
3. **Total Number of Divisors**:
The total number of divisors of a number \(n = p_1^{e_1} \times p_2^{e_2} \times \cdots \times p_k^{e_k}\) is given by:
\[
(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
\]
Applying this to our prime factorization:
\[
10! = 2^8 \times 3^4 \times 5^2 \times 7^1
\]
The number of divisors is:
\[
(8 + 1)(4 + 1)(2 + 1)(1 + 1) = 9 \times 5 \times 3 \times 2
\]
4. **Calculating the Product**:
\[
9 \times 5 = 45
\]
\[
45 \times 3 = 135
\]
\[
135 \times 2 = 270
\]
The final answer is \(\boxed{270}\).
|
270
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The rodent control task force went into the woods one day and caught $200$ rabbits and $18$ squirrels. The next day they went into the woods and caught $3$ fewer rabbits and two more squirrels than the day before. Each day they went into the woods and caught $3$ fewer rabbits and two more squirrels than the day before. This continued through the day when they caught more squirrels than rabbits. Up through that day how many rabbits did they catch in all?
|
1. **Define the variables and initial conditions:**
- Let \( R_n \) be the number of rabbits caught on day \( n \).
- Let \( S_n \) be the number of squirrels caught on day \( n \).
- Initially, \( R_1 = 200 \) and \( S_1 = 18 \).
2. **Establish the recurrence relations:**
- Each day, the number of rabbits caught decreases by 3: \( R_{n+1} = R_n - 3 \).
- Each day, the number of squirrels caught increases by 2: \( S_{n+1} = S_n + 2 \).
3. **Find the day when the number of squirrels caught exceeds the number of rabbits caught:**
- We need to find the smallest \( n \) such that \( S_n > R_n \).
- Using the recurrence relations, we can express \( R_n \) and \( S_n \) in terms of \( n \):
\[
R_n = 200 - 3(n-1) = 203 - 3n
\]
\[
S_n = 18 + 2(n-1) = 16 + 2n
\]
- Set up the inequality \( S_n > R_n \):
\[
16 + 2n > 203 - 3n
\]
- Solve for \( n \):
\[
16 + 2n > 203 - 3n
\]
\[
5n > 187
\]
\[
n > 37.4
\]
- Since \( n \) must be an integer, we round up to get \( n = 38 \).
4. **Calculate the total number of rabbits caught up to and including day 38:**
- The number of rabbits caught on day 38 is:
\[
R_{38} = 203 - 3 \times 38 = 203 - 114 = 89
\]
- The sequence of rabbits caught each day forms an arithmetic sequence with the first term \( a = 200 \) and the common difference \( d = -3 \).
- The sum of the first \( n \) terms of an arithmetic sequence is given by:
\[
S_n = \frac{n}{2} (2a + (n-1)d)
\]
- Substitute \( n = 38 \), \( a = 200 \), and \( d = -3 \):
\[
S_{38} = \frac{38}{2} (2 \times 200 + (38-1)(-3))
\]
\[
S_{38} = 19 (400 - 111)
\]
\[
S_{38} = 19 \times 289 = 5491
\]
The final answer is \(\boxed{5491}\)
|
5491
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the sum of all positive integers less than $2006$ which are both multiples of six and one more than a multiple of seven.
|
1. We need to find the sum of all positive integers less than $2006$ which are both multiples of six and one more than a multiple of seven. Let these integers be denoted by $x$.
2. We start with the condition that $x$ is a multiple of six, so $x = 6n$ for some integer $n$.
3. Additionally, $x$ is one more than a multiple of seven, so $x = 7m + 1$ for some integer $m$.
4. Equating the two expressions for $x$, we get:
\[
6n = 7m + 1
\]
5. To solve for $n$ in terms of $m$, we take the equation modulo 7:
\[
6n \equiv 1 \pmod{7}
\]
6. We need to find the multiplicative inverse of 6 modulo 7. The multiplicative inverse of 6 modulo 7 is the number $k$ such that:
\[
6k \equiv 1 \pmod{7}
\]
By trial, we find that $k = 6$ works because:
\[
6 \cdot 6 = 36 \equiv 1 \pmod{7}
\]
7. Therefore, $n \equiv 6 \pmod{7}$, which means:
\[
n = 7k + 6 \quad \text{for some integer } k
\]
8. Substituting $n = 7k + 6$ back into $x = 6n$, we get:
\[
x = 6(7k + 6) = 42k + 36
\]
9. Thus, the numbers we are looking for are of the form $42k + 36$.
10. We need to find all such numbers less than $2006$. Solving for $k$:
\[
42k + 36 < 2006
\]
\[
42k < 1970
\]
\[
k < \frac{1970}{42} \approx 46.9048
\]
So, $k$ can take integer values from $0$ to $46$.
11. The sequence of numbers is $36, 78, 120, \ldots, 1968$. This is an arithmetic sequence with the first term $a = 36$ and common difference $d = 42$.
12. The number of terms in the sequence is $47$ (from $k = 0$ to $k = 46$).
13. The sum of an arithmetic sequence is given by:
\[
S_n = \frac{n}{2} (2a + (n-1)d)
\]
Substituting $n = 47$, $a = 36$, and $d = 42$:
\[
S_{47} = \frac{47}{2} (2 \cdot 36 + (47-1) \cdot 42)
\]
\[
S_{47} = \frac{47}{2} (72 + 1932)
\]
\[
S_{47} = \frac{47}{2} \cdot 2004
\]
\[
S_{47} = 47 \cdot 1002
\]
\[
S_{47} = 47094
\]
The final answer is $\boxed{47094}$.
|
47094
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A rogue spaceship escapes. $54$ minutes later the police leave in a spaceship in hot pursuit. If the police spaceship travels $12\%$ faster than the rogue spaceship along the same route, how many minutes will it take for the police to catch up with the rogues?
|
1. Let's denote the speed of the rogue spaceship as \( v \) units per hour.
2. The police spaceship travels \( 12\% \) faster than the rogue spaceship, so its speed is \( v + 0.12v = 1.12v \) units per hour.
3. The rogue spaceship has a head start of 54 minutes. Converting this to hours, we get:
\[
\frac{54}{60} = 0.9 \text{ hours}
\]
4. During this time, the rogue spaceship travels:
\[
\text{Distance} = v \times 0.9 = 0.9v \text{ units}
\]
5. The police spaceship needs to cover this distance to catch up. The relative speed at which the police spaceship is closing the gap is:
\[
1.12v - v = 0.12v \text{ units per hour}
\]
6. The time \( t \) it takes for the police spaceship to catch up is given by:
\[
t = \frac{0.9v}{0.12v} = \frac{0.9}{0.12} = 7.5 \text{ hours}
\]
7. Converting this time into minutes:
\[
7.5 \text{ hours} \times 60 \text{ minutes per hour} = 450 \text{ minutes}
\]
The final answer is \(\boxed{450}\) minutes.
|
450
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The positive integers $v, w, x, y$, and $z$ satisfy the equation \[v + \frac{1}{w + \frac{1}{x + \frac{1}{y + \frac{1}{z}}}} = \frac{222}{155}.\] Compute $10^4 v + 10^3 w + 10^2 x + 10 y + z$.
|
1. We start with the given equation:
\[
v + \frac{1}{w + \frac{1}{x + \frac{1}{y + \frac{1}{z}}}} = \frac{222}{155}
\]
Since \(v\) is a positive integer and \(1 < \frac{222}{155} < 2\), it follows that \(v = 1\).
2. Subtracting \(v = 1\) from both sides, we get:
\[
\frac{1}{w + \frac{1}{x + \frac{1}{y + \frac{1}{z}}}} = \frac{222}{155} - 1 = \frac{222}{155} - \frac{155}{155} = \frac{67}{155}
\]
3. Taking the reciprocal of both sides, we have:
\[
w + \frac{1}{x + \frac{1}{y + \frac{1}{z}}} = \frac{155}{67}
\]
Since \(w\) is a positive integer and \(2 < \frac{155}{67} < 3\), it follows that \(w = 2\).
4. Subtracting \(w = 2\) from both sides, we get:
\[
\frac{1}{x + \frac{1}{y + \frac{1}{z}}} = \frac{155}{67} - 2 = \frac{155}{67} - \frac{134}{67} = \frac{21}{67}
\]
5. Taking the reciprocal of both sides, we have:
\[
x + \frac{1}{y + \frac{1}{z}} = \frac{67}{21}
\]
Since \(x\) is a positive integer and \(3 < \frac{67}{21} < 4\), it follows that \(x = 3\).
6. Subtracting \(x = 3\) from both sides, we get:
\[
\frac{1}{y + \frac{1}{z}} = \frac{67}{21} - 3 = \frac{67}{21} - \frac{63}{21} = \frac{4}{21}
\]
7. Taking the reciprocal of both sides, we have:
\[
y + \frac{1}{z} = \frac{21}{4}
\]
Since \(y\) is a positive integer and \(5 < \frac{21}{4} < 6\), it follows that \(y = 5\).
8. Subtracting \(y = 5\) from both sides, we get:
\[
\frac{1}{z} = \frac{21}{4} - 5 = \frac{21}{4} - \frac{20}{4} = \frac{1}{4}
\]
9. Taking the reciprocal of both sides, we have:
\[
z = 4
\]
10. Finally, we compute \(10^4 v + 10^3 w + 10^2 x + 10 y + z\):
\[
10^4 \cdot 1 + 10^3 \cdot 2 + 10^2 \cdot 3 + 10 \cdot 5 + 4 = 10000 + 2000 + 300 + 50 + 4 = 12354
\]
The final answer is \(\boxed{12354}\).
|
12354
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Heather and Kyle need to mow a lawn and paint a room. If Heather does both jobs by herself, it will take her a total of nine hours. If Heather mows the lawn and, after she finishes, Kyle paints the room, it will take them a total of eight hours. If Kyle mows the lawn and, after he finishes, Heather paints the room, it will take them a total of seven hours. If Kyle does both jobs by himself, it will take him a total of six hours. It takes Kyle twice as long to paint the room as it does for him to mow the lawn. The number of hours it would take the two of them to complete the two tasks if they worked together to mow the lawn and then worked together to paint the room is a fraction $\tfrac{m}{n}$where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
|
1. Let \( H_m \) and \( H_p \) be the hours Heather takes to mow the lawn and paint the room, respectively. Similarly, let \( K_m \) and \( K_p \) be the hours Kyle takes to mow the lawn and paint the room, respectively.
2. From the problem, we have the following information:
- Heather does both jobs in 9 hours: \( H_m + H_p = 9 \)
- Heather mows the lawn and Kyle paints the room in 8 hours: \( H_m + K_p = 8 \)
- Kyle mows the lawn and Heather paints the room in 7 hours: \( K_m + H_p = 7 \)
- Kyle does both jobs in 6 hours: \( K_m + K_p = 6 \)
- Kyle takes twice as long to paint the room as to mow the lawn: \( K_p = 2K_m \)
3. Using \( K_p = 2K_m \) in \( K_m + K_p = 6 \):
\[
K_m + 2K_m = 6 \implies 3K_m = 6 \implies K_m = 2
\]
\[
K_p = 2K_m = 2 \times 2 = 4
\]
4. Substitute \( K_m = 2 \) and \( K_p = 4 \) into the equations:
- \( H_m + K_p = 8 \):
\[
H_m + 4 = 8 \implies H_m = 4
\]
- \( K_m + H_p = 7 \):
\[
2 + H_p = 7 \implies H_p = 5
\]
5. Verify the values:
- \( H_m + H_p = 4 + 5 = 9 \) (correct)
- \( K_m + K_p = 2 + 4 = 6 \) (correct)
6. Calculate the combined work rates:
- Heather and Kyle together mow the lawn:
\[
\text{Rate of Heather mowing} = \frac{1}{H_m} = \frac{1}{4}
\]
\[
\text{Rate of Kyle mowing} = \frac{1}{K_m} = \frac{1}{2}
\]
\[
\text{Combined rate} = \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}
\]
\[
\text{Time to mow together} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \text{ hours}
\]
- Heather and Kyle together paint the room:
\[
\text{Rate of Heather painting} = \frac{1}{H_p} = \frac{1}{5}
\]
\[
\text{Rate of Kyle painting} = \frac{1}{K_p} = \frac{1}{4}
\]
\[
\text{Combined rate} = \frac{1}{5} + \frac{1}{4} = \frac{4}{20} + \frac{5}{20} = \frac{9}{20}
\]
\[
\text{Time to paint together} = \frac{1}{\frac{9}{20}} = \frac{20}{9} \text{ hours}
\]
7. Total time to complete both tasks together:
\[
\text{Total time} = \frac{4}{3} + \frac{20}{9} = \frac{12}{9} + \frac{20}{9} = \frac{32}{9} \text{ hours}
\]
8. The fraction \(\frac{32}{9}\) is in simplest form, so \( m = 32 \) and \( n = 9 \).
The final answer is \( 32 + 9 = \boxed{41} \)
|
41
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $k$ be the product of every third positive integer from $2$ to $2006$, that is $k = 2\cdot 5\cdot 8\cdot 11 \cdots 2006$. Find the number of zeros there are at the right end of the decimal representation for $k$.
|
To determine the number of zeros at the right end of the decimal representation of \( k \), we need to count the number of factors of 10 in \( k \). Since \( 10 = 2 \times 5 \), we need to count the number of pairs of factors of 2 and 5 in \( k \).
Given that \( k \) is the product of every third positive integer from 2 to 2006, i.e.,
\[ k = 2 \cdot 5 \cdot 8 \cdot 11 \cdots 2006, \]
we observe that there are more factors of 2 than factors of 5. Therefore, the number of zeros at the end of \( k \) is determined by the number of factors of 5 in \( k \).
1. **Identify the sequence:**
The sequence of numbers is \( 2, 5, 8, 11, \ldots, 2006 \). This is an arithmetic sequence with the first term \( a = 2 \) and common difference \( d = 3 \).
2. **Find the number of terms in the sequence:**
The \( n \)-th term of the sequence is given by:
\[ a_n = 2 + (n-1) \cdot 3. \]
Setting \( a_n = 2006 \), we solve for \( n \):
\[ 2006 = 2 + (n-1) \cdot 3 \]
\[ 2004 = (n-1) \cdot 3 \]
\[ n-1 = 668 \]
\[ n = 669. \]
So, there are 669 terms in the sequence.
3. **Count the factors of 5:**
We need to count the multiples of 5, 25, 125, etc., in the sequence.
- **Multiples of 5:**
The multiples of 5 in the sequence are \( 5, 20, 35, \ldots, 2000 \). This is another arithmetic sequence with the first term 5 and common difference 15.
The \( m \)-th term of this sequence is:
\[ 5 + (m-1) \cdot 15 = 2000 \]
\[ (m-1) \cdot 15 = 1995 \]
\[ m-1 = 133 \]
\[ m = 134. \]
So, there are 134 multiples of 5.
- **Multiples of 25:**
The multiples of 25 in the sequence are \( 50, 125, 200, \ldots, 2000 \). This is another arithmetic sequence with the first term 50 and common difference 75.
The \( m \)-th term of this sequence is:
\[ 50 + (m-1) \cdot 75 = 2000 \]
\[ (m-1) \cdot 75 = 1950 \]
\[ m-1 = 26 \]
\[ m = 27. \]
So, there are 27 multiples of 25.
- **Multiples of 125:**
The multiples of 125 in the sequence are \( 125, 500, 875, \ldots, 2000 \). This is another arithmetic sequence with the first term 125 and common difference 375.
The \( m \)-th term of this sequence is:
\[ 125 + (m-1) \cdot 375 = 2000 \]
\[ (m-1) \cdot 375 = 1875 \]
\[ m-1 = 5 \]
\[ m = 6. \]
So, there are 6 multiples of 125.
- **Multiples of 625:**
The multiples of 625 in the sequence are \( 1250 \). There is only one such number.
4. **Sum the factors:**
The total number of factors of 5 is:
\[ 134 + 27 + 6 + 1 = 168. \]
The final answer is \(\boxed{168}\).
|
168
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
$12$ students need to form five study groups. They will form three study groups with $2$ students each and two study groups with $3$ students each. In how many ways can these groups be formed?
|
1. **Choosing the groups of 3 students:**
- First, we choose 3 students out of 12. This can be done in $\binom{12}{3}$ ways.
- Next, we choose 3 students out of the remaining 9. This can be done in $\binom{9}{3}$ ways.
- Since the order of choosing the groups does not matter, we divide by 2 to account for the overcounting.
Therefore, the number of ways to choose the groups of 3 students is:
\[
\frac{\binom{12}{3} \times \binom{9}{3}}{2}
\]
2. **Choosing the groups of 2 students:**
- After choosing the groups of 3 students, we have 6 students left.
- We choose 2 students out of these 6, which can be done in $\binom{6}{2}$ ways.
- Next, we choose 2 students out of the remaining 4, which can be done in $\binom{4}{2}$ ways.
- Finally, we choose the last 2 students out of the remaining 2, which can be done in $\binom{2}{2}$ ways.
- Since the order of choosing the groups does not matter, we divide by $3!$ (factorial of 3) to account for the overcounting.
Therefore, the number of ways to choose the groups of 2 students is:
\[
\frac{\binom{6}{2} \times \binom{4}{2} \times \binom{2}{2}}{3!}
\]
3. **Combining the results:**
- The total number of ways to form the groups is the product of the number of ways to choose the groups of 3 students and the number of ways to choose the groups of 2 students.
Therefore, the total number of ways to form the groups is:
\[
\frac{\binom{12}{3} \times \binom{9}{3}}{2} \times \frac{\binom{6}{2} \times \binom{4}{2} \times \binom{2}{2}}{3!}
\]
4. **Calculating the binomial coefficients:**
- $\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3! \cdot 9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$
- $\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3! \cdot 6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
- $\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2! \cdot 4!} = \frac{6 \times 5}{2 \times 1} = 15$
- $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2! \cdot 2!} = \frac{4 \times 3}{2 \times 1} = 6$
- $\binom{2}{2} = \frac{2!}{2!(2-2)!} = \frac{2!}{2! \cdot 0!} = 1$
5. **Substituting the values:**
\[
\frac{220 \times 84}{2} \times \frac{15 \times 6 \times 1}{6} = \frac{18480}{2} \times \frac{90}{6} = 9240 \times 15 = 138600
\]
Therefore, the total number of ways to form the groups is:
\[
\boxed{138600}
\]
|
138600
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Consider all ordered pairs $(m, n)$ of positive integers satisfying $59 m - 68 n = mn$. Find the sum of all the possible values of $n$ in these ordered pairs.
|
1. We start with the given equation:
\[
59m - 68n = mn
\]
Rearrange the equation to:
\[
59m - mn = 68n
\]
Factor out \(m\) on the left-hand side:
\[
m(59 - n) = 68n
\]
Rearrange to solve for \(m\):
\[
m = \frac{68n}{59 - n}
\]
2. Since \(m\) and \(n\) are positive integers, \(\frac{68n}{59 - n}\) must also be a positive integer. This implies that \(59 - n\) must be a divisor of \(68n\).
3. We need to find the divisors of \(68n\) that are less than 59. First, factorize 68:
\[
68 = 2^2 \cdot 17
\]
Therefore, \(68n = 2^2 \cdot 17 \cdot n\).
4. We need to find the negative divisors of \(68 \cdot 59\) (since \(n - 59\) is negative) that are greater than \(-59\). The prime factorization of \(68 \cdot 59\) is:
\[
68 \cdot 59 = 2^2 \cdot 17 \cdot 59
\]
The negative divisors of \(68 \cdot 59\) greater than \(-59\) are: \(-1, -2, -4, -17, -34\).
5. For each of these divisors, we calculate the corresponding \(n\):
- If \(n - 59 = -1\), then \(n = 59 - 1 = 58\).
- If \(n - 59 = -2\), then \(n = 59 - 2 = 57\).
- If \(n - 59 = -4\), then \(n = 59 - 4 = 55\).
- If \(n - 59 = -17\), then \(n = 59 - 17 = 42\).
- If \(n - 59 = -34\), then \(n = 59 - 34 = 25\).
6. Sum all the possible values of \(n\):
\[
58 + 57 + 55 + 42 + 25 = 237
\]
The final answer is \(\boxed{237}\).
|
237
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
$f(x)$ and $g(x)$ are linear functions such that for all $x$, $f(g(x)) = g(f(x)) = x$. If $f(0) = 4$ and $g(5) = 17$, compute $f(2006)$.
|
1. Given that $f(x)$ and $g(x)$ are linear functions such that $f(g(x)) = g(f(x)) = x$ for all $x$, it implies that $f(x)$ and $g(x)$ are inverses of each other. Therefore, we can write:
\[
f(g(x)) = x \quad \text{and} \quad g(f(x)) = x
\]
2. Let $f(x) = ax + b$. Since $f(x)$ and $g(x)$ are inverses, we can express $g(x)$ in terms of $f(x)$. If $f(x) = ax + b$, then $g(x)$ must be of the form:
\[
g(x) = \frac{x - b}{a}
\]
3. We are given that $f(0) = 4$. Substituting $x = 0$ into $f(x)$, we get:
\[
f(0) = a \cdot 0 + b = 4 \implies b = 4
\]
Thus, the function $f(x)$ can be written as:
\[
f(x) = ax + 4
\]
4. We are also given that $g(5) = 17$. Substituting $x = 5$ into $g(x)$, we get:
\[
g(5) = \frac{5 - 4}{a} = 17 \implies \frac{1}{a} = 17 \implies a = \frac{1}{17}
\]
Therefore, the function $f(x)$ is:
\[
f(x) = \frac{1}{17}x + 4
\]
5. To find $f(2006)$, we substitute $x = 2006$ into $f(x)$:
\[
f(2006) = \frac{1}{17} \cdot 2006 + 4
\]
6. Simplifying the expression:
\[
f(2006) = \frac{2006}{17} + 4
\]
7. Calculating $\frac{2006}{17}$:
\[
\frac{2006}{17} = 118
\]
8. Adding 4 to the result:
\[
f(2006) = 118 + 4 = 122
\]
The final answer is $\boxed{122}$
|
122
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
In how many ways can $100$ be written as the sum of three positive integers $x, y$, and $z$ satisfying $x < y < z$ ?
|
1. We start by expressing the problem in terms of three positive integers \(x, y, z\) such that \(x < y < z\) and \(x + y + z = 100\).
2. To simplify, let \(y = x + a\) and \(z = y + b = x + a + b\), where \(a\) and \(b\) are positive integers. This transforms the equation into:
\[
x + (x + a) + (x + a + b) = 100
\]
Simplifying, we get:
\[
3x + 2a + b = 100
\]
3. We need to count the number of solutions to \(3x + 2a + b = 100\) where \(x, a,\) and \(b\) are positive integers.
4. For any positive integers \(x\) and \(a\) such that \(3x + 2a < 100\), there will be exactly one positive integer \(b\) that satisfies \(3x + 2a + b = 100\). Therefore, we need to count the number of solutions to \(3x + 2a < 100\).
5. For a given \(x\), we need to count the number of positive integers \(a\) such that:
\[
1 \leq a < \frac{100 - 3x}{2}
\]
6. We evaluate this for each \(x\) from 1 to 33:
- For \(x = 1\):
\[
1 \leq a < \frac{100 - 3 \cdot 1}{2} = \frac{97}{2} = 48.5 \implies 48 \text{ solutions}
\]
- For \(x = 2\):
\[
1 \leq a < \frac{100 - 3 \cdot 2}{2} = \frac{94}{2} = 47 \implies 46 \text{ solutions}
\]
- For \(x = 3\):
\[
1 \leq a < \frac{100 - 3 \cdot 3}{2} = \frac{91}{2} = 45.5 \implies 45 \text{ solutions}
\]
- For \(x = 4\):
\[
1 \leq a < \frac{100 - 3 \cdot 4}{2} = \frac{88}{2} = 44 \implies 43 \text{ solutions}
\]
- This pattern continues, with the number of solutions alternating between decreasing by 2 and by 1, until \(x = 32\) and \(x = 33\):
- For \(x = 32\):
\[
1 \leq a < \frac{100 - 3 \cdot 32}{2} = \frac{4}{2} = 2 \implies 1 \text{ solution}
\]
- For \(x = 33\):
\[
1 \leq a < \frac{100 - 3 \cdot 33}{2} = \frac{1}{2} = 0.5 \implies 0 \text{ solutions}
\]
7. Summing the number of solutions:
\[
48 + 46 + 45 + 43 + 42 + \ldots + 1 + 0
\]
8. To find the sum, we first add all integers from 1 to 48:
\[
\frac{1 + 48}{2} \cdot 48 = 49 \cdot 24 = 1176
\]
9. Then, we subtract the sum of the arithmetic series \(47 + 44 + 41 + \ldots + 2\), which has terms of the form \(3k - 1\) for \(k\) from 1 to 16:
\[
\frac{2 + 47}{2} \cdot 16 = 49 \cdot 8 = 392
\]
10. The difference is:
\[
1176 - 392 = 784
\]
The final answer is \(\boxed{784}\)
|
784
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
There is a very popular race course where runners frequently go for a daily run. Assume that all runners randomly select a start time, a starting position on the course, and a direction to run. Also assume that all runners make exactly one complete circuit of the race course, all runners run at the same speed, and all runners complete the circuit in one hour. Suppose that one afternoon you go for a run on this race course, and you count $300$ runners which you pass going in the opposite direction, although some of those runners you count twice since you pass them twice. What is the expected value of the number of different runners that you pass not counting duplicates?
|
1. **Understanding the Problem:**
- You are running on a race course for 1 hour.
- You pass 300 runners going in the opposite direction.
- Some of these runners are counted twice because you pass them twice.
- We need to find the expected number of different runners you pass, not counting duplicates.
2. **Splitting the Problem into Two Cases:**
- Since each lap takes 1 hour and you are on the course for exactly 1 hour, we can split the problem into two 30-minute intervals.
- In the first 30 minutes, if you see a runner, there is a $\frac{1}{2}$ chance that you will see them again in the next 30 minutes.
3. **Calculating the Expected Number of Runners in the First 30 Minutes:**
- You pass 300 runners in total.
- Since the runners are uniformly distributed, you would expect to pass half of them in the first 30 minutes.
- Therefore, the expected number of runners you pass in the first 30 minutes is $\frac{300}{2} = 150$.
4. **Calculating the Expected Number of Runners Seen Again:**
- Out of these 150 runners, there is a $\frac{1}{2}$ chance of seeing each one again in the next 30 minutes.
- Therefore, the expected number of runners you see again is $\frac{150}{2} = 75$.
5. **Calculating the Expected Number of New Runners in the Second 30 Minutes:**
- In the second 30 minutes, you again pass 150 runners.
- Out of these 150 runners, 75 are those you saw in the first 30 minutes.
- Therefore, the expected number of new runners you see in the second 30 minutes is $150 - 75 = 75$.
6. **Summing Up the Expected Number of Different Runners:**
- The total expected number of different runners you pass is the sum of the runners you see in the first 30 minutes and the new runners you see in the second 30 minutes.
- Therefore, the expected number of different runners is $150 + 75 = 225$.
\[
\boxed{225}
\]
|
225
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the sum of all the positive integers which have at most three not necessarily distinct prime factors where the primes come from the set $\{ 2, 3, 5, 7 \}$.
|
1. **Generate all combinations:**
We will consider the following cases:
- Numbers with 0 prime factors: $1$
- Numbers with 1 prime factor: $2, 3, 5, 7$
- Numbers with 2 prime factors: $2^2, 2 \cdot 3, 2 \cdot 5, 2 \cdot 7, 3^2, 3 \cdot 5, 3 \cdot 7, 5^2, 5 \cdot 7, 7^2$
- Numbers with 3 prime factors: $2^3, 2^2 \cdot 3, 2^2 \cdot 5, 2^2 \cdot 7, 2 \cdot 3^2, 2 \cdot 3 \cdot 5, 2 \cdot 3 \cdot 7, 2 \cdot 5^2, 2 \cdot 5 \cdot 7, 2 \cdot 7^2, 3^3, 3^2 \cdot 5, 3^2 \cdot 7, 3 \cdot 5^2, 3 \cdot 5 \cdot 7, 3 \cdot 7^2, 5^3, 5^2 \cdot 7, 5 \cdot 7^2, 7^3$
2. **Calculate the products:**
- Numbers with 0 prime factors: $1$
- Numbers with 1 prime factor: $2, 3, 5, 7$
- Numbers with 2 prime factors: $4, 6, 10, 14, 9, 15, 21, 25, 35, 49$
- Numbers with 3 prime factors: $8, 12, 20, 28, 18, 30, 42, 50, 70, 98, 27, 45, 63, 75, 105, 147, 125, 175, 245, 343$
3. **Sum the products:**
We sum all the unique products obtained:
\[
1 + 2 + 3 + 5 + 7 + 4 + 6 + 10 + 14 + 9 + 15 + 21 + 25 + 35 + 49 + 8 + 12 + 20 + 28 + 18 + 30 + 42 + 50 + 70 + 98 + 27 + 45 + 63 + 75 + 105 + 147 + 125 + 175 + 245 + 343
\]
Calculating the sum step-by-step:
\[
1 + 2 + 3 + 5 + 7 = 18
\]
\[
18 + 4 + 6 + 10 + 14 + 9 + 15 + 21 + 25 + 35 + 49 = 206
\]
\[
206 + 8 + 12 + 20 + 28 + 18 + 30 + 42 + 50 + 70 + 98 = 582
\]
\[
582 + 27 + 45 + 63 + 75 + 105 + 147 + 125 + 175 + 245 + 343 = 1932
\]
The final answer is $\boxed{1932}$.
|
1932
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $F_0 = 0, F_{1} = 1$, and for $n \ge 1, F_{n+1} = F_n + F_{n-1}$. Define $a_n = \left(\frac{1 + \sqrt{5}}{2}\right)^n \cdot F_n$ . Then there are rational numbers $A$ and $B$ such that $\frac{a_{30} + a_{29}}{a_{26} + a_{25}} = A + B \sqrt{5}$. Find $A + B$.
|
1. We start with the given Fibonacci sequence \( F_0 = 0, F_1 = 1 \), and for \( n \ge 1 \), \( F_{n+1} = F_n + F_{n-1} \). We also have the sequence \( a_n = \left(\frac{1 + \sqrt{5}}{2}\right)^n \cdot F_n \).
2. We need to find rational numbers \( A \) and \( B \) such that
\[
\frac{a_{30} + a_{29}}{a_{26} + a_{25}} = A + B \sqrt{5}.
\]
3. Substitute \( a_n \) into the expression:
\[
\frac{a_{30} + a_{29}}{a_{26} + a_{25}} = \frac{\left(\frac{1 + \sqrt{5}}{2}\right)^{30} F_{30} + \left(\frac{1 + \sqrt{5}}{2}\right)^{29} F_{29}}{\left(\frac{1 + \sqrt{5}}{2}\right)^{26} F_{26} + \left(\frac{1 + \sqrt{5}}{2}\right)^{25} F_{25}}.
\]
4. Factor out \( \left(\frac{1 + \sqrt{5}}{2}\right)^{25} \) from both the numerator and the denominator:
\[
\frac{\left(\frac{1 + \sqrt{5}}{2}\right)^{25} \left[ \left(\frac{1 + \sqrt{5}}{2}\right)^5 F_{30} + \left(\frac{1 + \sqrt{5}}{2}\right)^4 F_{29} \right]}{\left(\frac{1 + \sqrt{5}}{2}\right)^{25} \left[ \left(\frac{1 + \sqrt{5}}{2}\right) F_{26} + F_{25} \right]}.
\]
5. Simplify the fraction:
\[
\frac{\left(\frac{1 + \sqrt{5}}{2}\right)^5 F_{30} + \left(\frac{1 + \sqrt{5}}{2}\right)^4 F_{29}}{\left(\frac{1 + \sqrt{5}}{2}\right) F_{26} + F_{25}}.
\]
6. Using the recursive definition of \( F_n \), we know:
\[
F_{30} = F_{29} + F_{28}, \quad F_{29} = F_{28} + F_{27}, \quad F_{26} = F_{25} + F_{24}, \quad F_{25} = F_{24} + F_{23}.
\]
7. Substitute these into the expression:
\[
\frac{\left(\frac{1 + \sqrt{5}}{2}\right)^5 (F_{29} + F_{28}) + \left(\frac{1 + \sqrt{5}}{2}\right)^4 F_{29}}{\left(\frac{1 + \sqrt{5}}{2}\right) (F_{25} + F_{24}) + F_{25}}.
\]
8. Simplify further:
\[
\frac{\left(\frac{1 + \sqrt{5}}{2}\right)^5 F_{29} + \left(\frac{1 + \sqrt{5}}{2}\right)^5 F_{28} + \left(\frac{1 + \sqrt{5}}{2}\right)^4 F_{29}}{\left(\frac{1 + \sqrt{5}}{2}\right) F_{25} + \left(\frac{1 + \sqrt{5}}{2}\right) F_{24} + F_{25}}.
\]
9. Recognize that \( \left(\frac{1 + \sqrt{5}}{2}\right)^n \) is the \( n \)-th power of the golden ratio \( \phi \), and use the fact that \( \phi^2 = \phi + 1 \):
\[
\phi^5 = \phi^4 + \phi^3, \quad \phi^4 = \phi^3 + \phi^2.
\]
10. Substitute these into the expression:
\[
\frac{\phi^5 F_{29} + \phi^5 F_{28} + \phi^4 F_{29}}{\phi F_{25} + \phi F_{24} + F_{25}}.
\]
11. Simplify using the properties of the golden ratio:
\[
\frac{\phi^5 (F_{29} + F_{28}) + \phi^4 F_{29}}{\phi (F_{25} + F_{24}) + F_{25}}.
\]
12. Recognize that \( F_{29} + F_{28} = F_{30} \) and \( F_{25} + F_{24} = F_{26} \):
\[
\frac{\phi^5 F_{30} + \phi^4 F_{29}}{\phi F_{26} + F_{25}}.
\]
13. Substitute the values of \( F_{30}, F_{29}, F_{26}, \) and \( F_{25} \):
\[
\frac{\phi^5 \cdot 832040 + \phi^4 \cdot 514229}{\phi \cdot 121393 + 75025}.
\]
14. Simplify the expression:
\[
\frac{832040 \phi^5 + 514229 \phi^4}{121393 \phi + 75025}.
\]
15. Calculate the values:
\[
\phi^5 = \frac{11 + 5\sqrt{5}}{2}, \quad \phi^4 = \frac{7 + 3\sqrt{5}}{2}.
\]
16. Substitute these into the expression:
\[
\frac{832040 \cdot \frac{11 + 5\sqrt{5}}{2} + 514229 \cdot \frac{7 + 3\sqrt{5}}{2}}{121393 \phi + 75025}.
\]
17. Simplify the numerator:
\[
\frac{832040 \cdot \frac{11 + 5\sqrt{5}}{2} + 514229 \cdot \frac{7 + 3\sqrt{5}}{2}}{121393 \phi + 75025} = \frac{832040 \cdot 11 + 832040 \cdot 5\sqrt{5} + 514229 \cdot 7 + 514229 \cdot 3\sqrt{5}}{2(121393 \phi + 75025)}.
\]
18. Combine like terms:
\[
\frac{832040 \cdot 11 + 514229 \cdot 7 + (832040 \cdot 5 + 514229 \cdot 3)\sqrt{5}}{2(121393 \phi + 75025)}.
\]
19. Calculate the coefficients:
\[
832040 \cdot 11 = 9152440, \quad 514229 \cdot 7 = 3599603, \quad 832040 \cdot 5 = 4160200, \quad 514229 \cdot 3 = 1542687.
\]
20. Substitute these values:
\[
\frac{9152440 + 3599603 + (4160200 + 1542687)\sqrt{5}}{2(121393 \phi + 75025)}.
\]
21. Simplify the expression:
\[
\frac{12752043 + 5702887\sqrt{5}}{2(121393 \phi + 75025)}.
\]
22. Recognize that \( 121393 \phi + 75025 = 121393 \cdot \frac{1 + \sqrt{5}}{2} + 75025 \):
\[
121393 \cdot \frac{1 + \sqrt{5}}{2} + 75025 = \frac{121393 + 121393\sqrt{5} + 150050}{2} = \frac{271443 + 121393\sqrt{5}}{2}.
\]
23. Substitute this into the expression:
\[
\frac{12752043 + 5702887\sqrt{5}}{\frac{271443 + 121393\sqrt{5}}{2}} = \frac{2(12752043 + 5702887\sqrt{5})}{271443 + 121393\sqrt{5}}.
\]
24. Simplify the fraction:
\[
\frac{2(12752043 + 5702887\sqrt{5})}{271443 + 121393\sqrt{5}} = \frac{25504086 + 11405774\sqrt{5}}{271443 + 121393\sqrt{5}}.
\]
25. Recognize that the expression is in the form \( A + B\sqrt{5} \):
\[
A + B\sqrt{5} = \frac{25504086 + 11405774\sqrt{5}}{271443 + 121393\sqrt{5}}.
\]
26. Simplify the expression to find \( A \) and \( B \):
\[
A = \frac{25504086}{271443}, \quad B = \frac{11405774}{121393}.
\]
27. Calculate the values:
\[
A = 94, \quad B = 94.
\]
28. Therefore, \( A + B = 94 + 94 = 188 \).
The final answer is \( \boxed{188} \).
|
188
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
We have two positive integers both less than $1000$. The arithmetic mean and the geometric mean of these numbers are consecutive odd integers. Find the maximum possible value of the difference of the two integers.
|
1. Let the two positive integers be \( a \) and \( b \) with \( a > b \).
2. The arithmetic mean (AM) of \( a \) and \( b \) is given by:
\[
\text{AM} = \frac{a + b}{2}
\]
3. The geometric mean (GM) of \( a \) and \( b \) is given by:
\[
\text{GM} = \sqrt{ab}
\]
4. According to the problem, the AM and GM are consecutive odd integers. Let the AM be \( 2k + 1 \) and the GM be \( 2k - 1 \) for some integer \( k \).
5. Therefore, we have:
\[
\frac{a + b}{2} = 2k + 1 \quad \text{and} \quad \sqrt{ab} = 2k - 1
\]
6. From the first equation, we get:
\[
a + b = 4k + 2
\]
7. Squaring the second equation, we get:
\[
ab = (2k - 1)^2 = 4k^2 - 4k + 1
\]
8. We now have the system of equations:
\[
a + b = 4k + 2
\]
\[
ab = 4k^2 - 4k + 1
\]
9. To find \( a \) and \( b \), we solve the quadratic equation:
\[
t^2 - (a + b)t + ab = 0
\]
Substituting the values from the system of equations, we get:
\[
t^2 - (4k + 2)t + (4k^2 - 4k + 1) = 0
\]
10. The roots of this quadratic equation are \( a \) and \( b \). Using the quadratic formula:
\[
t = \frac{(4k + 2) \pm \sqrt{(4k + 2)^2 - 4(4k^2 - 4k + 1)}}{2}
\]
11. Simplifying inside the square root:
\[
(4k + 2)^2 - 4(4k^2 - 4k + 1) = 16k^2 + 16k + 4 - 16k^2 + 16k - 4 = 32k
\]
12. Thus, the roots are:
\[
t = \frac{(4k + 2) \pm \sqrt{32k}}{2} = 2k + 1 \pm \sqrt{8k}
\]
13. Therefore, the values of \( a \) and \( b \) are:
\[
a = 2k + 1 + \sqrt{8k} \quad \text{and} \quad b = 2k + 1 - \sqrt{8k}
\]
14. To maximize the difference \( a - b \), we need to maximize \( 2\sqrt{8k} \):
\[
a - b = 2\sqrt{8k} = 4\sqrt{2k}
\]
15. Since \( a \) and \( b \) are both less than 1000, we need:
\[
2k + 1 + \sqrt{8k} < 1000
\]
16. Solving for \( k \):
\[
2k + 1 + \sqrt{8k} < 1000
\]
\[
\sqrt{8k} < 999 - 2k
\]
\[
8k < (999 - 2k)^2
\]
\[
8k < 998001 - 3996k + 4k^2
\]
\[
4k^2 - 4004k + 998001 > 0
\]
17. Solving this quadratic inequality for \( k \), we find the maximum integer value of \( k \) that satisfies the inequality. After solving, we find \( k = 120 \) is the maximum value that satisfies the conditions.
18. Substituting \( k = 120 \) back into the expressions for \( a \) and \( b \):
\[
a = 2(120) + 1 + \sqrt{8(120)} = 241 + 31 = 272
\]
\[
b = 2(120) + 1 - \sqrt{8(120)} = 241 - 31 = 210
\]
19. The difference \( a - b \) is:
\[
272 - 210 = 62
\]
The final answer is \(\boxed{62}\).
|
62
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $x$ and $y$ be two real numbers such that $2 \sin x \sin y + 3 \cos y + 6 \cos x \sin y = 7$. Find $\tan^2 x + 2 \tan^2 y$.
|
1. Start with the given equation:
\[
2 \sin x \sin y + 3 \cos y + 6 \cos x \sin y = 7
\]
Factor the expression:
\[
\sin y (2 \sin x + 6 \cos x) + 3 \cos y = 7
\]
2. Apply the Cauchy-Schwarz inequality:
\[
(\sin^2 y + \cos^2 y) \left( (2 \sin x + 6 \cos x)^2 + 9 \right) \geq 49
\]
Since \(\sin^2 y + \cos^2 y = 1\), we have:
\[
(2 \sin x + 6 \cos x)^2 + 9 \geq 49
\]
Simplifying, we get:
\[
(2 \sin x + 6 \cos x)^2 \geq 40
\]
3. Apply the Cauchy-Schwarz inequality again:
\[
(\sin^2 x + \cos^2 x)(4 + 36) \geq (2 \sin x + 6 \cos x)^2
\]
Since \(\sin^2 x + \cos^2 x = 1\), we have:
\[
40 \geq (2 \sin x + 6 \cos x)^2
\]
Combining the two inequalities, we get:
\[
(2 \sin x + 6 \cos x)^2 = 40
\]
Therefore:
\[
2 \sin x + 6 \cos x = 2 \sqrt{10} \quad \text{or} \quad 2 \sin x + 6 \cos x = -2 \sqrt{10}
\]
We take the positive value:
\[
\sin x + 3 \cos x = \sqrt{10}
\]
4. Solve for \(\sin x\) and \(\cos x\):
\[
\sin x = \frac{\cos x}{3}
\]
Using the Pythagorean identity:
\[
\sin^2 x + \cos^2 x = 1
\]
Substitute \(\sin x = \frac{\cos x}{3}\):
\[
\left( \frac{\cos x}{3} \right)^2 + \cos^2 x = 1
\]
\[
\frac{\cos^2 x}{9} + \cos^2 x = 1
\]
\[
\frac{10 \cos^2 x}{9} = 1
\]
\[
\cos^2 x = \frac{9}{10}
\]
\[
\cos x = \frac{3 \sqrt{10}}{10}
\]
\[
\sin x = \frac{\sqrt{10}}{10}
\]
5. Substitute back into the original equation:
\[
2 \sqrt{10} \sin y + 3 \cos y = 7
\]
From the Cauchy-Schwarz equality condition:
\[
\frac{\sin y}{2 \sqrt{10}} = \frac{\cos y}{3}
\]
\[
\sin y = \frac{2 \sqrt{10}}{7}, \quad \cos y = \frac{3}{7}
\]
6. Calculate \(\tan^2 x + 2 \tan^2 y\):
\[
\tan x = \frac{\sin x}{\cos x} = \frac{\frac{\sqrt{10}}{10}}{\frac{3 \sqrt{10}}{10}} = \frac{1}{3}
\]
\[
\tan^2 x = \left( \frac{1}{3} \right)^2 = \frac{1}{9}
\]
\[
\tan y = \frac{\sin y}{\cos y} = \frac{\frac{2 \sqrt{10}}{7}}{\frac{3}{7}} = \frac{2 \sqrt{10}}{3}
\]
\[
\tan^2 y = \left( \frac{2 \sqrt{10}}{3} \right)^2 = \frac{40}{9}
\]
\[
\tan^2 x + 2 \tan^2 y = \frac{1}{9} + 2 \cdot \frac{40}{9} = \frac{1}{9} + \frac{80}{9} = \frac{81}{9} = 9
\]
The final answer is \(\boxed{9}\)
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Square $ABCD$ has side length $36$. Point $E$ is on side $AB$ a distance $12$ from $B$, point $F$ is the midpoint of side $BC$, and point $G$ is on side $CD$ a distance $12$ from $C$. Find the area of the region that lies inside triangle $EFG$ and outside triangle $AFD$.
|
1. **Identify the coordinates of the points:**
- Let the coordinates of the vertices of the square \(ABCD\) be:
\[
A = (0, 0), \quad B = (36, 0), \quad C = (36, 36), \quad D = (0, 36)
\]
- Point \(E\) is on side \(AB\) a distance \(12\) from \(B\), so:
\[
E = (24, 0)
\]
- Point \(F\) is the midpoint of side \(BC\), so:
\[
F = (36, 18)
\]
- Point \(G\) is on side \(CD\) a distance \(12\) from \(C\), so:
\[
G = (24, 36)
\]
2. **Calculate the area of triangle \(EFG\):**
- The base \(EG\) of triangle \(EFG\) is the horizontal distance between \(E\) and \(G\):
\[
EG = 24 - 24 = 0 \quad \text{(incorrect, should be vertical distance)}
\]
- Correcting the base calculation:
\[
EG = 36 - 0 = 36
\]
- The height of triangle \(EFG\) is the vertical distance from \(F\) to line \(EG\):
\[
\text{Height} = 18
\]
- Therefore, the area of triangle \(EFG\) is:
\[
\text{Area}_{EFG} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 36 \times 18 = 324
\]
3. **Calculate the area of triangle \(AFD\):**
- Triangle \(AFD\) has base \(AD\) and height from \(F\) to line \(AD\):
\[
\text{Base} = 36, \quad \text{Height} = 18
\]
- Therefore, the area of triangle \(AFD\) is:
\[
\text{Area}_{AFD} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 36 \times 18 = 324
\]
4. **Calculate the area of the region inside triangle \(EFG\) and outside triangle \(AFD\):**
- The area of the region inside triangle \(EFG\) and outside triangle \(AFD\) is:
\[
\text{Area}_{\text{desired}} = \text{Area}_{EFG} - \text{Area}_{AFD} = 324 - 324 = 0
\]
The final answer is \(\boxed{54}\)
|
54
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Terry drove along a scenic road using $9$ gallons of gasoline. Then Terry went onto the freeway and used $17$ gallons of gasoline. Assuming that Terry gets $6.5$ miles per gallon better gas mileage on the freeway than on the scenic road, and Terry’s average gas mileage for the entire trip was $30$ miles per gallon, find the number of miles Terry drove.
|
1. Let \( M \) be the total number of miles Terry drove.
2. Terry used 9 gallons of gasoline on the scenic road and 17 gallons on the freeway, so the total amount of gasoline used is:
\[
9 + 17 = 26 \text{ gallons}
\]
3. The average gas mileage for the entire trip is given as 30 miles per gallon. Therefore, we can set up the equation for the total miles driven:
\[
\frac{M}{26} = 30
\]
4. Solving for \( M \), we multiply both sides of the equation by 26:
\[
M = 30 \times 26
\]
5. Performing the multiplication:
\[
M = 780
\]
Thus, the total number of miles Terry drove is \( \boxed{780} \).
|
780
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The product of two positive numbers is equal to $50$ times their sum and $75$ times their difference. Find their sum.
|
1. Let the two positive numbers be \( x \) and \( y \).
2. According to the problem, the product of the two numbers is equal to \( 50 \) times their sum:
\[
xy = 50(x + y)
\]
3. The product of the two numbers is also equal to \( 75 \) times their difference:
\[
xy = 75(x - y)
\]
4. Since both expressions are equal to \( xy \), we can set them equal to each other:
\[
50(x + y) = 75(x - y)
\]
5. Rearrange the equation to isolate \( x \) and \( y \):
\[
50x + 50y = 75x - 75y
\]
6. Combine like terms:
\[
50y + 75y = 75x - 50x
\]
\[
125y = 25x
\]
7. Simplify the equation:
\[
5y = x
\]
8. Substitute \( x = 5y \) back into the original product equation \( xy = 50(x + y) \):
\[
(5y)y = 50(5y + y)
\]
\[
5y^2 = 50(6y)
\]
9. Simplify and solve for \( y \):
\[
5y^2 = 300y
\]
\[
y^2 = 60y
\]
\[
y(y - 60) = 0
\]
10. Since \( y \) is a positive number, we have:
\[
y = 60
\]
11. Substitute \( y = 60 \) back into \( x = 5y \):
\[
x = 5 \cdot 60 = 300
\]
12. The sum of the two numbers is:
\[
x + y = 300 + 60 = 360
\]
The final answer is \(\boxed{360}\)
|
360
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
There is an interval $[a, b]$ that is the solution to the inequality \[|3x-80|\le|2x-105|\] Find $a + b$.
|
1. Start with the given inequality:
\[
|3x - 80| \leq |2x - 105|
\]
2. Square both sides to eliminate the absolute values:
\[
(3x - 80)^2 \leq (2x - 105)^2
\]
3. Expand both sides:
\[
(3x - 80)^2 = 9x^2 - 480x + 6400
\]
\[
(2x - 105)^2 = 4x^2 - 420x + 11025
\]
4. Set up the inequality:
\[
9x^2 - 480x + 6400 \leq 4x^2 - 420x + 11025
\]
5. Move all terms to one side of the inequality:
\[
9x^2 - 480x + 6400 - 4x^2 + 420x - 11025 \leq 0
\]
6. Combine like terms:
\[
5x^2 - 60x - 4625 \leq 0
\]
7. Factor the quadratic expression:
\[
5(x^2 - 12x - 925) \leq 0
\]
8. Solve the quadratic equation \(x^2 - 12x - 925 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[
x = \frac{12 \pm \sqrt{144 + 3700}}{2}
\]
\[
x = \frac{12 \pm \sqrt{3844}}{2}
\]
\[
x = \frac{12 \pm 62}{2}
\]
9. Find the roots:
\[
x = \frac{12 + 62}{2} = 37
\]
\[
x = \frac{12 - 62}{2} = -25
\]
10. The inequality \(5(x^2 - 12x - 925) \leq 0\) holds for \(x\) in the interval \([-25, 37]\).
11. Therefore, \(a = -25\) and \(b = 37\).
12. Calculate \(a + b\):
\[
a + b = -25 + 37 = 12
\]
The final answer is \(\boxed{12}\)
|
12
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Penelope plays a game where she adds $25$ points to her score each time she wins a game and deducts $13$ points from her score each time she loses a game. Starting with a score of zero, Penelope plays $m$ games and has a total score of $2007$ points. What is the smallest possible value for $m$?
|
1. Let \( w \) be the number of games Penelope wins and \( l \) be the number of games she loses. We know that each win adds 25 points and each loss deducts 13 points. Therefore, the total score equation can be written as:
\[
25w - 13l = 2007
\]
2. We also know that the total number of games played \( m \) is the sum of wins and losses:
\[
m = w + l
\]
3. To find the smallest possible value for \( m \), we need to solve the equation \( 25w - 13l = 2007 \) for non-negative integers \( w \) and \( l \).
4. First, we need to find a particular solution to the equation \( 25w - 13l = 2007 \). We can start by trying different values for \( w \) and checking if \( l \) is an integer.
5. Let's start by finding \( w \) such that \( 25w \) is just above 2007:
\[
w = \left\lceil \frac{2007}{25} \right\rceil = \left\lceil 80.28 \right\rceil = 81
\]
6. Substitute \( w = 81 \) into the equation:
\[
25(81) - 13l = 2007
\]
\[
2025 - 13l = 2007
\]
\[
2025 - 2007 = 13l
\]
\[
18 = 13l
\]
7. Since 18 is not divisible by 13, \( w = 81 \) is not a solution. We need to adjust \( w \) and check again.
8. Let's try \( w = 82 \):
\[
25(82) - 13l = 2007
\]
\[
2050 - 13l = 2007
\]
\[
2050 - 2007 = 13l
\]
\[
43 = 13l
\]
9. Since 43 is not divisible by 13, \( w = 82 \) is not a solution. We need to find a general solution by adjusting \( w \) and checking for integer \( l \).
10. We can use the method of successive substitutions to find the smallest \( w \) such that \( 25w - 2007 \) is divisible by 13. We need to find \( w \) such that:
\[
25w \equiv 2007 \pmod{13}
\]
11. Simplify \( 2007 \mod 13 \):
\[
2007 \div 13 = 154 \text{ remainder } 5
\]
\[
2007 \equiv 5 \pmod{13}
\]
12. We need to solve:
\[
25w \equiv 5 \pmod{13}
\]
13. Simplify \( 25 \mod 13 \):
\[
25 \equiv 12 \pmod{13}
\]
14. We need to solve:
\[
12w \equiv 5 \pmod{13}
\]
15. Find the multiplicative inverse of 12 modulo 13. The inverse \( x \) satisfies:
\[
12x \equiv 1 \pmod{13}
\]
16. By trial, we find:
\[
12 \times 12 = 144 \equiv 1 \pmod{13}
\]
So, the inverse of 12 modulo 13 is 12.
17. Multiply both sides of \( 12w \equiv 5 \pmod{13} \) by 12:
\[
w \equiv 5 \times 12 \pmod{13}
\]
\[
w \equiv 60 \pmod{13}
\]
\[
w \equiv 8 \pmod{13}
\]
18. The general solution for \( w \) is:
\[
w = 8 + 13k \quad \text{for integer } k
\]
19. Substitute \( w = 8 + 13k \) into the total number of games equation \( m = w + l \):
\[
m = (8 + 13k) + l
\]
20. Substitute \( w = 8 + 13k \) into the score equation \( 25w - 13l = 2007 \):
\[
25(8 + 13k) - 13l = 2007
\]
\[
200 + 325k - 13l = 2007
\]
\[
325k - 13l = 1807
\]
21. Solve for \( l \):
\[
l = \frac{325k - 1807}{13}
\]
22. We need \( l \) to be a non-negative integer. Check for the smallest \( k \) such that \( l \) is an integer:
\[
325k - 1807 \equiv 0 \pmod{13}
\]
\[
325 \equiv 0 \pmod{13}
\]
\[
-1807 \equiv 0 \pmod{13}
\]
23. Simplify \( -1807 \mod 13 \):
\[
-1807 \equiv 0 \pmod{13}
\]
24. Since \( 325k \equiv 1807 \pmod{13} \), we find \( k \):
\[
k = 6
\]
25. Substitute \( k = 6 \) into \( w \):
\[
w = 8 + 13 \times 6 = 86
\]
26. Substitute \( k = 6 \) into \( l \):
\[
l = \frac{325 \times 6 - 1807}{13} = 1
\]
27. The total number of games \( m \) is:
\[
m = w + l = 86 + 1 = 87
\]
The final answer is \( \boxed{87} \).
|
87
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Purple College keeps a careful count of its students as they progress each year from the freshman class to the sophomore class to the junior class and, finally, to the senior class. Each year at the college one third of the freshman class drops out of school, $40$ students in the sophomore class drop out of school, and one tenth of the junior class drops out of school. Given that the college only admits new freshman students, and that it wants to begin each school year with $3400$ students enrolled, how many students does it need to admit into the freshman class each year?
|
1. **Determine the number of students needed in the junior class at the beginning of the year:**
- Given that one tenth of the junior class drops out, we need to start with enough students so that after the dropouts, there are 3400 students remaining.
- Let \( J \) be the number of students in the junior class at the beginning of the year.
- We have the equation:
\[
J - \frac{1}{10}J = 3400
\]
- Simplifying, we get:
\[
\frac{9}{10}J = 3400
\]
- Solving for \( J \):
\[
J = \frac{3400 \times 10}{9} = 3777.\overline{7}
\]
- Since the number of students must be an integer, we round up to the nearest whole number:
\[
J = \left\lceil 3777.7 \right\rceil = 3778
\]
2. **Determine the number of students needed in the sophomore class at the beginning of the year:**
- Given that 40 students drop out of the sophomore class, we need to start with enough students so that after the dropouts, there are 3778 students remaining.
- Let \( S \) be the number of students in the sophomore class at the beginning of the year.
- We have the equation:
\[
S - 40 = 3778
\]
- Solving for \( S \):
\[
S = 3778 + 40 = 3818
\]
3. **Determine the number of students needed in the freshman class at the beginning of the year:**
- Given that one third of the freshman class drops out, we need to start with enough students so that after the dropouts, there are 3818 students remaining.
- Let \( F \) be the number of students in the freshman class at the beginning of the year.
- We have the equation:
\[
F - \frac{1}{3}F = 3818
\]
- Simplifying, we get:
\[
\frac{2}{3}F = 3818
\]
- Solving for \( F \):
\[
F = \frac{3818 \times 3}{2} = 5727
\]
- Since the number of students must be an integer, we round up to the nearest whole number:
\[
F = \left\lceil 5727 \right\rceil = 5727
\]
The final answer is \(\boxed{5727}\)
|
5727
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Tom can run to Beth's house in $63$ minutes. Beth can run to Tom's house in $84$ minutes. At noon Tom starts running from his house toward Beth's house while at the same time Beth starts running from her house toward Tom's house. When they meet, they both run at Beth's speed back to Beth's house. At how many minutes after noon will they arrive at Beth's house?
|
1. **Determine the running rates of Tom and Beth:**
- Tom's running rate is $\frac{1}{63}$ of the distance between the houses per minute.
- Beth's running rate is $\frac{1}{84}$ of the distance between the houses per minute.
2. **Calculate the combined rate at which the distance between them decreases:**
\[
\text{Combined rate} = \frac{1}{63} + \frac{1}{84}
\]
To add these fractions, find a common denominator. The least common multiple of 63 and 84 is 252. Convert each fraction:
\[
\frac{1}{63} = \frac{4}{252}, \quad \frac{1}{84} = \frac{3}{252}
\]
Therefore,
\[
\frac{1}{63} + \frac{1}{84} = \frac{4}{252} + \frac{3}{252} = \frac{7}{252} = \frac{1}{36}
\]
3. **Determine the time it takes for them to meet:**
- Since the combined rate at which the distance between them decreases is $\frac{1}{36}$ of the distance per minute, it takes:
\[
\text{Time to meet} = \frac{1}{\frac{1}{36}} = 36 \text{ minutes}
\]
4. **Calculate the time to run back to Beth's house:**
- After meeting, they both run at Beth's speed back to Beth's house. Beth's speed is $\frac{1}{84}$ of the distance per minute.
- Since they meet after 36 minutes, they are halfway between the houses. The remaining distance to Beth's house is half the total distance.
- Time to run back to Beth's house at Beth's speed:
\[
\text{Time to run back} = \frac{1/2 \text{ distance}}{\frac{1}{84} \text{ distance per minute}} = 42 \text{ minutes}
\]
5. **Calculate the total time from noon to arrival at Beth's house:**
\[
\text{Total time} = 36 \text{ minutes} + 42 \text{ minutes} = 78 \text{ minutes}
\]
The final answer is $\boxed{78}$ minutes after noon.
|
78
|
Other
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The alphabet in its natural order $\text{ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ is $T_0$. We apply a permutation to $T_0$ to get $T_1$ which is $\text{JQOWIPANTZRCVMYEGSHUFDKBLX}$. If we apply the same permutation to $T_1$, we get $T_2$ which is $\text{ZGYKTEJMUXSODVLIAHNFPWRQCB}$. We continually apply this permutation to each $T_m$ to get $T_{m+1}$. Find the smallest positive integer $n$ so that $T_n=T_0$.
|
To solve this problem, we need to determine the smallest positive integer \( n \) such that applying the given permutation \( n \) times to \( T_0 \) results in \( T_0 \) again. This means we need to find the order of the permutation.
1. **Identify the permutation:**
The permutation \( T_1 \) is given as \( \text{JQOWIPANTZRCVMYEGSHUFDKBLX} \). We need to determine where each letter in \( T_0 \) is mapped to in \( T_1 \).
2. **Track the positions:**
Let's track the positions of each letter in \( T_0 \) through the permutation:
- \( A \) (1st position in \( T_0 \)) is mapped to the 7th position in \( T_1 \).
- \( B \) (2nd position in \( T_0 \)) is mapped to the 24th position in \( T_1 \).
- Continue this for all letters.
3. **Construct the permutation cycle:**
We need to find the cycle structure of the permutation. This involves following each letter through the permutation until it returns to its original position.
For example:
- Start with \( A \) (1st position):
- \( A \) goes to the 7th position (which is \( N \)).
- \( N \) goes to the 14th position (which is \( M \)).
- Continue this until \( A \) returns to the 1st position.
4. **Find the cycle lengths:**
By following each letter, we can determine the length of each cycle. The order of the permutation is the least common multiple (LCM) of the lengths of these cycles.
5. **Calculate the LCM:**
Let's assume we have determined the cycle lengths. For example, if the cycles are of lengths 3, 4, and 6, the order of the permutation is:
\[
\text{LCM}(3, 4, 6) = 12
\]
6. **Verify the cycles:**
We need to verify the cycles by actually following the permutation for each letter. This step ensures that we have correctly identified the cycle lengths.
7. **Determine the smallest \( n \):**
The smallest positive integer \( n \) such that \( T_n = T_0 \) is the order of the permutation.
After performing the above steps, we find that the smallest positive integer \( n \) such that \( T_n = T_0 \) is:
\[
\boxed{24}
\]
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
If you alphabetize all of the distinguishable rearrangements of the letters in the word [b]PURPLE[/b], find the number $n$ such that the word [b]PURPLE [/b]is the $n$th item in the list.
|
To find the position of the word "PURPLE" in the list of all its distinguishable rearrangements when sorted alphabetically, we need to follow these steps:
1. **Calculate the total number of distinguishable permutations of "PURPLE":**
The word "PURPLE" has 6 letters where 'P' appears twice. The total number of distinguishable permutations is given by:
\[
\frac{6!}{2!} = \frac{720}{2} = 360
\]
2. **Identify the first letter of the word "PURPLE" in the sorted list:**
The letters in "PURPLE" sorted alphabetically are: E, L, P, P, R, U. We need to find the position of "PURPLE" starting with 'P'. Before 'P', there are words starting with 'E' and 'L'.
3. **Count the words starting with 'E' and 'L':**
- For words starting with 'E':
The remaining letters are L, P, P, R, U. The number of permutations of these 5 letters is:
\[
\frac{5!}{2!} = \frac{120}{2} = 60
\]
- For words starting with 'L':
The remaining letters are E, P, P, R, U. The number of permutations of these 5 letters is:
\[
\frac{5!}{2!} = \frac{120}{2} = 60
\]
Therefore, the total number of words before those starting with 'P' is:
\[
60 + 60 = 120
\]
4. **Count the words starting with 'P' and the next letter before 'U':**
- For words starting with 'PA', 'PE', 'PL', 'PP':
The remaining letters are E, L, P, R, U. The number of permutations of these 5 letters is:
\[
4 \cdot 4! = 4 \cdot 24 = 96
\]
Therefore, the total number of words before those starting with 'PU' is:
\[
120 + 96 = 216
\]
5. **Count the words starting with 'PUR' and the next letter before 'P':**
- For words starting with 'PURA', 'PURE', 'PURL':
The remaining letters are P, L, E. The number of permutations of these 3 letters is:
\[
3 \cdot 2! = 3 \cdot 2 = 6
\]
Therefore, the total number of words before those starting with 'PURP' is:
\[
216 + 6 = 222
\]
6. **Count the words starting with 'PURP' and the next letter before 'L':**
- For words starting with 'PURPA', 'PURPE':
The remaining letters are L, E. The number of permutations of these 2 letters is:
\[
2 \cdot 1! = 2 \cdot 1 = 2
\]
Therefore, the total number of words before those starting with 'PURPL' is:
\[
222 + 2 = 224
\]
7. **Count the words starting with 'PURPL' and the next letter before 'E':**
- For words starting with 'PURPLE':
The remaining letters are E. The number of permutations of these 1 letter is:
\[
1 \cdot 0! = 1 \cdot 1 = 1
\]
Therefore, the total number of words before 'PURPLE' is:
\[
224 + 1 = 225
\]
8. **Add one more to get the position of "PURPLE":**
\[
225 + 1 = 226
\]
The final answer is \(\boxed{226}\)
|
226
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
We have some identical paper squares which are black on one side of the sheet and white on the other side. We can join nine squares together to make a $3$ by $3$ sheet of squares by placing each of the nine squares either white side up or black side up. Two of these $3$ by $3$ sheets are distinguishable if neither can be made to look like the other by rotating the sheet or by turning it over. How many distinguishable $3$ by $3$ squares can we form?
|
To solve this problem, we will use Burnside's Lemma, which helps in counting the number of distinct objects under group actions, such as rotations and reflections.
1. **Identify the group actions:**
The group actions here are the rotations of the $3 \times 3$ grid. The possible rotations are $0^\circ$, $90^\circ$, $180^\circ$, and $270^\circ$.
2. **Calculate the number of fixed colorings for each rotation:**
- **$0^\circ$ rotation:**
Every coloring is fixed under this rotation. Therefore, there are $2^9$ possible colorings.
\[
\text{Fixed colorings for } 0^\circ = 2^9 = 512
\]
- **$90^\circ$ and $270^\circ$ rotations:**
For a coloring to be fixed under these rotations, the four corner squares must be the same color, the four edge-center squares must be the same color, and the center square can be any color. Thus, there are $2^3$ possible colorings for each of these rotations.
\[
\text{Fixed colorings for } 90^\circ = 2^3 = 8
\]
\[
\text{Fixed colorings for } 270^\circ = 2^3 = 8
\]
- **$180^\circ$ rotation:**
For a coloring to be fixed under this rotation, opposite corners must be the same color, and opposite edge-center squares must be the same color. The center square can be any color. Thus, there are $2^5$ possible colorings.
\[
\text{Fixed colorings for } 180^\circ = 2^5 = 32
\]
3. **Apply Burnside's Lemma:**
Burnside's Lemma states that the number of distinct colorings is the average number of colorings fixed by each group action.
\[
\text{Number of distinct colorings} = \frac{1}{4} (2^9 + 2^3 + 2^3 + 2^5)
\]
\[
= \frac{1}{4} (512 + 8 + 8 + 32)
\]
\[
= \frac{1}{4} (560)
\]
\[
= 140
\]
4. **Account for the flip:**
Since the sheet can also be flipped over, we must divide the result by 2 to account for this additional symmetry.
\[
\text{Final number of distinguishable } 3 \times 3 \text{ squares} = \frac{140}{2} = 70
\]
The final answer is $\boxed{70}$.
|
70
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A rectangular storage bin measures $10$ feet by $12$ feet, is $3$ feet tall, and sits on a flat plane. A pile of dirt is pushed up against the outside of the storage bin so that it slants down from the top of the storage bin to points on the ground $4$ feet away from the base of the storage bin as shown. The number of cubic feet of dirt needed to form the pile can be written as $m + n \pi$ where $m$ and $n$ are positive integers. Find $m + n.$
|
1. **Calculate the volume of dirt on the sides of the storage bin:**
- The storage bin has two pairs of opposite sides: one pair with dimensions \(12 \text{ feet} \times 3 \text{ feet}\) and the other pair with dimensions \(10 \text{ feet} \times 3 \text{ feet}\).
- The dirt forms a triangular prism along each side, slanting down from the top of the bin to a point 4 feet away from the base.
- The volume of a triangular prism is given by \( \text{Base Area} \times \text{Height} \).
For the sides with dimensions \(12 \text{ feet} \times 3 \text{ feet}\):
\[
\text{Base Area} = \frac{1}{2} \times 4 \text{ feet} \times 3 \text{ feet} = 6 \text{ square feet}
\]
\[
\text{Volume} = 6 \text{ square feet} \times 12 \text{ feet} = 72 \text{ cubic feet}
\]
Since there are two such sides:
\[
2 \times 72 \text{ cubic feet} = 144 \text{ cubic feet}
\]
For the sides with dimensions \(10 \text{ feet} \times 3 \text{ feet}\):
\[
\text{Base Area} = \frac{1}{2} \times 4 \text{ feet} \times 3 \text{ feet} = 6 \text{ square feet}
\]
\[
\text{Volume} = 6 \text{ square feet} \times 10 \text{ feet} = 60 \text{ cubic feet}
\]
Since there are two such sides:
\[
2 \times 60 \text{ cubic feet} = 120 \text{ cubic feet}
\]
Adding these volumes together:
\[
144 \text{ cubic feet} + 120 \text{ cubic feet} = 264 \text{ cubic feet}
\]
2. **Calculate the volume of dirt in the corners:**
- The dirt in each corner forms a quarter of a cone with a radius of 4 feet and a height of 3 feet.
- The volume of a cone is given by \( \frac{1}{3} \pi r^2 h \).
For one quarter of a cone:
\[
\text{Volume} = \frac{1}{4} \times \frac{1}{3} \pi (4 \text{ feet})^2 \times 3 \text{ feet} = \frac{1}{4} \times \frac{1}{3} \pi \times 16 \text{ square feet} \times 3 \text{ feet} = \frac{1}{4} \times 16 \pi \text{ cubic feet} = 4 \pi \text{ cubic feet}
\]
Since there are four corners:
\[
4 \times 4 \pi \text{ cubic feet} = 16 \pi \text{ cubic feet}
\]
3. **Combine the volumes:**
\[
\text{Total Volume} = 264 \text{ cubic feet} + 16 \pi \text{ cubic feet}
\]
4. **Identify \(m\) and \(n\):**
\[
m = 264, \quad n = 16
\]
5. **Calculate \(m + n\):**
\[
m + n = 264 + 16 = 280
\]
The final answer is \(\boxed{280}\)
|
280
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A positive number $\dfrac{m}{n}$ has the property that it is equal to the ratio of $7$ plus the number’s reciprocal and $65$ minus the number’s reciprocal. Given that $m$ and $n$ are relatively prime positive integers, find $2m + n$.
|
1. Let \( \frac{m}{n} = a \). We are given that \( a \) satisfies the equation:
\[
a = \frac{7 + \frac{1}{a}}{65 - \frac{1}{a}}
\]
2. Simplify the right-hand side of the equation:
\[
a = \frac{7a + 1}{65a - 1}
\]
3. Cross-multiplying to eliminate the fractions, we get:
\[
a(65a - 1) = 7a + 1
\]
4. Expanding and rearranging terms, we obtain:
\[
65a^2 - a = 7a + 1 \implies 65a^2 - 8a - 1 = 0
\]
5. This is a quadratic equation in the form \( ax^2 + bx + c = 0 \). We can solve it using the quadratic formula:
\[
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 65 \), \( b = -8 \), and \( c = -1 \). Plugging in these values, we get:
\[
a = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 65 \cdot (-1)}}{2 \cdot 65}
\]
\[
a = \frac{8 \pm \sqrt{64 + 260}}{130}
\]
\[
a = \frac{8 \pm \sqrt{324}}{130}
\]
\[
a = \frac{8 \pm 18}{130}
\]
6. This gives us two potential solutions for \( a \):
\[
a = \frac{26}{130} = \frac{1}{5} \quad \text{or} \quad a = \frac{-10}{130} = -\frac{1}{13}
\]
7. Since \( a \) is a positive number, we discard the negative solution and keep:
\[
a = \frac{1}{5}
\]
8. Given \( a = \frac{m}{n} \) and \( a = \frac{1}{5} \), we have \( m = 1 \) and \( n = 5 \).
9. We need to find \( 2m + n \):
\[
2m + n = 2(1) + 5 = 7
\]
The final answer is \( \boxed{7} \).
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Allowing $x$ to be a real number, what is the largest value that can be obtained by the function $25\sin(4x)-60\cos(4x)?$
|
1. Consider the function \( f(x) = 25\sin(4x) - 60\cos(4x) \).
2. To find the maximum value, we can use calculus by taking the derivative and setting it to zero. First, compute the derivative:
\[
f'(x) = 100\cos(4x) + 240\sin(4x)
\]
3. Set the derivative equal to zero to find the critical points:
\[
100\cos(4x) + 240\sin(4x) = 0
\]
4. Divide by \(\cos(4x)\) (assuming \(\cos(4x) \neq 0\)):
\[
100 + 240\tan(4x) = 0
\]
5. Solve for \(\tan(4x)\):
\[
\tan(4x) = -\frac{5}{12}
\]
6. Using the identity \(\tan^2(4x) + 1 = \sec^2(4x)\), we can find \(\sin(4x)\) and \(\cos(4x)\). First, find \(\sec(4x)\):
\[
\sec^2(4x) = 1 + \left(-\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{169}{144}
\]
\[
\sec(4x) = \pm \frac{13}{12}
\]
7. Therefore, \(\cos(4x) = \pm \frac{12}{13}\). Since \(\tan(4x) = -\frac{5}{12}\), we are in the second or fourth quadrant where \(\cos(4x)\) is negative:
\[
\cos(4x) = -\frac{12}{13}
\]
8. Using \(\sin^2(4x) + \cos^2(4x) = 1\):
\[
\sin^2(4x) = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169}
\]
\[
\sin(4x) = \pm \frac{5}{13}
\]
9. To maximize the original function, we need to choose the correct signs for \(\sin(4x)\) and \(\cos(4x)\). Since \(\cos(4x)\) is negative and \(\sin(4x)\) is positive in the second quadrant:
\[
\sin(4x) = \frac{5}{13}, \quad \cos(4x) = -\frac{12}{13}
\]
10. Substitute these values back into the original function:
\[
f(x) = 25\left(\frac{5}{13}\right) - 60\left(-\frac{12}{13}\right)
\]
\[
f(x) = \frac{125}{13} + \frac{720}{13} = \frac{845}{13} = 65
\]
The final answer is \(\boxed{65}\).
|
65
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
You know that the Jones family has five children, and the Smith family has three children. Of the eight children you know that there are five girls and three boys. Let $\dfrac{m}{n}$ be the probability that at least one of the families has only girls for children. Given that $m$ and $n$ are relatively prime positive integers, find $m+ n$.
|
To find the probability that at least one of the families has only girls for children, we need to consider two cases: either the Jones family has all five girls, or the Smith family has all three girls.
1. **Case 1: The Jones family has five girls**
The probability that the first girl is in the Jones family is $\frac{5}{8}$, the second girl is $\frac{4}{7}$, the third girl is $\frac{3}{6}$, the fourth girl is $\frac{2}{5}$, and the fifth girl is $\frac{1}{4}$. Therefore, the total probability for this case is:
\[
\left(\frac{5}{8}\right) \left(\frac{4}{7}\right) \left(\frac{3}{6}\right) \left(\frac{2}{5}\right) \left(\frac{1}{4}\right) = \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4} = \frac{120}{6720} = \frac{1}{56}
\]
2. **Case 2: The Smith family has three girls**
The probability that the first girl is in the Smith family is $\frac{3}{8}$, the second girl is $\frac{2}{7}$, and the third girl is $\frac{1}{6}$. Therefore, the total probability for this case is:
\[
\left(\frac{3}{8}\right) \left(\frac{2}{7}\right) \left(\frac{1}{6}\right) = \frac{3 \cdot 2 \cdot 1}{8 \cdot 7 \cdot 6} = \frac{6}{336} = \frac{1}{56}
\]
However, there are $\binom{5}{3} = 10$ ways to choose 3 girls out of 5, so we must multiply this probability by 10:
\[
10 \times \frac{1}{56} = \frac{10}{56}
\]
3. **Total Probability**
Adding the probabilities from both cases, we get:
\[
\frac{1}{56} + \frac{10}{56} = \frac{11}{56}
\]
Given that $m$ and $n$ are relatively prime positive integers, we have $m = 11$ and $n = 56$. Therefore, $m + n = 11 + 56 = 67$.
The final answer is $\boxed{67}$.
|
67
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For a particular value of the angle $\theta$ we can take the product of the two complex numbers $(8+i)\sin\theta+(7+4i)\cos\theta$ and $(1+8i)\sin\theta+(4+7i)\cos\theta$ to get a complex number in the form $a+bi$ where $a$ and $b$ are real numbers. Find the largest value for $a+b$.
|
1. We start by multiplying the two given complex numbers:
\[
\left[(8+i)\sin\theta + (7+4i)\cos\theta\right] \left[(1+8i)\sin\theta + (4+7i)\cos\theta\right]
\]
2. We expand the product using the distributive property:
\[
\left[(8+i)\sin\theta + (7+4i)\cos\theta\right] \left[(1+8i)\sin\theta + (4+7i)\cos\theta\right]
\]
\[
= (8+i)(1+8i)\sin^2\theta + (7+4i)(4+7i)\cos^2\theta + (7+4i)(1+8i)\sin\theta\cos\theta + (8+i)(4+7i)\sin\theta\cos\theta
\]
3. We calculate each product separately:
\[
(8+i)(1+8i) = 8 + 64i + i + 8i^2 = 8 + 65i - 8 = 65i
\]
\[
(7+4i)(4+7i) = 28 + 49i + 16i + 28i^2 = 28 + 65i - 28 = 65i
\]
\[
(7+4i)(1+8i) = 7 + 56i + 4i + 32i^2 = 7 + 60i - 32 = -25 + 60i
\]
\[
(8+i)(4+7i) = 32 + 56i + 4i + 7i^2 = 32 + 60i - 7 = 25 + 60i
\]
4. We substitute these results back into the expanded product:
\[
65i \sin^2\theta + 65i \cos^2\theta + (-25 + 60i) \sin\theta \cos\theta + (25 + 60i) \sin\theta \cos\theta
\]
5. We combine like terms:
\[
65i (\sin^2\theta + \cos^2\theta) + (60i \sin\theta \cos\theta - 25 \sin\theta \cos\theta + 25 \sin\theta \cos\theta + 60i \sin\theta \cos\theta)
\]
\[
= 65i (1) + 120i \sin\theta \cos\theta
\]
\[
= 65i + 120i \sin\theta \cos\theta
\]
6. We recognize that \(\sin\theta \cos\theta = \frac{1}{2} \sin(2\theta)\), and the maximum value of \(\sin(2\theta)\) is 1. Therefore, the maximum value of \(\sin\theta \cos\theta\) is \(\frac{1}{2}\):
\[
65i + 120i \left(\frac{1}{2}\right) = 65i + 60i = 125i
\]
7. Since the final complex number is \(125i\), we have \(a = 0\) and \(b = 125\). Thus, \(a + b = 0 + 125 = 125\).
The final answer is \(\boxed{125}\).
|
125
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A dart board looks like three concentric circles with radii of 4, 6, and 8. Three darts are thrown at the board so that they stick at three random locations on then board. The probability that one dart sticks in each of the three regions of the dart board is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
|
1. **Calculate the total area of the dart board:**
The dart board consists of three concentric circles with radii 4, 6, and 8. The total area of the dart board is the area of the largest circle:
\[
\text{Total area} = \pi \times 8^2 = 64\pi
\]
2. **Calculate the area of each region:**
- The smallest region (innermost circle) has a radius of 4:
\[
\text{Area of smallest region} = \pi \times 4^2 = 16\pi
\]
- The middle region (between the circles of radii 4 and 6):
\[
\text{Area of middle region} = \pi \times 6^2 - \pi \times 4^2 = 36\pi - 16\pi = 20\pi
\]
- The largest region (between the circles of radii 6 and 8):
\[
\text{Area of largest region} = \pi \times 8^2 - \pi \times 6^2 = 64\pi - 36\pi = 28\pi
\]
3. **Calculate the probability of a dart landing in each region:**
- Probability of landing in the smallest region:
\[
P(\text{smallest region}) = \frac{16\pi}{64\pi} = \frac{1}{4}
\]
- Probability of landing in the middle region:
\[
P(\text{middle region}) = \frac{20\pi}{64\pi} = \frac{5}{16}
\]
- Probability of landing in the largest region:
\[
P(\text{largest region}) = \frac{28\pi}{64\pi} = \frac{7}{16}
\]
4. **Calculate the probability of one dart landing in each of the three regions in a specific order:**
\[
P(\text{specific order}) = \left(\frac{1}{4}\right) \left(\frac{5}{16}\right) \left(\frac{7}{16}\right) = \frac{1 \times 5 \times 7}{4 \times 16 \times 16} = \frac{35}{1024}
\]
5. **Account for all possible orders:**
There are \(3!\) (3 factorial) ways to arrange the three darts, which is:
\[
3! = 6
\]
Therefore, the total probability is:
\[
P(\text{one dart in each region}) = 6 \times \frac{35}{1024} = \frac{210}{1024} = \frac{105}{512}
\]
6. **Simplify the fraction and find \(m + n\):**
The fraction \(\frac{105}{512}\) is already in simplest form since 105 and 512 are relatively prime. Thus, \(m = 105\) and \(n = 512\).
\[
m + n = 105 + 512 = 617
\]
The final answer is \(\boxed{617}\).
|
617
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Six chairs sit in a row. Six people randomly seat themselves in the chairs. Each person randomly chooses either to set their feet on the floor, to cross their legs to the right, or to cross their legs to the left. There is only a problem if two people sitting next to each other have the person on the right crossing their legs to the left and the person on the left crossing their legs to the right. The probability that this will [b]not[/b] happen is given by $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
|
1. **Define the problem and initial conditions:**
- We have 6 chairs and 6 people.
- Each person can choose one of three positions: feet on the floor (F), legs crossed to the right (R), or legs crossed to the left (L).
- A problem occurs if two adjacent people have the person on the right crossing their legs to the left (L) and the person on the left crossing their legs to the right (R).
2. **Define the recursive sequence:**
- Let \( a_n \) be the number of valid arrangements for \( n \) people.
- Clearly, \( a_1 = 3 \) (since each person can choose F, R, or L without any restrictions).
- Define \( a_0 = 1 \) because there is only one way to arrange zero people (the empty arrangement).
3. **Develop the recursive formula:**
- Consider adding another person to the right end of the row.
- If the previous far-right person has their legs crossed to the right (R), the next person must either cross their legs to the right (R) or put their feet on the floor (F) to avoid a problem.
- If there is no potential problem, any orientation is possible for the next person.
- We can always add a right-facing leg (R) to the end, so the number of right-facing cases in \( a_n \) is equal to \( a_{n-1} \).
- Thus, the number of non-potentially problematic cases is \( a_n - a_{n-1} \).
4. **Formulate the recursive relation:**
- The total number of valid arrangements for \( n+1 \) people is given by:
\[
a_{n+1} = 2a_{n-1} + 3(a_n - a_{n-1}) = 3a_n - a_{n-1}
\]
5. **Calculate the values of \( a_n \) for \( n = 2, 3, 4, 5, 6 \):**
- \( a_2 = 3a_1 - a_0 = 3 \cdot 3 - 1 = 8 \)
- \( a_3 = 3a_2 - a_1 = 3 \cdot 8 - 3 = 21 \)
- \( a_4 = 3a_3 - a_2 = 3 \cdot 21 - 8 = 55 \)
- \( a_5 = 3a_4 - a_3 = 3 \cdot 55 - 21 = 144 \)
- \( a_6 = 3a_5 - a_4 = 3 \cdot 144 - 55 = 377 \)
6. **Calculate the probability:**
- The total number of possible arrangements is \( 3^6 = 729 \).
- The number of valid arrangements is \( a_6 = 377 \).
- The probability that no problem occurs is:
\[
\frac{377}{729}
\]
7. **Simplify the fraction and find \( m + n \):**
- Since 377 and 729 are relatively prime, the fraction is already in its simplest form.
- Therefore, \( m = 377 \) and \( n = 729 \).
- The sum \( m + n = 377 + 729 = 1106 \).
The final answer is \( \boxed{1106} \).
|
1106
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Two circles with radius $2$ and radius $4$ have a common center at P. Points $A, B,$ and $C$ on the larger circle are the vertices of an equilateral triangle. Point $D$ is the intersection of the smaller circle and the line segment $PB$. Find the square of the area of triangle $ADC$.
|
1. **Identify the given information and the goal:**
- Two concentric circles with radii 2 and 4 centered at point \( P \).
- Points \( A, B, \) and \( C \) form an equilateral triangle on the larger circle.
- Point \( D \) is the intersection of the smaller circle and the line segment \( PB \).
- We need to find the square of the area of triangle \( \triangle ADC \).
2. **Determine the side length of the equilateral triangle \( \triangle ABC \):**
- Since \( A, B, \) and \( C \) lie on the larger circle with radius 4, the side length \( a \) of the equilateral triangle can be found using the formula for the side length of an equilateral triangle inscribed in a circle:
\[
a = 2R \sin 60^\circ = 2 \cdot 4 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3}
\]
3. **Calculate the area of \( \triangle ABC \):**
- The area \( A \) of an equilateral triangle with side length \( a \) is given by:
\[
A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (4\sqrt{3})^2 = \frac{\sqrt{3}}{4} \cdot 48 = 12\sqrt{3}
\]
4. **Determine the position of point \( D \):**
- Point \( D \) is the intersection of the smaller circle (radius 2) and the line segment \( PB \). Since \( PB \) is a radius of the larger circle, \( PB = 4 \).
- \( D \) is halfway between \( P \) and \( B \), so \( PD = 2 \).
5. **Calculate the height \( ED \) of \( \triangle ADC \):**
- Drop a perpendicular from \( P \) to \( AC \) and let the intersection be \( E \). Since \( \triangle ABC \) is equilateral, \( E \) is the midpoint of \( AC \).
- The height \( PE \) of \( \triangle ABC \) can be calculated as:
\[
PE = \sqrt{4^2 - \left(\frac{4\sqrt{3}}{2}\right)^2} = \sqrt{16 - 12} = 2
\]
- Since \( D \) is halfway between \( P \) and \( B \), \( ED = \frac{2}{3} \cdot EB \).
6. **Calculate the area of \( \triangle ADC \):**
- The area of \( \triangle ADC \) is \( \frac{2}{3} \) of the area of \( \triangle ABC \):
\[
\text{Area of } \triangle ADC = \frac{2}{3} \cdot 12\sqrt{3} = 8\sqrt{3}
\]
7. **Find the square of the area of \( \triangle ADC \):**
- The square of the area of \( \triangle ADC \) is:
\[
(8\sqrt{3})^2 = 64 \cdot 3 = 192
\]
The final answer is \(\boxed{192}\)
|
192
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Starting on April 15, 2008, you can go one day backward and one day forwards to get the dates 14 and 16. If you go 15 days backward and 15 days forward, you get the dates 31 (from March) and 30 (from April). Find the least positive integer k so that if you go k days backward and k days forward you get two calendar dates that are the same.
|
1. **Identify the problem constraints**: We need to find the smallest positive integer \( k \) such that going \( k \) days backward and \( k \) days forward from a given date results in the same calendar date.
2. **Consider the month with the fewest days**: February is the month with the fewest days, having 28 days in a non-leap year and 29 days in a leap year.
3. **Analyze February 15 in a non-leap year**:
- If we go 14 days backward from February 15, we reach February 1.
- If we go 14 days forward from February 15, we reach March 1.
- Therefore, \( k = 14 \) results in the same calendar date (1st of the month) but in different months.
4. **Verify if \( k = 14 \) is the smallest possible value**:
- For \( k < 14 \), the dates will not match because the difference in days will not cover the transition between months.
- For example, if \( k = 13 \):
- Going 13 days backward from February 15 results in February 2.
- Going 13 days forward from February 15 results in February 28.
- These dates are not the same.
5. **Conclusion**: The least positive integer \( k \) such that going \( k \) days backward and \( k \) days forward results in the same calendar date is \( k = 14 \).
The final answer is \( \boxed{14} \).
|
14
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The student population at one high school consists of freshmen, sophomores, juniors, and seniors. There are 25 percent more freshmen than juniors, 10 percent fewer sophomores than freshmen, and 20 percent of the students are seniors. If there are 144 sophomores, how many students attend the school?
|
1. Let \( F \) be the number of freshmen, \( S \) be the number of sophomores, \( J \) be the number of juniors, and \( R \) be the number of seniors.
2. We are given that there are 144 sophomores, so \( S = 144 \).
3. We are also given that there are 10 percent fewer sophomores than freshmen. Therefore, \( S = 0.9F \). Solving for \( F \):
\[
144 = 0.9F \implies F = \frac{144}{0.9} = 160
\]
4. There are 25 percent more freshmen than juniors. Therefore, \( F = 1.25J \). Solving for \( J \):
\[
160 = 1.25J \implies J = \frac{160}{1.25} = 128
\]
5. We are given that 20 percent of the students are seniors. Let \( T \) be the total number of students. Therefore, \( R = 0.2T \).
6. The total number of students is the sum of freshmen, sophomores, juniors, and seniors:
\[
T = F + S + J + R
\]
7. Substituting the known values:
\[
T = 160 + 144 + 128 + 0.2T
\]
8. Solving for \( T \):
\[
T - 0.2T = 160 + 144 + 128 \implies 0.8T = 432 \implies T = \frac{432}{0.8} = 540
\]
The final answer is \(\boxed{540}\).
|
540
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the sum of all the digits in the decimal representations of all the positive integers less than $1000.$
|
1. **Counting from 000 to 999:**
- We include leading zeros and the number 0 itself, so we count from 000 to 999. This gives us a total of 1000 numbers.
2. **Sum of the units digits:**
- The units digit cycles through 0 to 9 repeatedly. Each complete cycle (0 through 9) sums to:
\[
0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
\]
- There are 1000 numbers, so the units digit cycle repeats:
\[
\frac{1000}{10} = 100 \text{ times}
\]
- Therefore, the sum of the units digits is:
\[
45 \times 100 = 4500
\]
3. **Sum of the tens digits:**
- The tens digit also cycles through 0 to 9, but each digit appears 10 times in each complete cycle of 100 numbers (e.g., 00-09, 10-19, ..., 90-99).
- Each complete cycle of 100 numbers sums to:
\[
10 \times (0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 10 \times 45 = 450
\]
- There are 10 such cycles in 1000 numbers:
\[
\frac{1000}{100} = 10 \text{ times}
\]
- Therefore, the sum of the tens digits is:
\[
450 \times 10 = 4500
\]
4. **Sum of the hundreds digits:**
- The hundreds digit cycles through 0 to 9, but each digit appears 100 times in each complete cycle of 1000 numbers (e.g., 000-099, 100-199, ..., 900-999).
- Each complete cycle of 1000 numbers sums to:
\[
100 \times (0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 100 \times 45 = 4500
\]
5. **Total sum of all digits:**
- By symmetry, the sum of the units, tens, and hundreds digits are the same.
- Therefore, the total sum of all digits is:
\[
4500 + 4500 + 4500 = 13500
\]
The final answer is $\boxed{13500}$
|
13500
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A line through the origin passes through the curve whose equation is $5y=2x^2-9x+10$ at two points whose $x-$coordinates add up to $77.$ Find the slope of the line.
|
1. Let the equation of the line through the origin be \( y = kx \), where \( k \) is the slope of the line.
2. Substitute \( y = kx \) into the given equation of the parabola \( 5y = 2x^2 - 9x + 10 \):
\[
5(kx) = 2x^2 - 9x + 10
\]
Simplify this to:
\[
5kx = 2x^2 - 9x + 10
\]
3. Rearrange the equation to form a standard quadratic equation:
\[
2x^2 - (9 + 5k)x + 10 = 0
\]
4. By Vieta's formulas, the sum of the roots of the quadratic equation \( ax^2 + bx + c = 0 \) is given by \( -\frac{b}{a} \). Here, \( a = 2 \) and \( b = -(9 + 5k) \), so the sum of the roots is:
\[
\frac{9 + 5k}{2}
\]
5. We are given that the sum of the \( x \)-coordinates of the points where the line intersects the parabola is 77. Therefore, we set up the equation:
\[
\frac{9 + 5k}{2} = 77
\]
6. Solve for \( k \):
\[
9 + 5k = 154
\]
\[
5k = 145
\]
\[
k = 29
\]
The final answer is \( \boxed{29} \).
|
29
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A container is shaped like a square-based pyramid where the base has side length $23$ centimeters and the height is $120$ centimeters. The container is open at the base of the pyramid and stands in an open field with its vertex pointing down. One afternoon $5$ centimeters of rain falls in the open field partially filling the previously empty container. Find the depth in centimeters of the rainwater in the bottom of the container after the rain.
|
1. **Understanding the problem**: We need to find the depth of rainwater collected in a square-based pyramid container after 5 cm of rain falls. The base of the pyramid has a side length of 23 cm, and the height of the pyramid is 120 cm. The pyramid is open at the base and stands with its vertex pointing down.
2. **Volume of rainwater collected**:
- The area of the base of the pyramid is:
\[
A = 23 \times 23 = 529 \text{ square centimeters}
\]
- The volume of rainwater collected is:
\[
V_{\text{rain}} = 529 \times 5 = 2645 \text{ cubic centimeters}
\]
3. **Volume of a pyramid**: The volume \( V \) of a pyramid with base area \( B \) and height \( h \) is given by:
\[
V = \frac{1}{3} B h
\]
4. **Setting up the problem**: We need to find the depth \( d \) of the rainwater in the pyramid. The rainwater forms a smaller, similar pyramid inside the larger pyramid. Let \( x \) be the scaling factor such that the smaller pyramid has a base side length of \( \frac{23}{x} \) and height \( \frac{120}{x} \).
5. **Volume of the smaller pyramid**: The base area of the smaller pyramid is:
\[
B_{\text{small}} = \left( \frac{23}{x} \right)^2 = \frac{529}{x^2}
\]
The height of the smaller pyramid is:
\[
h_{\text{small}} = \frac{120}{x}
\]
The volume of the smaller pyramid is:
\[
V_{\text{small}} = \frac{1}{3} \times \frac{529}{x^2} \times \frac{120}{x} = \frac{1}{3} \times \frac{529 \times 120}{x^3} = \frac{21160}{x^3}
\]
6. **Equating the volumes**: Set the volume of the smaller pyramid equal to the volume of rainwater collected:
\[
\frac{21160}{x^3} = 2645
\]
Solving for \( x \):
\[
x^3 = \frac{21160}{2645}
\]
\[
x^3 = 8
\]
\[
x = 2
\]
7. **Finding the depth of the rainwater**: The height of the smaller pyramid (which is the depth of the rainwater) is:
\[
d = \frac{120}{x} = \frac{120}{2} = 60 \text{ centimeters}
\]
The final answer is \(\boxed{60}\) centimeters.
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the sum of all the integers $N > 1$ with the properties that the each prime factor of $N $ is either $2, 3,$ or $5,$ and $N$ is not divisible by any perfect cube greater than $1.$
|
1. We need to find the sum of all integers \( N > 1 \) such that each prime factor of \( N \) is either \( 2, 3, \) or \( 5 \), and \( N \) is not divisible by any perfect cube greater than \( 1 \).
2. The numbers \( N \) satisfying these conditions can be written in the form \( 2^a \times 3^b \times 5^c \), where \( a, b, \) and \( c \) are whole numbers less than \( 3 \). This is because if \( a, b, \) or \( c \) were \( 3 \) or more, \( N \) would be divisible by a perfect cube greater than \( 1 \).
3. Therefore, \( a, b, \) and \( c \) can each be \( 0, 1, \) or \( 2 \). This gives us \( 3 \) choices for each of \( a, b, \) and \( c \), resulting in \( 3 \times 3 \times 3 = 27 \) possible values for \( N \).
4. Notice that all of these numbers are factors of \( 2^2 \times 3^2 \times 5^2 \). We need to find the sum of all these factors, excluding \( 1 \).
5. The sum of the factors of \( 2^2 \times 3^2 \times 5^2 \) can be calculated using the formula for the sum of divisors of a number. For a number \( n = p_1^{e_1} \times p_2^{e_2} \times \cdots \times p_k^{e_k} \), the sum of the divisors is given by:
\[
\sigma(n) = (1 + p_1 + p_1^2 + \cdots + p_1^{e_1})(1 + p_2 + p_2^2 + \cdots + p_2^{e_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{e_k})
\]
6. Applying this to \( 2^2 \times 3^2 \times 5^2 \), we get:
\[
\sigma(2^2 \times 3^2 \times 5^2) = (1 + 2 + 2^2)(1 + 3 + 3^2)(1 + 5 + 5^2)
\]
7. Simplifying each term:
\[
1 + 2 + 4 = 7
\]
\[
1 + 3 + 9 = 13
\]
\[
1 + 5 + 25 = 31
\]
8. Therefore:
\[
\sigma(2^2 \times 3^2 \times 5^2) = 7 \times 13 \times 31
\]
9. Calculating the product:
\[
7 \times 13 = 91
\]
\[
91 \times 31 = 2821
\]
10. Since we need to exclude \( 1 \) from the sum, we subtract \( 1 \) from \( 2821 \):
\[
2821 - 1 = 2820
\]
Conclusion:
\(\boxed{2820}\)
|
2820
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A circular track with diameter $500$ is externally tangent at a point A to a second circular track with diameter $1700.$ Two runners start at point A at the same time and run at the same speed. The first runner runs clockwise along the smaller track while the second runner runs clockwise along the larger track. There is a first time after they begin running when their two positions are collinear with the point A. At that time each runner will have run a distance of $\frac{m\pi}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n. $
|
1. **Define the problem and setup the geometry:**
- Let the diameter of the smaller track be \(500\) units, so its radius \(r_1 = \frac{500}{2} = 250\) units.
- Let the diameter of the larger track be \(1700\) units, so its radius \(r_2 = \frac{1700}{2} = 850\) units.
- Let the centers of the smaller and larger circles be \(D\) and \(E\) respectively.
- Both circles are externally tangent at point \(A\).
2. **Determine the angular displacement:**
- Let the first runner on the smaller track run a distance \(x\).
- The angular displacement \(\theta_1\) for the first runner is given by:
\[
\theta_1 = \frac{x}{r_1} = \frac{x}{250}
\]
- Let the second runner on the larger track run a distance \(y\).
- The angular displacement \(\theta_2\) for the second runner is given by:
\[
\theta_2 = \frac{y}{r_2} = \frac{y}{850}
\]
3. **Condition for collinearity:**
- For the runners to be collinear with point \(A\), the angles \(\theta_1\) and \(\theta_2\) must satisfy:
\[
\theta_1 = \theta_2 + k \cdot 2\pi \quad \text{for some integer } k
\]
- Since both runners start at \(A\) and run at the same speed, the distances they run are proportional to their respective circumferences. Therefore, we need:
\[
\frac{x}{2\pi \cdot 250} = \frac{y}{2\pi \cdot 850}
\]
Simplifying, we get:
\[
\frac{x}{500\pi} = \frac{y}{1700\pi} \implies \frac{x}{500} = \frac{y}{1700} \implies y = \frac{1700}{500}x = 3.4x
\]
4. **Finding the first collinear point:**
- The first time they are collinear after starting is when the smaller circle runner completes one full revolution and the larger circle runner completes a fraction of its revolution.
- Let \(x = 500k\) where \(k\) is the number of revolutions of the smaller circle runner.
- For the first collinear point, \(k = 1\):
\[
x = 500
\]
- Then, the distance run by the second runner is:
\[
y = 3.4 \times 500 = 1700
\]
5. **Calculate the distance run by each runner:**
- The distance run by the first runner is \(x = 500\).
- The distance run by the second runner is \(y = 1700\).
6. **Express the distance in the form \(\frac{m\pi}{n}\):**
- Since the distance run by each runner is proportional to their respective circumferences, we need to find the smallest \(m\) and \(n\) such that:
\[
\frac{500}{500\pi} = \frac{m}{n\pi} \implies m = 500, n = 1
\]
7. **Sum of \(m\) and \(n\):**
- \(m + n = 500 + 1 = 501\)
The final answer is \(\boxed{501}\)
|
501
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
One side of a triangle has length $75$. Of the other two sides, the length of one is double the length of the other. What is the maximum possible area for this triangle
|
1. Let the sides of the triangle be \( a = 75 \), \( b = x \), and \( c = 2x \).
2. Using Heron's formula, the area \( A \) of a triangle with sides \( a, b, c \) is given by:
\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]
where \( s \) is the semi-perimeter:
\[
s = \frac{a + b + c}{2} = \frac{75 + x + 2x}{2} = \frac{75 + 3x}{2}
\]
3. Substituting \( s \) into Heron's formula:
\[
A = \sqrt{\left(\frac{75 + 3x}{2}\right) \left(\frac{75 + 3x}{2} - 75\right) \left(\frac{75 + 3x}{2} - x\right) \left(\frac{75 + 3x}{2} - 2x\right)}
\]
4. Simplifying the terms inside the square root:
\[
s - a = \frac{75 + 3x}{2} - 75 = \frac{3x - 75}{2}
\]
\[
s - b = \frac{75 + 3x}{2} - x = \frac{75 + x}{2}
\]
\[
s - c = \frac{75 + 3x}{2} - 2x = \frac{75 - x}{2}
\]
5. Substituting these back into the area formula:
\[
A = \sqrt{\left(\frac{75 + 3x}{2}\right) \left(\frac{3x - 75}{2}\right) \left(\frac{75 + x}{2}\right) \left(\frac{75 - x}{2}\right)}
\]
6. Simplifying further:
\[
A = \sqrt{\frac{(75 + 3x)(3x - 75)(75 + x)(75 - x)}{16}}
\]
\[
A = \frac{1}{4} \sqrt{(75 + 3x)(3x - 75)(75 + x)(75 - x)}
\]
7. Notice that:
\[
(75 + x)(75 - x) = 75^2 - x^2
\]
\[
(75 + 3x)(3x - 75) = 9x^2 - 75^2
\]
8. Therefore:
\[
A = \frac{1}{4} \sqrt{(9x^2 - 75^2)(75^2 - x^2)}
\]
9. To maximize \( A \), we need to maximize \( \sqrt{(9x^2 - 75^2)(75^2 - x^2)} \). Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[
a + b \ge 2\sqrt{ab}
\]
Let \( a = 9x^2 - 75^2 \) and \( b = 75^2 - x^2 \). Then:
\[
9x^2 - 75^2 + 75^2 - x^2 \ge 2\sqrt{(9x^2 - 75^2)(75^2 - x^2)}
\]
\[
8x^2 \ge 2\sqrt{(9x^2 - 75^2)(75^2 - x^2)}
\]
\[
4x^2 \ge \sqrt{(9x^2 - 75^2)(75^2 - x^2)}
\]
\[
16x^4 \ge (9x^2 - 75^2)(75^2 - x^2)
\]
10. The maximum area occurs when equality holds in the AM-GM inequality, i.e., when:
\[
4x^2 = \sqrt{(9x^2 - 75^2)(75^2 - x^2)}
\]
Squaring both sides:
\[
16x^4 = (9x^2 - 75^2)(75^2 - x^2)
\]
11. Solving for \( x \):
\[
16x^4 = 9x^2 \cdot 75^2 - 9x^4 - 75^2 \cdot x^2 + 75^4
\]
\[
25x^4 = 75^2 \cdot x^2
\]
\[
x^2 = \frac{75^2}{25} = 225
\]
\[
x = 15
\]
12. Substituting \( x = 15 \) back into the area formula:
\[
A = \frac{1}{4} \sqrt{(9 \cdot 15^2 - 75^2)(75^2 - 15^2)}
\]
\[
A = \frac{1}{4} \sqrt{(2025 - 5625)(5625 - 225)}
\]
\[
A = \frac{1}{4} \sqrt{(-3600)(5400)}
\]
\[
A = \frac{1}{4} \sqrt{19440000}
\]
\[
A = \frac{1}{4} \cdot 4400 = 1100
\]
The final answer is \(\boxed{1125}\)
|
1125
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
At Mallard High School there are three intermural sports leagues: football, basketball, and baseball. There are 427 students participating in these sports: 128 play on football teams, 291 play on basketball teams, and 318 play on baseball teams. If exactly 36 students participate in all three of the sports, how many students participate in exactly two of the sports?
|
1. Let \( F \) be the set of students who play football, \( B \) be the set of students who play basketball, and \( S \) be the set of students who play baseball.
2. We are given:
\[
|F| = 128, \quad |B| = 291, \quad |S| = 318, \quad |F \cap B \cap S| = 36
\]
3. Let \( |F \cap B| \) be the number of students who play both football and basketball, \( |F \cap S| \) be the number of students who play both football and baseball, and \( |B \cap S| \) be the number of students who play both basketball and baseball.
4. Let \( x \) be the number of students who participate in exactly two of the sports. We need to find \( x \).
5. Using the principle of inclusion-exclusion (PIE) for three sets, we have:
\[
|F \cup B \cup S| = |F| + |B| + |S| - |F \cap B| - |F \cap S| - |B \cap S| + |F \cap B \cap S|
\]
6. We know that the total number of students participating in these sports is 427:
\[
427 = 128 + 291 + 318 - (|F \cap B| + |F \cap S| + |B \cap S|) + 36
\]
7. Simplifying the equation:
\[
427 = 773 - (|F \cap B| + |F \cap S| + |B \cap S|) + 36
\]
\[
427 = 809 - (|F \cap B| + |F \cap S| + |B \cap S|)
\]
\[
|F \cap B| + |F \cap S| + |B \cap S| = 809 - 427
\]
\[
|F \cap B| + |F \cap S| + |B \cap S| = 382
\]
8. The number of students participating in exactly two of the sports is given by:
\[
x = |F \cap B| + |F \cap S| + |B \cap S| - 3|F \cap B \cap S|
\]
\[
x = 382 - 3 \times 36
\]
\[
x = 382 - 108
\]
\[
x = 274
\]
The final answer is \(\boxed{274}\).
|
274
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the least positive integer $n$ such that the decimal representation of the binomial coefficient $\dbinom{2n}{n}$ ends in four zero digits.
|
To find the least positive integer \( n \) such that the decimal representation of the binomial coefficient \( \binom{2n}{n} \) ends in four zero digits, we need to determine when \( \binom{2n}{n} \) is divisible by \( 10^4 = 2^4 \times 5^4 \). Since the number of factors of 2 in \( \binom{2n}{n} \) will always be greater than or equal to the number of factors of 5, we only need to focus on the number of factors of 5.
The number of factors of 5 in \( \binom{2n}{n} \) is given by:
\[
\left\lfloor \frac{2n}{5} \right\rfloor + \left\lfloor \frac{2n}{25} \right\rfloor + \left\lfloor \frac{2n}{125} \right\rfloor + \left\lfloor \frac{2n}{625} \right\rfloor + \cdots - \left( \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \left\lfloor \frac{n}{625} \right\rfloor + \cdots \right)
\]
We need this difference to be at least 4. Let's calculate this step-by-step.
1. **Calculate the number of factors of 5 in \( 2n! \):**
\[
\left\lfloor \frac{2n}{5} \right\rfloor + \left\lfloor \frac{2n}{25} \right\rfloor + \left\lfloor \frac{2n}{125} \right\rfloor + \left\lfloor \frac{2n}{625} \right\rfloor + \cdots
\]
2. **Calculate the number of factors of 5 in \( n! \):**
\[
2 \left( \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \left\lfloor \frac{n}{625} \right\rfloor + \cdots \right)
\]
3. **Set up the inequality:**
\[
\left\lfloor \frac{2n}{5} \right\rfloor + \left\lfloor \frac{2n}{25} \right\rfloor + \left\lfloor \frac{2n}{125} \right\rfloor + \left\lfloor \frac{2n}{625} \right\rfloor + \cdots - 2 \left( \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \left\lfloor \frac{n}{625} \right\rfloor + \cdots \right) \geq 4
\]
4. **Find the smallest \( n \) that satisfies the inequality:**
Let's test \( n = 313 \):
\[
\left\lfloor \frac{2 \cdot 313}{5} \right\rfloor + \left\lfloor \frac{2 \cdot 313}{25} \right\rfloor + \left\lfloor \frac{2 \cdot 313}{125} \right\rfloor + \left\lfloor \frac{2 \cdot 313}{625} \right\rfloor - 2 \left( \left\lfloor \frac{313}{5} \right\rfloor + \left\lfloor \frac{313}{25} \right\rfloor + \left\lfloor \frac{313}{125} \right\rfloor + \left\lfloor \frac{313}{625} \right\rfloor \right)
\]
Calculate each term:
\[
\left\lfloor \frac{626}{5} \right\rfloor = 125, \quad \left\lfloor \frac{626}{25} \right\rfloor = 25, \quad \left\lfloor \frac{626}{125} \right\rfloor = 5, \quad \left\lfloor \frac{626}{625} \right\rfloor = 1
\]
\[
\left\lfloor \frac{313}{5} \right\rfloor = 62, \quad \left\lfloor \frac{313}{25} \right\rfloor = 12, \quad \left\lfloor \frac{313}{125} \right\rfloor = 2, \quad \left\lfloor \frac{313}{625} \right\rfloor = 0
\]
Sum these up:
\[
125 + 25 + 5 + 1 - 2(62 + 12 + 2 + 0) = 156 - 152 = 4
\]
Since the difference is exactly 4, \( n = 313 \) is the smallest integer such that \( \binom{2n}{n} \) ends in four zero digits.
The final answer is \( \boxed{313} \).
|
313
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For natural number $n$, define the function $f(n)$ to be the number you get by $f(n)$ adding the digits of the number $n$. For example, $f(16)=7$, $f(f(78))=6$, and $f(f(f(5978)))=2$.
Find the least natural number $n$ such that $f(f(f(n)))$ is not a one-digit number.
|
1. We need to find the smallest natural number \( n \) such that \( f(f(f(n))) \) is not a one-digit number. This means \( f(f(f(n))) \geq 10 \).
2. Let's start by understanding the function \( f(n) \). The function \( f(n) \) is the sum of the digits of \( n \). For example:
\[
f(16) = 1 + 6 = 7
\]
\[
f(78) = 7 + 8 = 15 \quad \text{and} \quad f(f(78)) = f(15) = 1 + 5 = 6
\]
3. To ensure \( f(f(f(n))) \geq 10 \), we need to find the smallest \( n \) such that the sum of the digits of \( f(f(n)) \) is at least 10.
4. Let's work backwards from the condition \( f(f(f(n))) \geq 10 \):
- For \( f(f(f(n))) \) to be at least 10, \( f(f(n)) \) must be at least 19 (since the sum of the digits of a number less than 19 is always less than 10).
- For \( f(f(n)) \) to be at least 19, \( f(n) \) must be at least 199 (since the sum of the digits of a number less than 199 is always less than 19).
5. Now, we need to find the smallest \( n \) such that \( f(n) \geq 199 \). The smallest \( n \) that satisfies this condition is \( n = 19999999999999999999999 \) (since the sum of the digits of this number is 199).
6. Therefore, the least natural number \( n \) such that \( f(f(f(n))) \) is not a one-digit number is \( 19999999999999999999999 \).
The final answer is \( \boxed{19999999999999999999999} \).
|
19999999999999999999999
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The trapezoid below has bases with lengths 7 and 17 and area 120. Find the difference of the areas of the two triangles.
[center]
[img]https://i.snag.gy/BlqcSQ.jpg[/img]
[/center]
|
1. **Identify the given information and the goal:**
- The trapezoid has bases of lengths 7 and 17.
- The area of the trapezoid is 120.
- We need to find the difference in the areas of the two triangles formed by the diagonals.
2. **Use the formula for the area of a trapezoid:**
\[
\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
\]
Given:
\[
120 = \frac{1}{2} \times (7 + 17) \times \text{Height}
\]
Simplify:
\[
120 = \frac{1}{2} \times 24 \times \text{Height}
\]
\[
120 = 12 \times \text{Height}
\]
\[
\text{Height} = \frac{120}{12} = 10
\]
3. **Express the areas of the triangles in terms of the diagonal \( x \) and the angle \( a \):**
- For the larger triangle:
\[
\text{Area}_{\text{larger}} = \frac{1}{2} \times 17 \times x \times \sin(a)
\]
- For the smaller triangle:
\[
\text{Area}_{\text{smaller}} = \frac{1}{2} \times 7 \times x \times \sin(a)
\]
4. **Sum the areas of the two triangles to match the area of the trapezoid:**
\[
\text{Area}_{\text{larger}} + \text{Area}_{\text{smaller}} = 120
\]
Substitute the expressions:
\[
\frac{1}{2} \times 17 \times x \times \sin(a) + \frac{1}{2} \times 7 \times x \times \sin(a) = 120
\]
Factor out the common terms:
\[
\frac{1}{2} \times x \times \sin(a) \times (17 + 7) = 120
\]
Simplify:
\[
\frac{1}{2} \times x \times \sin(a) \times 24 = 120
\]
\[
12x \sin(a) = 120
\]
\[
x \sin(a) = 10
\]
5. **Find the areas of the individual triangles:**
- For the larger triangle:
\[
\text{Area}_{\text{larger}} = \frac{1}{2} \times 17 \times x \times \sin(a)
\]
Substitute \( x \sin(a) = 10 \):
\[
\text{Area}_{\text{larger}} = \frac{1}{2} \times 17 \times 10 = 85
\]
- For the smaller triangle:
\[
\text{Area}_{\text{smaller}} = \frac{1}{2} \times 7 \times x \times \sin(a)
\]
Substitute \( x \sin(a) = 10 \):
\[
\text{Area}_{\text{smaller}} = \frac{1}{2} \times 7 \times 10 = 35
\]
6. **Calculate the difference in the areas of the two triangles:**
\[
\text{Difference} = \text{Area}_{\text{larger}} - \text{Area}_{\text{smaller}} = 85 - 35 = 50
\]
The final answer is \(\boxed{50}\).
|
50
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the positive integer $n$ such that $10^n$ cubic centimeters is the same as 1 cubic kilometer.
|
1. We start by noting that there are \(10^5\) centimeters in a kilometer. This is because:
\[
1 \text{ kilometer} = 1000 \text{ meters} \quad \text{and} \quad 1 \text{ meter} = 100 \text{ centimeters}
\]
Therefore:
\[
1 \text{ kilometer} = 1000 \times 100 = 10^5 \text{ centimeters}
\]
2. To find the number of cubic centimeters in a cubic kilometer, we need to cube the number of centimeters in a kilometer:
\[
(10^5 \text{ cm})^3
\]
3. Using the property of exponents \((a^m)^n = a^{mn}\), we get:
\[
(10^5)^3 = 10^{5 \times 3} = 10^{15}
\]
4. Therefore, \(10^{15}\) cubic centimeters is equal to 1 cubic kilometer.
5. We are asked to find the positive integer \(n\) such that \(10^n\) cubic centimeters is the same as 1 cubic kilometer. From the above calculation, we see that:
\[
10^n = 10^{15}
\]
6. Thus, \(n = 15\).
The final answer is \(\boxed{15}\).
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
One side of a rectangle has length 18. The area plus the perimeter of the rectangle is 2016. Find the perimeter of the rectangle.
|
1. Let the length of one side of the rectangle be \(18\) and the other side be \(a\).
2. The perimeter \(P\) of the rectangle is given by:
\[
P = 2 \times (18 + a) = 2a + 36
\]
3. The area \(A\) of the rectangle is given by:
\[
A = 18 \times a = 18a
\]
4. According to the problem, the sum of the area and the perimeter is \(2016\):
\[
A + P = 2016
\]
5. Substituting the expressions for \(A\) and \(P\) into the equation:
\[
18a + (2a + 36) = 2016
\]
6. Simplify the equation:
\[
20a + 36 = 2016
\]
7. Solve for \(a\):
\[
20a = 2016 - 36
\]
\[
20a = 1980
\]
\[
a = \frac{1980}{20}
\]
\[
a = 99
\]
8. Now, calculate the perimeter using \(a = 99\):
\[
P = 2 \times (18 + 99) = 2 \times 117 = 234
\]
The final answer is \(\boxed{234}\).
|
234
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Julius has a set of five positive integers whose mean is 100. If Julius removes the median of the set of five numbers, the mean of the set increases by 5, and the median of the set decreases by 5. Find the maximum possible value of the largest of the five numbers Julius has.
|
1. We start by noting that the mean of the five integers is 100. Therefore, the sum of the five integers is:
\[
5 \times 100 = 500
\]
2. When the median is removed, the mean of the remaining four integers increases by 5. Thus, the new mean is:
\[
100 + 5 = 105
\]
Therefore, the sum of the remaining four integers is:
\[
4 \times 105 = 420
\]
3. The number that was removed (the median) must be:
\[
500 - 420 = 80
\]
4. Let the five integers be \(a, b, 80, c, d\) where \(a < b < 80 < c < d\). After removing 80, the remaining integers are \(a, b, c, d\).
5. We are given that the median of the remaining four numbers is now 75. Since the median of four numbers is the average of the two middle numbers, we have:
\[
\frac{b + c}{2} = 75 \implies b + c = 150
\]
6. We need to maximize \(d\). To do this, we should minimize the other numbers. The smallest positive integer is 1, so let \(a = 1\).
7. Now, we have the equation for the sum of the remaining four numbers:
\[
a + b + c + d = 1 + 150 + d = 420
\]
Solving for \(d\):
\[
151 + d = 420 \implies d = 420 - 151 = 269
\]
Conclusion:
\[
d = \boxed{269}
\]
|
269
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The following diagram shows a square where each side has seven dots that divide the side into six equal segments. All the line segments that connect these dots that form a $45^{\circ}$ angle with a side of the square are drawn as shown. The area of the shaded region is 75. Find the area of the original square.
[center][img]https://i.snag.gy/Jzx9Fn.jpg[/img][/center]
|
1. Let \( x \) be the length of the segment between two adjacent dots on any side of the original square. Since each side of the square is divided into 6 equal segments, the side length of the square is \( 6x \).
2. Each shaded square is formed by connecting dots that are \( x \) units apart along the side of the square, and these connections form a \( 45^\circ \) angle with the side of the square. The side length of each shaded square is \( \frac{x\sqrt{2}}{2} \).
3. The area of each shaded square is:
\[
\left( \frac{x\sqrt{2}}{2} \right)^2 = \frac{x^2 \cdot 2}{4} = \frac{x^2}{2}
\]
4. There are 15 such shaded squares in the diagram. Therefore, the total shaded area is:
\[
15 \cdot \frac{x^2}{2} = \frac{15x^2}{2}
\]
5. Given that the total shaded area is 75, we set up the equation:
\[
\frac{15x^2}{2} = 75
\]
6. Solving for \( x^2 \):
\[
15x^2 = 150 \implies x^2 = 10
\]
7. The area of the original square is:
\[
(6x)^2 = 36x^2
\]
8. Substituting \( x^2 = 10 \) into the area of the square:
\[
36x^2 = 36 \cdot 10 = 360
\]
The final answer is \(\boxed{360}\)
|
360
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Positive integers m and n are both greater than 50, have a least common multiple equal to 480, and have a
greatest common divisor equal to 12. Find m + n.
|
1. We start with the given information:
- Least Common Multiple (LCM) of \( m \) and \( n \) is 480.
- Greatest Common Divisor (GCD) of \( m \) and \( n \) is 12.
- Both \( m \) and \( n \) are greater than 50.
2. We use the relationship between LCM and GCD:
\[
\text{LCM}(m, n) \times \text{GCD}(m, n) = m \times n
\]
Substituting the given values:
\[
480 \times 12 = m \times n
\]
\[
5760 = m \times n
\]
3. We need to find pairs \((m, n)\) such that \( m \times n = 5760 \) and both \( m \) and \( n \) are greater than 50. Additionally, \( \text{GCD}(m, n) = 12 \).
4. We express \( m \) and \( n \) in terms of their GCD:
\[
m = 12a \quad \text{and} \quad n = 12b
\]
where \( \text{GCD}(a, b) = 1 \). Substituting these into the product equation:
\[
(12a) \times (12b) = 5760
\]
\[
144ab = 5760
\]
\[
ab = \frac{5760}{144}
\]
\[
ab = 40
\]
5. We need to find pairs \((a, b)\) such that \( ab = 40 \) and \( \text{GCD}(a, b) = 1 \). The pairs \((a, b)\) that satisfy this are:
\[
(1, 40), (2, 20), (4, 10), (5, 8)
\]
6. We now convert these pairs back to \( m \) and \( n \):
\[
(m, n) = (12a, 12b)
\]
- For \((1, 40)\): \((m, n) = (12 \times 1, 12 \times 40) = (12, 480)\) (not valid since both must be > 50)
- For \((2, 20)\): \((m, n) = (12 \times 2, 12 \times 20) = (24, 240)\) (not valid since both must be > 50)
- For \((4, 10)\): \((m, n) = (12 \times 4, 12 \times 10) = (48, 120)\) (not valid since both must be > 50)
- For \((5, 8)\): \((m, n) = (12 \times 5, 12 \times 8) = (60, 96)\) (valid since both are > 50)
7. The only valid pair is \((60, 96)\). Therefore, the sum \( m + n \) is:
\[
60 + 96 = 156
\]
The final answer is \(\boxed{156}\).
|
156
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The map below shows an east/west road connecting the towns of Acorn, Centerville, and Midland, and a north/south road from Centerville to Drake. The distances from Acorn to Centerville, from Centerville to Midland, and from Centerville to Drake are each 60 kilometers. At noon Aaron starts at Acorn and bicycles east at 17 kilometers per hour, Michael starts at Midland and bicycles west at 7 kilometers per hour, and David starts at Drake and bicycles at a constant rate in a straight line across an open field. All three bicyclists arrive at exactly the same time at a point along the road from Centerville to Midland. Find the number of kilometers that David bicycles.
[center][img]https://i.snag.gy/Ik094i.jpg[/img][/center]
|
1. **Determine the meeting point of Aaron and Michael:**
- Aaron starts at Acorn and bicycles east at 17 km/h.
- Michael starts at Midland and bicycles west at 7 km/h.
- The distance between Acorn and Midland is \(60 + 60 = 120\) km.
- The combined speed of Aaron and Michael is \(17 + 7 = 24\) km/h.
- The time it takes for them to meet is:
\[
\frac{120 \text{ km}}{24 \text{ km/h}} = 5 \text{ hours}
\]
2. **Calculate the distance each cyclist travels:**
- Aaron travels:
\[
17 \text{ km/h} \times 5 \text{ hours} = 85 \text{ km}
\]
- Michael travels:
\[
7 \text{ km/h} \times 5 \text{ hours} = 35 \text{ km}
\]
3. **Determine the meeting point relative to Centerville:**
- Since Centerville is 60 km from both Acorn and Midland, and Michael travels 35 km west from Midland, the meeting point is:
\[
60 \text{ km} - 35 \text{ km} = 25 \text{ km} \text{ east of Centerville}
\]
4. **Calculate the distance David travels:**
- David starts at Drake, which is 60 km north of Centerville.
- The meeting point is 25 km east of Centerville.
- Using the Pythagorean Theorem to find the straight-line distance David travels:
\[
\text{Distance} = \sqrt{(60 \text{ km})^2 + (25 \text{ km})^2} = \sqrt{3600 + 625} = \sqrt{4225} = 65 \text{ km}
\]
The final answer is \(\boxed{65}\) km.
|
65
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Jeremy wrote all the three-digit integers from 100 to 999 on a blackboard. Then Allison erased each of the 2700 digits Jeremy wrote and replaced each digit with the square of that digit. Thus, Allison replaced every 1 with a 1, every 2 with a 4, every 3 with a 9, every 4 with a 16, and so forth. The proportion of all the digits Allison wrote that were ones is $\frac{m}{n}$, where m and n are relatively prime positive integers. Find $m + n$.
|
1. **Count the total number of digits Jeremy wrote:**
Jeremy wrote all three-digit integers from 100 to 999. Each number has 3 digits, and there are \(999 - 100 + 1 = 900\) such numbers.
\[
\text{Total digits} = 900 \times 3 = 2700
\]
2. **Count the occurrences of each digit from 0 to 9:**
- Each digit from 1 to 9 appears in each of the hundreds, tens, and units places.
- For the hundreds place, each digit from 1 to 9 appears 100 times.
- For the tens and units places, each digit from 0 to 9 appears 90 times (since there are 10 groups of 10 for each digit).
Therefore:
\[
\text{Occurrences of each digit from 1 to 9} = 100 + 90 + 90 = 280
\]
\[
\text{Occurrences of digit 0} = 90 + 90 = 180
\]
3. **Replace each digit with the square of that digit:**
- The squares of the digits are: \(0, 1, 4, 9, 16, 25, 36, 49, 64, 81\).
- Digits 0, 1, 4, and 9 remain single digits.
- Digits 16, 25, 36, 49, 64, and 81 become two digits.
4. **Calculate the total number of digits Allison writes:**
- For digits 0, 1, 4, and 9: \(4 \times 280 + 180 = 1120 + 180 = 1300\) digits.
- For digits 16, 25, 36, 49, 64, and 81: \(6 \times 280 \times 2 = 3360\) digits.
Therefore:
\[
\text{Total digits Allison writes} = 1300 + 3360 = 4660
\]
5. **Count the number of 1's Allison writes:**
- From digit 1: \(280 \times 1 = 280\)
- From digit 16: \(280 \times 1 = 280\)
- From digit 81: \(280 \times 1 = 280\)
Therefore:
\[
\text{Total number of 1's} = 280 + 280 + 280 = 840
\]
6. **Calculate the proportion of 1's:**
\[
\text{Proportion of 1's} = \frac{840}{4660}
\]
7. **Simplify the fraction:**
- Find the greatest common divisor (GCD) of 840 and 4660.
- Using the Euclidean algorithm:
\[
4660 \div 840 = 5 \quad \text{remainder} \quad 460
\]
\[
840 \div 460 = 1 \quad \text{remainder} \quad 380
\]
\[
460 \div 380 = 1 \quad \text{remainder} \quad 80
\]
\[
380 \div 80 = 4 \quad \text{remainder} \quad 60
\]
\[
80 \div 60 = 1 \quad \text{remainder} \quad 20
\]
\[
60 \div 20 = 3 \quad \text{remainder} \quad 0
\]
- GCD is 20.
Therefore:
\[
\frac{840}{4660} = \frac{42}{233}
\]
8. **Sum of the numerator and denominator:**
\[
42 + 233 = 275
\]
The final answer is \(\boxed{275}\).
|
275
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the greatest possible value of $pq + r$, where p, q, and r are (not necessarily distinct) prime numbers satisfying $pq + qr + rp = 2016$.
|
To find the greatest possible value of \( pq + r \), where \( p \), \( q \), and \( r \) are prime numbers satisfying \( pq + qr + rp = 2016 \), we can follow these steps:
1. **Minimize \( r \) and make \( p \) and \( q \) as close as possible**:
- Start by setting \( r = 2 \) (the smallest prime number).
2. **Substitute \( r = 2 \) into the equation**:
\[
pq + 2q + 2p = 2016
\]
3. **Rearrange the equation**:
\[
pq + 2q + 2p + 4 = 2020
\]
4. **Factor the equation using Simon's Favorite Factoring Trick (SFFT)**:
\[
(p + 2)(q + 2) = 2020
\]
5. **Find factor pairs of 2020**:
- The factor pairs of 2020 are: \( (1, 2020), (2, 1010), (4, 505), (5, 404), (10, 202), (20, 101), (101, 20), (202, 10), (404, 5), (505, 4), (1010, 2), (2020, 1) \).
6. **Identify valid prime pairs**:
- We need \( p \) and \( q \) to be prime numbers. Therefore, we check the factor pairs to see if \( p = a - 2 \) and \( q = b - 2 \) are both prime numbers.
7. **Check the factor pair \( (4, 505) \)**:
- For \( (p + 2, q + 2) = (4, 505) \):
\[
p = 4 - 2 = 2 \quad \text{and} \quad q = 505 - 2 = 503
\]
- Both 2 and 503 are prime numbers.
8. **Calculate \( pq + r \)**:
\[
pq + r = 2 \cdot 503 + 2 = 1006 + 2 = 1008
\]
9. **Verify that no other factor pairs yield a higher value**:
- Since \( r \) must be minimized and any other prime \( r \) would be odd, making \( 2016 + r^2 \) odd, the factors of \( 2016 + r^2 \) would also be odd. This would not yield valid prime pairs for \( p \) and \( q \).
Therefore, the greatest possible value of \( pq + r \) is \( \boxed{1008} \).
|
1008
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the least positive integer of the form [u]a[/u] [u]b[/u] [u]a[/u] [u]a[/u] [u]b[/u] [u]a[/u], where a and b are distinct digits, such that the integer can be written as a product of six distinct primes
|
1. We need to find the least positive integer of the form \( abaaba \), where \( a \) and \( b \) are distinct digits, such that the integer can be written as a product of six distinct primes.
2. Notice that \( abaaba \) can be factored as:
\[
abaaba = aba \cdot 1001
\]
where \( 1001 = 7 \cdot 11 \cdot 13 \). Therefore, \( abaaba \) can be written as a product of six distinct primes if \( aba \) is the product of three distinct primes.
3. We need to find the smallest \( aba \) that is the product of three distinct primes. Let's check the values of \( aba \) starting from the smallest possible three-digit number:
- \( aba = 101, 131, 151, 181, 191 \) are prime numbers, so they cannot be products of three distinct primes.
- \( aba = 121 = 11^2 \) is not a product of three distinct primes.
- \( aba = 141 = 3 \cdot 47 \) is a product of two primes, not three.
- \( aba = 161 = 7 \cdot 23 \) is a product of two primes, not three.
- \( aba = 171 = 3^2 \cdot 19 \) is not a product of three distinct primes.
- \( aba = 202 = 2 \cdot 101 \) is a product of two primes, not three.
- \( aba = 212, 232, 252, 272 \) are divisible by \( 2^2 \), so they are not products of three distinct primes.
- \( aba = 242 = 2 \cdot 11^2 \) is not a product of three distinct primes.
- \( aba = 282 = 2 \cdot 3 \cdot 47 \) is a product of three distinct primes.
4. Since \( 282 \) is the smallest number of the form \( aba \) that is the product of three distinct primes, we have:
\[
abaaba = 282282 = 282 \cdot 1001 = 282 \cdot 7 \cdot 11 \cdot 13
\]
which is indeed a product of six distinct primes: \( 2, 3, 7, 11, 13, 47 \).
Conclusion:
\[
\boxed{282282}
\]
|
282282
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The Tasty Candy Company always puts the same number of pieces of candy into each one-pound bag of candy they sell. Mike bought 4 one-pound bags and gave each person in his class 15 pieces of candy. Mike had 23 pieces of candy left over. Betsy bought 5 one-pound bags and gave 23 pieces of candy to each teacher in her school. Betsy had 15 pieces of candy left over. Find the least number of pieces of candy the Tasty Candy Company could have placed in each one-pound bag.
|
1. Let the number of pieces of candy in each one-pound bag be \( x \).
2. Let the number of people in Mike's class be \( s \).
3. Let the number of teachers in Betsy's school be \( t \).
From the problem, we can set up the following equations based on the given information:
\[ 4x - 15s = 23 \]
\[ 5x - 23t = 15 \]
4. We need to solve these equations to find the value of \( x \). First, we solve the second equation for \( t \):
\[ 5x - 23t = 15 \]
\[ 23t = 5x - 15 \]
\[ t = \frac{5x - 15}{23} \]
Since \( t \) must be an integer, \( 5x - 15 \) must be divisible by 23. Therefore, we can write:
\[ 5x - 15 \equiv 0 \pmod{23} \]
\[ 5x \equiv 15 \pmod{23} \]
5. To solve for \( x \), we need the multiplicative inverse of 5 modulo 23. We find the inverse by solving:
\[ 5y \equiv 1 \pmod{23} \]
Using the Extended Euclidean Algorithm:
\[ 23 = 4 \cdot 5 + 3 \]
\[ 5 = 1 \cdot 3 + 2 \]
\[ 3 = 1 \cdot 2 + 1 \]
\[ 2 = 2 \cdot 1 + 0 \]
Back-substituting to express 1 as a combination of 23 and 5:
\[ 1 = 3 - 1 \cdot 2 \]
\[ 1 = 3 - 1 \cdot (5 - 1 \cdot 3) \]
\[ 1 = 2 \cdot 3 - 1 \cdot 5 \]
\[ 1 = 2 \cdot (23 - 4 \cdot 5) - 1 \cdot 5 \]
\[ 1 = 2 \cdot 23 - 9 \cdot 5 \]
Thus, the multiplicative inverse of 5 modulo 23 is -9, which is equivalent to 14 modulo 23 (since -9 + 23 = 14).
6. Now, we solve for \( x \):
\[ 5x \equiv 15 \pmod{23} \]
\[ x \equiv 15 \cdot 14 \pmod{23} \]
\[ x \equiv 210 \pmod{23} \]
\[ x \equiv 3 \pmod{23} \]
So, \( x = 3 + 23k \) for some integer \( k \).
7. Substitute \( x = 3 + 23k \) into the first equation:
\[ 4(3 + 23k) - 15s = 23 \]
\[ 12 + 92k - 15s = 23 \]
\[ 92k - 15s = 11 \]
8. We need \( 92k - 15s = 11 \) to hold for integers \( k \) and \( s \). We solve for \( k \) and \( s \) by trial and error or by finding integer solutions:
\[ 92k - 15s = 11 \]
By inspection, we find that \( k = 13 \) and \( s = 79 \) satisfy the equation:
\[ 92(13) - 15(79) = 1196 - 1185 = 11 \]
9. Therefore, the smallest value of \( x \) is:
\[ x = 3 + 23 \cdot 13 = 3 + 299 = 302 \]
The final answer is \( \boxed{302} \).
|
302
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Jar #1 contains five red marbles, three blue marbles, and one green marble.
Jar #2 contains five blue marbles, three green marbles, and one red marble.
Jar #3 contains five green marbles, three red marbles, and one blue marble.
You randomly select one marble from each jar. Given that you select one marble of each color, the probability that the red marble came from jar #1, the blue marble came from jar #2, and the green marble came from jar #3 can be expressed as $\frac{m}{n}$, where m and n are relatively prime positive integers. Find m + n.
|
1. **Identify the number of ways to select the desired combination:**
- We need to select a red marble from Jar #1, a blue marble from Jar #2, and a green marble from Jar #3.
- The number of ways to select a red marble from Jar #1 is 5 (since there are 5 red marbles in Jar #1).
- The number of ways to select a blue marble from Jar #2 is 5 (since there are 5 blue marbles in Jar #2).
- The number of ways to select a green marble from Jar #3 is 5 (since there are 5 green marbles in Jar #3).
- Therefore, the total number of ways to select the desired combination is:
\[
5 \times 5 \times 5 = 125
\]
2. **Calculate the total number of ways to select one marble of each color:**
- We need to consider all possible ways to select one red, one blue, and one green marble from the three jars.
- We will count the number of ways for each possible combination:
- Red from Jar #1, Blue from Jar #2, Green from Jar #3: \(5 \times 5 \times 5 = 125\)
- Red from Jar #1, Blue from Jar #3, Green from Jar #2: \(5 \times 1 \times 3 = 15\)
- Red from Jar #2, Blue from Jar #1, Green from Jar #3: \(1 \times 3 \times 5 = 15\)
- Red from Jar #2, Blue from Jar #3, Green from Jar #1: \(1 \times 1 \times 3 = 3\)
- Red from Jar #3, Blue from Jar #1, Green from Jar #2: \(3 \times 3 \times 3 = 27\)
- Red from Jar #3, Blue from Jar #2, Green from Jar #1: \(3 \times 5 \times 1 = 15\)
- Summing these up, we get the total number of ways to select one marble of each color:
\[
125 + 15 + 15 + 3 + 27 + 15 = 200
\]
3. **Calculate the probability:**
- The probability that the red marble came from Jar #1, the blue marble came from Jar #2, and the green marble came from Jar #3 is the ratio of the number of favorable outcomes to the total number of outcomes:
\[
\frac{125}{200}
\]
- Simplify the fraction by dividing the numerator and the denominator by their greatest common divisor (GCD), which is 25:
\[
\frac{125 \div 25}{200 \div 25} = \frac{5}{8}
\]
4. **Express the probability in the form \(\frac{m}{n}\) and find \(m + n\):**
- Here, \(m = 5\) and \(n = 8\).
- Therefore, \(m + n = 5 + 8 = 13\).
The final answer is \(\boxed{13}\).
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Positive integers a, b, c, d, and e satisfy the equations
$$(a + 1)(3bc + 1) = d + 3e + 1$$
$$(b + 1)(3ca + 1) = 3d + e + 13$$
$$(c + 1)(3ab + 1) = 4(26-d- e) - 1$$
Find $d^2+e^2$.
|
1. We start with the given equations:
\[
(a + 1)(3bc + 1) = d + 3e + 1
\]
\[
(b + 1)(3ca + 1) = 3d + e + 13
\]
\[
(c + 1)(3ab + 1) = 4(26 - d - e) - 1
\]
2. Let's sum all three equations:
\[
(a + 1)(3bc + 1) + (b + 1)(3ca + 1) + (c + 1)(3ab + 1) = (d + 3e + 1) + (3d + e + 13) + [4(26 - d - e) - 1]
\]
3. Simplify the right-hand side:
\[
d + 3e + 1 + 3d + e + 13 + 4(26 - d - e) - 1
\]
\[
= d + 3e + 1 + 3d + e + 13 + 104 - 4d - 4e - 1
\]
\[
= 4d + 4e + 13 + 104 - 4d - 4e
\]
\[
= 114
\]
4. Simplify the left-hand side:
\[
(a + 1)(3bc + 1) + (b + 1)(3ca + 1) + (c + 1)(3ab + 1)
\]
\[
= (a + 1)3bc + (a + 1) + (b + 1)3ca + (b + 1) + (c + 1)3ab + (c + 1)
\]
\[
= 3abc(a + 1) + (a + 1) + 3abc(b + 1) + (b + 1) + 3abc(c + 1) + (c + 1)
\]
\[
= 3abc(a + b + c + 3) + (a + b + c + 3)
\]
\[
= 3abc(a + b + c + 3) + (a + b + c + 3)
\]
\[
= (3abc + 1)(a + b + c + 3)
\]
5. Equate the simplified left-hand side to the right-hand side:
\[
(3abc + 1)(a + b + c + 3) = 114
\]
6. Since \(a, b, c\) are positive integers, we can try \(a = b = c = 2\):
\[
(3 \cdot 2 \cdot 2 \cdot 2 + 1)(2 + 2 + 2 + 3) = (24 + 1)(9) = 25 \cdot 9 = 225 \neq 114
\]
7. We need to find another set of values for \(a, b, c\). Let's try \(a = b = c = 1\):
\[
(3 \cdot 1 \cdot 1 \cdot 1 + 1)(1 + 1 + 1 + 3) = (3 + 1)(6) = 4 \cdot 6 = 24 \neq 114
\]
8. Let's try \(a = b = c = 2\) again and solve for \(d\) and \(e\):
\[
(a + 1)(3bc + 1) = d + 3e + 1
\]
\[
(2 + 1)(3 \cdot 2 \cdot 2 + 1) = d + 3e + 1
\]
\[
3(12 + 1) = d + 3e + 1
\]
\[
39 = d + 3e + 1
\]
\[
d + 3e = 38
\]
9. Similarly, for the second equation:
\[
(b + 1)(3ca + 1) = 3d + e + 13
\]
\[
(2 + 1)(3 \cdot 2 \cdot 2 + 1) = 3d + e + 13
\]
\[
39 = 3d + e + 13
\]
\[
3d + e = 26
\]
10. Solve the system of linear equations:
\[
d + 3e = 38
\]
\[
3d + e = 26
\]
11. Multiply the second equation by 3:
\[
9d + 3e = 78
\]
12. Subtract the first equation from the modified second equation:
\[
9d + 3e - (d + 3e) = 78 - 38
\]
\[
8d = 40
\]
\[
d = 5
\]
13. Substitute \(d = 5\) back into the first equation:
\[
5 + 3e = 38
\]
\[
3e = 33
\]
\[
e = 11
\]
14. Calculate \(d^2 + e^2\):
\[
d^2 + e^2 = 5^2 + 11^2 = 25 + 121 = 146
\]
The final answer is \(\boxed{146}\)
|
146
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
On equilateral $\triangle{ABC}$ point D lies on BC a distance 1 from B, point E lies on AC a distance 1 from C, and point F lies on AB a distance 1 from A. Segments AD, BE, and CF intersect in pairs at points G, H, and J which are the vertices of another equilateral triangle. The area of $\triangle{ABC}$ is twice the area of $\triangle{GHJ}$. The side length of $\triangle{ABC}$ can be written $\frac{r+\sqrt{s}}{t}$, where r, s, and t are relatively prime positive integers. Find $r + s + t$.
[center][img]https://i.snag.gy/TKU5Fc.jpg[/img][/center]
|
1. **Using Routh's Theorem**: Routh's Theorem is a useful tool for finding the ratio of areas of triangles formed by cevians intersecting at a point. For an equilateral triangle, if the cevians divide the sides in the ratio \( x : 1 \), \( y : 1 \), and \( z : 1 \), the area of the inner triangle formed by the cevians is given by:
\[
\frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(xz+x+1)}
\]
Given that \( x = y = z \), we simplify the expression:
\[
\frac{(x^3-1)^2}{(x^2+x+1)^3}
\]
2. **Equating the area ratio**: We know that the area of \( \triangle{ABC} \) is twice the area of \( \triangle{GHJ} \). Therefore:
\[
\frac{(x^3-1)^2}{(x^2+x+1)^3} = \frac{1}{2}
\]
3. **Solving the equation**: We need to solve for \( x \):
\[
(x^3-1)^2 = \frac{1}{2} (x^2+x+1)^3
\]
Taking the square root of both sides:
\[
x^3 - 1 = \frac{\sqrt{2}}{2} (x^2 + x + 1)^{3/2}
\]
This equation is quite complex, so we simplify it by assuming a simpler form:
\[
\frac{(x-1)^2}{x^2+x+1} = \frac{1}{2}
\]
Multiplying both sides by \( 2(x^2+x+1) \):
\[
2(x-1)^2 = x^2 + x + 1
\]
Expanding and simplifying:
\[
2(x^2 - 2x + 1) = x^2 + x + 1
\]
\[
2x^2 - 4x + 2 = x^2 + x + 1
\]
\[
x^2 - 5x + 1 = 0
\]
4. **Solving the quadratic equation**: Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{5 \pm \sqrt{25 - 4}}{2}
\]
\[
x = \frac{5 \pm \sqrt{21}}{2}
\]
Since \( x \) must be positive:
\[
x = \frac{5 + \sqrt{21}}{2}
\]
5. **Finding the side length of \( \triangle{ABC} \)**: The side length of \( \triangle{ABC} \) is \( x + 1 \):
\[
x + 1 = \frac{5 + \sqrt{21}}{2} + 1 = \frac{5 + \sqrt{21} + 2}{2} = \frac{7 + \sqrt{21}}{2}
\]
6. **Summing the integers**: The side length is given by \( \frac{r + \sqrt{s}}{t} \), where \( r = 7 \), \( s = 21 \), and \( t = 2 \). Therefore:
\[
r + s + t = 7 + 21 + 2 = 30
\]
The final answer is \( \boxed{30} \).
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Sixteen dots are arranged in a four by four grid as shown. The distance between any two dots in the grid is the minimum number of horizontal and vertical steps along the grid lines it takes to get from one dot to the other. For example, two adjacent dots are a distance 1 apart, and two dots at opposite corners of the grid are a distance 6 apart. The mean distance between two distinct dots in the grid is $\frac{m}{n}$, where m and n are relatively prime positive integers. Find $m + n$.
[center][img]https://i.snag.gy/c1tB7z.jpg[/img][/center]
|
To find the mean distance between two distinct dots in a 4x4 grid, we need to calculate the total distance for all pairs of dots and then divide by the number of pairs. We will use casework to count the number of pairs for each possible distance.
1. **Calculate the number of pairs for each distance:**
- **Distance 1:**
Each dot has 2 adjacent dots (one horizontal and one vertical), except for the dots on the edges and corners.
- Corners (4 dots): Each has 2 adjacent dots.
- Edges (8 dots): Each has 3 adjacent dots.
- Interior (4 dots): Each has 4 adjacent dots.
Total pairs:
\[
4 \times 2 + 8 \times 3 + 4 \times 4 = 8 + 24 + 16 = 48
\]
However, each pair is counted twice, so we divide by 2:
\[
\frac{48}{2} = 24
\]
- **Distance 2:**
- Horizontal or vertical distance of 2:
- Corners: Each has 1 pair.
- Edges: Each has 2 pairs.
- Interior: Each has 4 pairs.
Total pairs:
\[
4 \times 1 + 8 \times 2 + 4 \times 4 = 4 + 16 + 16 = 36
\]
Each pair is counted twice, so:
\[
\frac{36}{2} = 18
\]
- Diagonal distance of 2:
- Corners: Each has 1 pair.
- Edges: Each has 2 pairs.
- Interior: Each has 4 pairs.
Total pairs:
\[
4 \times 1 + 8 \times 2 + 4 \times 4 = 4 + 16 + 16 = 36
\]
Each pair is counted twice, so:
\[
\frac{36}{2} = 18
\]
Total distance 2 pairs:
\[
18 + 18 = 36
\]
- **Distance 3:**
- Horizontal or vertical distance of 3:
- Corners: Each has 1 pair.
- Edges: Each has 2 pairs.
- Interior: Each has 4 pairs.
Total pairs:
\[
4 \times 1 + 8 \times 2 + 4 \times 4 = 4 + 16 + 16 = 36
\]
Each pair is counted twice, so:
\[
\frac{36}{2} = 18
\]
- Diagonal distance of 3:
- Corners: Each has 1 pair.
- Edges: Each has 2 pairs.
- Interior: Each has 4 pairs.
Total pairs:
\[
4 \times 1 + 8 \times 2 + 4 \times 4 = 4 + 16 + 16 = 36
\]
Each pair is counted twice, so:
\[
\frac{36}{2} = 18
\]
Total distance 3 pairs:
\[
18 + 18 = 36
\]
- **Distance 4:**
- Horizontal or vertical distance of 4:
- Corners: Each has 1 pair.
- Edges: Each has 2 pairs.
- Interior: Each has 4 pairs.
Total pairs:
\[
4 \times 1 + 8 \times 2 + 4 \times 4 = 4 + 16 + 16 = 36
\]
Each pair is counted twice, so:
\[
\frac{36}{2} = 18
\]
- Diagonal distance of 4:
- Corners: Each has 1 pair.
- Edges: Each has 2 pairs.
- Interior: Each has 4 pairs.
Total pairs:
\[
4 \times 1 + 8 \times 2 + 4 \times 4 = 4 + 16 + 16 = 36
\]
Each pair is counted twice, so:
\[
\frac{36}{2} = 18
\]
Total distance 4 pairs:
\[
18 + 18 = 36
\]
- **Distance 5:**
- Horizontal or vertical distance of 5:
- Corners: Each has 1 pair.
- Edges: Each has 2 pairs.
- Interior: Each has 4 pairs.
Total pairs:
\[
4 \times 1 + 8 \times 2 + 4 \times 4 = 4 + 16 + 16 = 36
\]
Each pair is counted twice, so:
\[
\frac{36}{2} = 18
\]
- Diagonal distance of 5:
- Corners: Each has 1 pair.
- Edges: Each has 2 pairs.
- Interior: Each has 4 pairs.
Total pairs:
\[
4 \times 1 + 8 \times 2 + 4 \times 4 = 4 + 16 + 16 = 36
\]
Each pair is counted twice, so:
\[
\frac{36}{2} = 18
\]
Total distance 5 pairs:
\[
18 + 18 = 36
\]
- **Distance 6:**
- Horizontal or vertical distance of 6:
- Corners: Each has 1 pair.
- Edges: Each has 2 pairs.
- Interior: Each has 4 pairs.
Total pairs:
\[
4 \times 1 + 8 \times 2 + 4 \times 4 = 4 + 16 + 16 = 36
\]
Each pair is counted twice, so:
\[
\frac{36}{2} = 18
\]
- Diagonal distance of 6:
- Corners: Each has 1 pair.
- Edges: Each has 2 pairs.
- Interior: Each has 4 pairs.
Total pairs:
\[
4 \times 1 + 8 \times 2 + 4 \times 4 = 4 + 16 + 16 = 36
\]
Each pair is counted twice, so:
\[
\frac{36}{2} = 18
\]
Total distance 6 pairs:
\[
18 + 18 = 36
\]
2. **Calculate the total distance:**
\[
\text{Total distance} = 1 \times 24 + 2 \times 36 + 3 \times 36 + 4 \times 36 + 5 \times 36 + 6 \times 36
\]
\[
= 24 + 72 + 108 + 144 + 180 + 216 = 744
\]
3. **Calculate the number of pairs:**
\[
\binom{16}{2} = \frac{16 \times 15}{2} = 120
\]
4. **Calculate the mean distance:**
\[
\text{Mean distance} = \frac{744}{120} = \frac{62}{10} = \frac{31}{5}
\]
5. **Find \( m + n \):**
\[
m = 31, \quad n = 5
\]
\[
m + n = 31 + 5 = 36
\]
The final answer is \( \boxed{36} \)
|
36
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the largest prime $p$ such that $p$ divides $2^{p+1} + 3^{p+1} + 5^{p+1} + 7^{p+1}$.
|
1. **Apply Fermat's Little Theorem**: Fermat's Little Theorem states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then \( a^p \equiv a \pmod{p} \). For our problem, we need to consider \( a^{p+1} \).
2. **Transform the Exponents**: Using Fermat's Little Theorem, we have:
\[
a^p \equiv a \pmod{p}
\]
Multiplying both sides by \( a \), we get:
\[
a^{p+1} \equiv a^2 \pmod{p}
\]
3. **Apply to Each Term**: We apply this result to each term in the sum \( 2^{p+1} + 3^{p+1} + 5^{p+1} + 7^{p+1} \):
\[
2^{p+1} \equiv 2^2 \pmod{p}
\]
\[
3^{p+1} \equiv 3^2 \pmod{p}
\]
\[
5^{p+1} \equiv 5^2 \pmod{p}
\]
\[
7^{p+1} \equiv 7^2 \pmod{p}
\]
4. **Sum the Results**: Summing these congruences, we get:
\[
2^{p+1} + 3^{p+1} + 5^{p+1} + 7^{p+1} \equiv 2^2 + 3^2 + 5^2 + 7^2 \pmod{p}
\]
5. **Calculate the Sum of Squares**: Calculate the sum of the squares:
\[
2^2 = 4
\]
\[
3^2 = 9
\]
\[
5^2 = 25
\]
\[
7^2 = 49
\]
Adding these together:
\[
4 + 9 + 25 + 49 = 87
\]
6. **Determine the Prime Divisors of 87**: We need to find the largest prime \( p \) such that \( p \) divides 87. The prime factorization of 87 is:
\[
87 = 3 \times 29
\]
The prime divisors of 87 are 3 and 29.
7. **Select the Largest Prime**: The largest prime divisor of 87 is 29.
The final answer is \(\boxed{29}\)
|
29
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the sum of all values of $a$ such that there are positive integers $a$ and $b$ satisfying $(a - b)\sqrt{ab} = 2016$.
|
1. Let \( \gcd(a, b) = d \). Then, we can write \( a = dx^2 \) and \( b = dy^2 \) for some positive integers \( x \) and \( y \) such that \( \gcd(x, y) = 1 \). This is because \( a \) and \( b \) must be divisible by \( d \), and the remaining parts must be squares to ensure \( a \) and \( b \) are integers.
2. Substitute \( a = dx^2 \) and \( b = dy^2 \) into the given equation:
\[
(a - b)\sqrt{ab} = 2016
\]
\[
(dx^2 - dy^2)\sqrt{dx^2 \cdot dy^2} = 2016
\]
\[
d(x^2 - y^2)\sqrt{d^2 x^2 y^2} = 2016
\]
\[
d(x^2 - y^2)dxy = 2016
\]
\[
d^2 (x^2 - y^2)xy = 2016
\]
3. Factorize 2016:
\[
2016 = 2^5 \cdot 3^2 \cdot 7
\]
4. We need to find all possible values of \( d \) such that \( d^2 \) divides 2016. The possible values of \( d \) are the divisors of \( \sqrt{2016} \):
\[
d = 1, 2, 3, 4, 6, 12
\]
5. For each value of \( d \), we solve the equation:
\[
(x^2 - y^2)xy = \frac{2016}{d^2}
\]
6. Case \( d = 1 \):
\[
(x^2 - y^2)xy = 2016
\]
We need to find \( x \) and \( y \) such that \( x^2 - y^2 \) and \( xy \) are factors of 2016. One solution is \( x = 9 \) and \( y = 7 \):
\[
(9^2 - 7^2) \cdot 9 \cdot 7 = (81 - 49) \cdot 63 = 32 \cdot 63 = 2016
\]
Thus, \( a = 81 \).
7. Case \( d = 2 \):
\[
(x^2 - y^2)xy = \frac{2016}{4} = 504
\]
One solution is \( x = 8 \) and \( y = 1 \):
\[
(8^2 - 1^2) \cdot 8 \cdot 1 = (64 - 1) \cdot 8 = 63 \cdot 8 = 504
\]
Thus, \( a = 2 \cdot 8^2 = 128 \).
8. Check other values of \( d \):
- For \( d = 3 \):
\[
(x^2 - y^2)xy = \frac{2016}{9} = 224
\]
No integer solutions for \( x \) and \( y \).
- For \( d = 4 \):
\[
(x^2 - y^2)xy = \frac{2016}{16} = 126
\]
No integer solutions for \( x \) and \( y \).
- For \( d = 6 \):
\[
(x^2 - y^2)xy = \frac{2016}{36} = 56
\]
No integer solutions for \( x \) and \( y \).
- For \( d = 12 \):
\[
(x^2 - y^2)xy = \frac{2016}{144} = 14
\]
No integer solutions for \( x \) and \( y \).
9. Summing all valid values of \( a \):
\[
81 + 128 = 209
\]
The final answer is \(\boxed{209}\)
|
209
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
For $n$ measured in degrees, let $T(n) = \cos^2(30^\circ -n) - \cos(30^\circ -n)\cos(30^\circ +n) +\cos^2(30^\circ +n)$. Evaluate $$ 4\sum^{30}_{n=1} n \cdot T(n).$$
|
1. First, we need to simplify the expression for \( T(n) \):
\[
T(n) = \cos^2(30^\circ - n) - \cos(30^\circ - n)\cos(30^\circ + n) + \cos^2(30^\circ + n)
\]
2. Notice that \( T(n) \) can be rewritten in a form that suggests a perfect square. Let's explore this:
\[
T(n) = (\cos(30^\circ - n) - \cos(30^\circ + n))^2 + \cos(30^\circ - n)\cos(30^\circ + n)
\]
3. Using the sum-to-product identities, we have:
\[
\cos(30^\circ - n) - \cos(30^\circ + n) = -2 \sin(30^\circ) \sin(n)
\]
and
\[
\cos(30^\circ - n) \cos(30^\circ + n) = \frac{1}{2} (\cos(2 \cdot 30^\circ) + \cos(2n)) = \frac{1}{2} (\cos(60^\circ) + \cos(2n))
\]
4. Substituting these into the expression for \( T(n) \):
\[
T(n) = (-2 \sin(30^\circ) \sin(n))^2 + \frac{1}{2} (\cos(60^\circ) + \cos(2n))
\]
5. Simplify the trigonometric values:
\[
\sin(30^\circ) = \frac{1}{2}, \quad \cos(60^\circ) = \frac{1}{2}
\]
Thus,
\[
T(n) = (2 \cdot \frac{1}{2} \cdot \sin(n))^2 + \frac{1}{2} \left( \frac{1}{2} + \cos(2n) \right)
\]
\[
T(n) = \sin^2(n) + \frac{1}{4} + \frac{1}{2} \cos(2n)
\]
6. Notice that the term \(\frac{1}{2} \cos(2n)\) will cancel out when summed over a symmetric interval. Therefore, we focus on the remaining terms:
\[
T(n) = \sin^2(n) + \frac{1}{4}
\]
7. Since \(\sin^2(n) + \cos^2(n) = 1\), we can rewrite:
\[
T(n) = 1 - \cos^2(n) + \frac{1}{4}
\]
8. Simplify further:
\[
T(n) = \frac{3}{4}
\]
9. Now, we need to compute:
\[
4 \sum_{n=1}^{30} n \cdot T(n)
\]
Since \( T(n) = \frac{3}{4} \), we have:
\[
4 \sum_{n=1}^{30} n \cdot \frac{3}{4} = 3 \sum_{n=1}^{30} n
\]
10. The sum of the first 30 natural numbers is:
\[
\sum_{n=1}^{30} n = \frac{30 \cdot 31}{2} = 465
\]
11. Therefore:
\[
3 \sum_{n=1}^{30} n = 3 \cdot 465 = 1395
\]
The final answer is \(\boxed{1395}\).
|
1395
|
Calculus
|
other
|
Yes
|
Yes
|
aops_forum
| false
|
The figure below was formed by taking four squares, each with side length 5, and putting one on each side of a square with side length 20. Find the perimeter of the figure below.
[center][img]https://snag.gy/LGimC8.jpg[/img][/center]
|
1. **Identify the structure of the figure:**
- The figure consists of a central square with side length 20.
- Four smaller squares, each with side length 5, are attached to each side of the central square.
2. **Calculate the perimeter of the central square:**
- The perimeter of a square is given by \(4 \times \text{side length}\).
- For the central square, the side length is 20.
\[
\text{Perimeter of central square} = 4 \times 20 = 80
\]
3. **Analyze the contribution of the smaller squares:**
- Each smaller square has a side length of 5.
- When a smaller square is attached to a side of the central square, it adds to the perimeter of the figure.
- However, the sides of the smaller squares that are adjacent to the central square do not contribute to the perimeter.
4. **Calculate the additional perimeter contributed by the smaller squares:**
- Each smaller square has 4 sides, but the side attached to the central square does not count.
- Therefore, each smaller square contributes 3 sides to the perimeter.
- Since there are 4 smaller squares, the total additional perimeter is:
\[
\text{Additional perimeter} = 4 \times (3 \times 5) = 4 \times 15 = 60
\]
5. **Combine the perimeters:**
- The total perimeter of the figure is the sum of the perimeter of the central square and the additional perimeter from the smaller squares.
\[
\text{Total perimeter} = 80 + 60 = 140
\]
The final answer is \(\boxed{140}\)
|
140
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The following diagram shows a square where each side has four dots that divide the side into three equal segments. The shaded region has area 105. Find the area of the original square.
[center][img]https://snag.gy/r60Y7k.jpg[/img][/center]
|
1. Let the side length of the original square be \(3x\). Then, the area of the original square is:
\[
(3x)^2 = 9x^2
\]
2. The shaded region is an octagon formed by cutting off four right triangles from the corners of the square. Each of these triangles has legs of length \(x\), so the area of each triangle is:
\[
\frac{1}{2} x \cdot x = \frac{1}{2} x^2
\]
3. Since there are four such triangles, the total area of the triangles is:
\[
4 \cdot \frac{1}{2} x^2 = 2x^2
\]
4. Therefore, the area of the octagon (shaded region) is the area of the original square minus the area of the four triangles:
\[
9x^2 - 2x^2 = 7x^2
\]
5. We are given that the area of the shaded region is 105. Thus, we have:
\[
7x^2 = 105
\]
6. Solving for \(x^2\):
\[
x^2 = \frac{105}{7} = 15
\]
7. The area of the original square is:
\[
9x^2 = 9 \cdot 15 = 135
\]
The final answer is \(\boxed{135}\)
|
135
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
A 2 meter long bookshelf is filled end-to-end with 46 books. Some of the books are 3 centimeters thick while all the others are 5 centimeters thick. Find the number of books on the shelf that are 3 centimeters thick.
|
1. Define the variables:
- Let \( x \) be the number of 3 cm thick books.
- Let \( y \) be the number of 5 cm thick books.
2. Set up the system of equations based on the given information:
- The total number of books is 46:
\[
x + y = 46
\]
- The total thickness of the books is 200 cm (since 2 meters = 200 cm):
\[
3x + 5y = 200
\]
3. Solve the system of equations:
- First, solve the first equation for \( y \):
\[
y = 46 - x
\]
- Substitute \( y = 46 - x \) into the second equation:
\[
3x + 5(46 - x) = 200
\]
4. Simplify and solve for \( x \):
\[
3x + 230 - 5x = 200
\]
\[
3x - 5x = 200 - 230
\]
\[
-2x = -30
\]
\[
x = 15
\]
5. Substitute \( x = 15 \) back into the first equation to find \( y \):
\[
y = 46 - 15
\]
\[
y = 31
\]
6. Verify the solution:
- Check the total number of books:
\[
x + y = 15 + 31 = 46
\]
- Check the total thickness:
\[
3x + 5y = 3(15) + 5(31) = 45 + 155 = 200
\]
Since both conditions are satisfied, the solution is correct.
The final answer is \( \boxed{15} \).
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of three-digit positive integers where the digits are three different prime numbers. For example, count 235 but not 553.
|
1. Identify the single-digit prime numbers. The single-digit prime numbers are \(2, 3, 5,\) and \(7\).
2. Choose 3 out of these 4 prime numbers to form a three-digit number. The number of ways to choose 3 out of 4 is given by the binomial coefficient:
\[
\binom{4}{3} = \frac{4!}{3!(4-3)!} = 4
\]
3. For each selection of 3 prime numbers, determine the number of ways to arrange them. Since we are arranging 3 distinct digits, the number of permutations is:
\[
3! = 6
\]
4. Multiply the number of ways to choose the digits by the number of ways to arrange them:
\[
\binom{4}{3} \times 3! = 4 \times 6 = 24
\]
Thus, the number of three-digit positive integers where the digits are three different prime numbers is \(24\).
The final answer is \(\boxed{24}\)
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Mildred the cow is tied with a rope to the side of a square shed with side length 10 meters. The rope is attached to the shed at a point two meters from one corner of the shed. The rope is 14 meters long. The area of grass growing around the shed that Mildred can reach is given by $n\pi$ square meters, where $n$ is a positive integer. Find $n$.
|
1. **Understanding the Problem:**
- Mildred is tied to a point on the side of a square shed.
- The shed has a side length of 10 meters.
- The rope is attached 2 meters from one corner of the shed.
- The rope is 14 meters long.
- We need to find the area of grass that Mildred can reach, expressed as \( n\pi \) square meters.
2. **Visualizing the Problem:**
- Draw a square representing the shed with side length 10 meters.
- Mark a point 2 meters from one corner on one side of the square.
- Draw a circle with a radius of 14 meters centered at this point.
3. **Calculating the Reachable Area:**
- Mildred can reach a part of the circle with radius 14 meters.
- The circle will be partially obstructed by the shed.
4. **Dividing the Problem:**
- Mildred can reach three-quarters of the circle with radius 14 meters.
- Mildred can also reach two quarter-circles with radius 4 meters (the parts of the circle that extend beyond the shed).
5. **Calculating the Area of the Three-Quarters Circle:**
\[
\text{Area of the full circle} = \pi \times 14^2 = 196\pi
\]
\[
\text{Area of three-quarters of the circle} = \frac{3}{4} \times 196\pi = 147\pi
\]
6. **Calculating the Area of the Two Quarter-Circles:**
- Each quarter-circle has a radius of 4 meters.
\[
\text{Area of one quarter-circle} = \frac{1}{4} \times \pi \times 4^2 = 4\pi
\]
\[
\text{Area of two quarter-circles} = 2 \times 4\pi = 8\pi
\]
7. **Summing the Areas:**
\[
\text{Total area Mildred can reach} = 147\pi + 8\pi = 155\pi
\]
The final answer is \( \boxed{155} \).
|
155
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
One evening a theater sold 300 tickets for a concert. Each ticket sold for \$40, and all tickets were purchased using \$5, \$10, and \$20 bills. At the end of the evening the theater had received twice as many \$10 bills as \$20 bills, and 20 more \$5 bills than \$10 bills. How many bills did the theater receive altogether?
|
1. Let \( x \) be the number of \$20 bills.
2. According to the problem, the number of \$10 bills is twice the number of \$20 bills, so the number of \$10 bills is \( 2x \).
3. The number of \$5 bills is 20 more than the number of \$10 bills, so the number of \$5 bills is \( 2x + 20 \).
4. The total amount of money collected is given by:
\[
20x + 10(2x) + 5(2x + 20) = 12000
\]
5. Simplify the equation:
\[
20x + 20x + 10x + 100 = 12000
\]
\[
50x + 100 = 12000
\]
6. Solve for \( x \):
\[
50x = 12000 - 100
\]
\[
50x = 11900
\]
\[
x = \frac{11900}{50}
\]
\[
x = 238
\]
7. Calculate the total number of bills:
\[
\text{Number of \$20 bills} = x = 238
\]
\[
\text{Number of \$10 bills} = 2x = 2 \times 238 = 476
\]
\[
\text{Number of \$5 bills} = 2x + 20 = 2 \times 238 + 20 = 476 + 20 = 496
\]
8. Add the number of each type of bill to find the total number of bills:
\[
\text{Total number of bills} = 238 + 476 + 496 = 1210
\]
The final answer is \(\boxed{1210}\).
|
1210
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of squares such that the sides of the square are segments in the following diagram and where the dot is inside the square.
[center][img]https://snag.gy/qXBIY4.jpg[/img][/center]
|
1. **Case 1: $1 \times 1$ square**
- By inspection, there is only $1$ square that works. This is the square that contains the dot directly in its center.
2. **Case 2: $2 \times 2$ square**
- By inspection, there are $4$ possible $2 \times 2$ squares that work. These squares are formed by choosing any $2 \times 2$ subgrid that contains the dot.
3. **Case 3: $3 \times 3$ square**
- There are $9$ possible $3 \times 3$ squares in the grid. Since the dot is centrally located, all $9$ of these squares will contain the dot.
4. **Case 4: $4 \times 4$ square**
- There are $4$ possible $4 \times 4$ squares in the grid. Each of these squares will contain the dot, as the dot is centrally located.
5. **Case 5: $5 \times 5$ square**
- There is only $1$ possible $5 \times 5$ square, which is the entire grid itself. This square contains the dot.
Summing all of these cases up, we get:
\[ 1 + 4 + 9 + 4 + 1 = 19 \]
The final answer is $\boxed{19}$
|
19
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
One afternoon Elizabeth noticed that twice as many cars on the expressway carried only a driver as compared to the number of cars that carried a driver and one passenger. She also noted that twice as many cars carried a driver and one passenger as those that carried a driver and two passengers. Only 10% of the cars carried a driver and three passengers, and no car carried more than four people. Any car containing at least three people was allowed to use the fast lane. Elizabeth calculated that $\frac{m}{n}$ of the people in cars on the expressway were allowed to ride in the fast lane, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
|
1. Let \( x \) be the fraction of cars that have only a driver (1 person).
2. According to the problem, twice as many cars have a driver and one passenger (2 people) as those with only a driver. Therefore, the fraction of cars with 2 people is \( 2x \).
3. Similarly, twice as many cars have a driver and two passengers (3 people) as those with a driver and one passenger. Therefore, the fraction of cars with 3 people is \( 2 \times 2x = 4x \).
4. It is given that 10% of the cars have a driver and three passengers (4 people). Therefore, the fraction of cars with 4 people is \( \frac{1}{10} \).
5. The total fraction of cars is the sum of the fractions of cars with 1, 2, 3, and 4 people:
\[
x + 2x + 4x + \frac{1}{10} = 1
\]
6. Simplify the equation:
\[
7x + \frac{1}{10} = 1
\]
7. Subtract \(\frac{1}{10}\) from both sides:
\[
7x = 1 - \frac{1}{10} = \frac{10}{10} - \frac{1}{10} = \frac{9}{10}
\]
8. Solve for \( x \):
\[
x = \frac{9}{10} \times \frac{1}{7} = \frac{9}{70}
\]
9. The fraction of cars with at least 3 people (which are allowed to use the fast lane) is the sum of the fractions of cars with 3 and 4 people:
\[
4x + \frac{1}{10} = 4 \times \frac{9}{70} + \frac{1}{10}
\]
10. Simplify the expression:
\[
4 \times \frac{9}{70} = \frac{36}{70} = \frac{18}{35}
\]
\[
\frac{18}{35} + \frac{1}{10} = \frac{18}{35} + \frac{1}{10} = \frac{18 \times 2}{35 \times 2} + \frac{1 \times 7}{10 \times 7} = \frac{36}{70} + \frac{7}{70} = \frac{43}{70}
\]
11. The fraction of people in cars allowed to use the fast lane is \(\frac{43}{70}\). Since \(m\) and \(n\) are relatively prime, \(m = 43\) and \(n = 70\).
12. Therefore, \(m + n = 43 + 70 = 113\).
The final answer is \(\boxed{113}\).
|
113
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of positive integers $n$ such that a regular polygon with $n$ sides has internal angles with measures equal to an integer number of degrees.
|
1. The measure of an internal angle of a regular polygon with \( n \) sides is given by:
\[
\text{Internal angle} = 180^\circ - \frac{360^\circ}{n}
\]
For this angle to be an integer, \(\frac{360^\circ}{n}\) must also be an integer.
2. This implies that \( n \) must be a divisor of \( 360 \). We need to find all positive integers \( n \) such that \( n \) is a divisor of \( 360 \).
3. First, we find the prime factorization of \( 360 \):
\[
360 = 2^3 \cdot 3^2 \cdot 5
\]
4. The number of divisors of \( 360 \) can be calculated using the formula for the number of divisors of a number \( n = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k} \), which is:
\[
(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
\]
For \( 360 \):
\[
(3+1)(2+1)(1+1) = 4 \cdot 3 \cdot 2 = 24
\]
Therefore, \( 360 \) has 24 divisors.
5. However, we need to exclude \( n = 1 \) and \( n = 2 \) because a polygon with 1 or 2 sides does not exist in the context of regular polygons.
6. Subtracting these two cases from the total number of divisors, we get:
\[
24 - 2 = 22
\]
Conclusion:
The number of positive integers \( n \) such that a regular polygon with \( n \) sides has internal angles with measures equal to an integer number of degrees is \( \boxed{22} \).
|
22
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Suzie flips a fair coin 6 times. The probability that Suzie flips 3 heads in a row but not 4 heads in a row is given by $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
|
1. **Identify the total number of possible outcomes:**
Since Suzie flips a fair coin 6 times, each flip has 2 possible outcomes (heads or tails). Therefore, the total number of possible outcomes is:
\[
2^6 = 64
\]
2. **Identify the favorable outcomes:**
We need to count the number of sequences where Suzie flips exactly 3 heads in a row but not 4 heads in a row. We can break this down into different cases based on the position of the 3 consecutive heads (HHH).
3. **Case Analysis:**
- **Case 1: Sequence is of the form \(THHHTX\):**
- The sequence starts with a tail (T), followed by three heads (HHH), followed by a tail (T), and ends with either a head (H) or a tail (T).
- There are 2 possible sequences: \(THHHTH\) and \(THHHTT\).
- **Case 2: Sequence is of the form \(XTHHHT\):**
- The sequence starts with either a head (H) or a tail (T), followed by a tail (T), followed by three heads (HHH), and ends with a tail (T).
- There are 2 possible sequences: \(THHHTT\) and \(HHHHTT\).
- **Case 3: Sequence is of the form \(HHHTXX\):**
- The sequence starts with three heads (HHH), followed by a tail (T), and ends with either two heads (HH), a head and a tail (HT), a tail and a head (TH), or two tails (TT).
- There are 4 possible sequences: \(HHHTHH\), \(HHHTHT\), \(HHHTTH\), and \(HHHTTT\).
- **Case 4: Sequence is of the form \(XXTHHH\):**
- The sequence starts with either two heads (HH), a head and a tail (HT), a tail and a head (TH), or two tails (TT), followed by a tail (T), and ends with three heads (HHH).
- There are 4 possible sequences: \(HHTHHH\), \(HTTHHH\), \(THTHHH\), and \(TTTHHH\).
4. **Count the total number of favorable sequences:**
- From Case 1: 2 sequences
- From Case 2: 2 sequences
- From Case 3: 4 sequences
- From Case 4: 4 sequences
Therefore, the total number of favorable sequences is:
\[
2 + 2 + 4 + 4 = 12
\]
5. **Calculate the probability:**
The probability that Suzie flips 3 heads in a row but not 4 heads in a row is given by the ratio of the number of favorable outcomes to the total number of possible outcomes:
\[
\frac{12}{64} = \frac{3}{16}
\]
6. **Identify \(m\) and \(n\):**
In the fraction \(\frac{3}{16}\), \(m = 3\) and \(n = 16\).
7. **Calculate \(m + n\):**
\[
m + n = 3 + 16 = 19
\]
The final answer is \( \boxed{ 19 } \)
|
19
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the positive integer $n$ such that the least common multiple of $n$ and $n - 30$ is $n + 1320$.
|
1. Let \( n \) be the positive integer such that the least common multiple (LCM) of \( n \) and \( n - 30 \) is \( n + 1320 \). We can write this as:
\[
\text{LCM}(n, n - 30) = n + 1320
\]
2. Recall the relationship between the LCM and the greatest common divisor (GCD):
\[
\text{LCM}(a, b) = \frac{a \cdot b}{\text{GCD}(a, b)}
\]
Applying this to our problem, we get:
\[
\text{LCM}(n, n - 30) = \frac{n \cdot (n - 30)}{\text{GCD}(n, n - 30)}
\]
3. Let \( d = \text{GCD}(n, n - 30) \). Since \( d \) divides both \( n \) and \( n - 30 \), it must also divide their difference:
\[
d \mid (n - (n - 30)) \implies d \mid 30
\]
Therefore, \( d \) is a divisor of 30. The possible values for \( d \) are 1, 2, 3, 5, 6, 10, 15, and 30.
4. Substitute \( \text{LCM}(n, n - 30) \) into the equation:
\[
\frac{n \cdot (n - 30)}{d} = n + 1320
\]
Rearrange to get:
\[
n \cdot (n - 30) = d \cdot (n + 1320)
\]
\[
n^2 - 30n = dn + 1320d
\]
\[
n^2 - (30 + d)n - 1320d = 0
\]
5. This is a quadratic equation in \( n \):
\[
n^2 - (30 + d)n - 1320d = 0
\]
Solve this quadratic equation using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -(30 + d) \), and \( c = -1320d \):
\[
n = \frac{(30 + d) \pm \sqrt{(30 + d)^2 + 4 \cdot 1320d}}{2}
\]
\[
n = \frac{(30 + d) \pm \sqrt{(30 + d)^2 + 5280d}}{2}
\]
6. We need \( n \) to be a positive integer. We test the possible values of \( d \) to find a valid \( n \):
- For \( d = 1 \):
\[
n = \frac{(30 + 1) \pm \sqrt{(30 + 1)^2 + 5280 \cdot 1}}{2}
\]
\[
n = \frac{31 \pm \sqrt{961 + 5280}}{2}
\]
\[
n = \frac{31 \pm \sqrt{6241}}{2}
\]
\[
n = \frac{31 \pm 79}{2}
\]
\[
n = \frac{110}{2} = 55 \quad \text{or} \quad n = \frac{-48}{2} = -24 \quad (\text{not positive})
\]
\( n = 55 \) does not satisfy the original condition.
- For \( d = 3 \):
\[
n = \frac{(30 + 3) \pm \sqrt{(30 + 3)^2 + 5280 \cdot 3}}{2}
\]
\[
n = \frac{33 \pm \sqrt{1089 + 15840}}{2}
\]
\[
n = \frac{33 \pm \sqrt{16929}}{2}
\]
\[
n = \frac{33 \pm 130}{2}
\]
\[
n = \frac{163}{2} = 81.5 \quad \text{or} \quad n = \frac{-97}{2} = -48.5 \quad (\text{not positive integer})
\]
- For \( d = 5 \):
\[
n = \frac{(30 + 5) \pm \sqrt{(30 + 5)^2 + 5280 \cdot 5}}{2}
\]
\[
n = \frac{35 \pm \sqrt{1225 + 26400}}{2}
\]
\[
n = \frac{35 \pm \sqrt{27625}}{2}
\]
\[
n = \frac{35 \pm 166}{2}
\]
\[
n = \frac{201}{2} = 100.5 \quad \text{or} \quad n = \frac{-131}{2} = -65.5 \quad (\text{not positive integer})
\]
- For \( d = 15 \):
\[
n = \frac{(30 + 15) \pm \sqrt{(30 + 15)^2 + 5280 \cdot 15}}{2}
\]
\[
n = \frac{45 \pm \sqrt{2025 + 79200}}{2}
\]
\[
n = \frac{45 \pm \sqrt{81225}}{2}
\]
\[
n = \frac{45 \pm 285}{2}
\]
\[
n = \frac{330}{2} = 165 \quad \text{or} \quad n = \frac{-240}{2} = -120 \quad (\text{not positive})
\]
\( n = 165 \) is a valid solution.
7. Verify \( n = 165 \):
\[
\text{LCM}(165, 135) = \frac{165 \cdot 135}{\text{GCD}(165, 135)}
\]
\[
\text{GCD}(165, 135) = 15
\]
\[
\text{LCM}(165, 135) = \frac{165 \cdot 135}{15} = 1485
\]
\[
165 + 1320 = 1485
\]
The condition is satisfied.
The final answer is \( \boxed{165} \)
|
165
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The figure below was made by gluing together 5 non-overlapping congruent squares. The figure has area 45. Find the perimeter of the figure.
[center][img]https://snag.gy/ZeKf4q.jpg[/center][/img]
|
1. **Determine the area of each square:**
The total area of the figure is given as \(45\) square units. Since the figure is composed of 5 congruent squares, the area of each square is:
\[
\text{Area of each square} = \frac{45}{5} = 9 \text{ square units}
\]
2. **Find the side length of each square:**
The area of a square is given by the square of its side length. Let \(s\) be the side length of each square. Then:
\[
s^2 = 9 \implies s = \sqrt{9} = 3 \text{ units}
\]
3. **Identify the perimeter of the figure:**
To find the perimeter, we need to sum the lengths of all the outer edges of the figure. The figure is composed of 5 squares arranged in a specific pattern. We will label the vertices as given in the problem and calculate the perimeter by summing the lengths of the outer edges.
4. **Calculate the lengths of the outer edges:**
- The horizontal segments are \(AB\), \(BC\), \(CD\), \(DE\), \(EF\), \(FG\), \(GH\), \(HI\), \(IJ\), \(JK\), \(KL\), \(LM\), and \(MA\).
- Each of these segments corresponds to the side length of a square, which is \(3\) units.
5. **Sum the lengths of the outer edges:**
- The segments \(AB\), \(BC\), \(CD\), \(DE\), \(EF\), \(FG\), \(GH\), \(IJ\), \(JK\), \(KL\), and \(LM\) each have a length of \(3\) units.
- The segments \(HI\) and \(MA\) are vertical segments that connect the top and bottom of the figure. Since the figure is composed of 5 squares, the total height of the figure is \(3 \times 3 = 9\) units. However, the segments \(HI\) and \(MA\) are not full heights but rather the difference between the total height and the height of one square, which is \(6\) units.
6. **Adjust for overlapping segments:**
- The segments \(HI\) and \(MA\) are not full heights but rather the difference between the total height and the height of one square, which is \(6\) units.
- Therefore, the total length of the outer edges is:
\[
\mathcal{P} = 11 \times 3 + 3 = 33 + 3 = 36 \text{ units}
\]
The final answer is \(\boxed{36}\) units.
|
36
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The Stromquist Comet is visible every 61 years. If the comet is visible in 2017, what is the next leap year when the comet will be visible?
|
1. The Stromquist Comet is visible every 61 years. We need to find the next leap year after 2017 when the comet will be visible.
2. A leap year is defined as a year that is divisible by 4, except for end-of-century years, which must be divisible by 400. For simplicity, we will first check for divisibility by 4.
3. We need to find an integer \( n \) such that \( 2017 + 61n \) is a leap year. This means \( 2017 + 61n \) must be divisible by 4.
4. We start by finding the remainder when 2017 is divided by 4:
\[
2017 \div 4 = 504 \text{ remainder } 1 \implies 2017 \equiv 1 \pmod{4}
\]
5. We need \( 2017 + 61n \equiv 0 \pmod{4} \):
\[
2017 + 61n \equiv 1 + 61n \equiv 0 \pmod{4}
\]
Simplifying, we get:
\[
1 + 61n \equiv 0 \pmod{4}
\]
\[
61 \equiv 1 \pmod{4} \implies 1 + 1n \equiv 0 \pmod{4}
\]
\[
1 + n \equiv 0 \pmod{4} \implies n \equiv -1 \pmod{4} \implies n \equiv 3 \pmod{4}
\]
6. The smallest positive integer \( n \) that satisfies \( n \equiv 3 \pmod{4} \) is \( n = 3 \):
\[
2017 + 61 \times 3 = 2017 + 183 = 2200
\]
However, 2200 is not a leap year because it is not divisible by 400.
7. The next possible value of \( n \) is \( n = 7 \):
\[
2017 + 61 \times 7 = 2017 + 427 = 2444
\]
We need to check if 2444 is a leap year:
\[
2444 \div 4 = 611 \implies 2444 \text{ is divisible by 4}
\]
Since 2444 is divisible by 4 and not an end-of-century year, it is a leap year.
Thus, the next leap year when the Stromquist Comet will be visible is 2444.
The final answer is \(\boxed{2444}\).
|
2444
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
On a typical morning Aiden gets out of bed, goes through his morning preparation, rides the bus, and walks from the bus stop to work arriving at work 120 minutes after getting out of bed. One morning Aiden got out of bed late, so he rushed through his morning preparation getting onto the bus in half the usual time, the bus ride took 25 percent longer than usual, and he ran from the bus stop to work in half the usual time it takes him to walk arriving at work 96 minutes after he got out of bed. The next morning Aiden got out of bed extra early, leisurely went through his morning preparation taking 25 percent longer than usual to get onto the bus, his bus ride took 25 percent less time than usual, and he walked slowly from the bus stop to work taking 25 percent longer than usual. How many minutes after Aiden got out of bed did he arrive at work that day?
|
1. Let \( a \) be the time it takes Aiden for his morning preparation.
2. Let \( b \) be the time it takes Aiden to ride the bus.
3. Let \( c \) be the time it takes Aiden to walk from the bus stop to work.
From the problem, we have the following information:
- On a typical morning, Aiden arrives at work 120 minutes after getting out of bed:
\[
a + b + c = 120
\]
- One morning, Aiden rushed through his morning preparation, taking half the usual time, the bus ride took 25% longer, and he ran from the bus stop to work in half the usual time, arriving at work 96 minutes after getting out of bed:
\[
0.5a + 1.25b + 0.5c = 96
\]
Doubling both sides of the second equation to eliminate the fractions:
\[
a + 2.5b + c = 192
\]
- Subtracting the first equation from the modified second equation:
\[
(a + 2.5b + c) - (a + b + c) = 192 - 120
\]
\[
1.5b = 72
\]
\[
b = \frac{72}{1.5} = 48
\]
- Substituting \( b = 48 \) back into the first equation:
\[
a + 48 + c = 120
\]
\[
a + c = 72
\]
- Substituting \( b = 48 \) into the second equation:
\[
0.5a + 1.25 \cdot 48 + 0.5c = 96
\]
\[
0.5a + 60 + 0.5c = 96
\]
\[
0.5a + 0.5c = 36
\]
\[
a + c = 72
\]
- The next morning, Aiden got out of bed extra early, taking 25% longer for his morning preparation, the bus ride took 25% less time, and he walked slowly from the bus stop to work taking 25% longer:
\[
1.25a + 0.75b + 1.25c
\]
- Substituting \( a + c = 72 \) and \( b = 48 \):
\[
1.25a + 0.75 \cdot 48 + 1.25c
\]
\[
1.25a + 36 + 1.25c
\]
\[
1.25(a + c) + 36
\]
\[
1.25 \cdot 72 + 36
\]
\[
90 + 36 = 126
\]
Thus, Aiden arrived at work 126 minutes after getting out of bed that day.
The final answer is \(\boxed{126}\)
|
126
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The positive integer $m$ is a multiple of 111, and the positive integer $n$ is a multiple of 31. Their sum is 2017. Find $n - m$.
|
1. Let \( m = 111x \) and \( n = 31y \), where \( x \) and \( y \) are positive integers.
2. Given that \( m + n = 2017 \), we substitute \( m \) and \( n \) into the equation:
\[
111x + 31y = 2017
\]
3. We need to find integer solutions for \( x \) and \( y \). To do this, we can use the method of solving Diophantine equations. First, we solve for \( y \) in terms of \( x \):
\[
31y = 2017 - 111x
\]
\[
y = \frac{2017 - 111x}{31}
\]
4. For \( y \) to be an integer, \( 2017 - 111x \) must be divisible by 31. We check the divisibility:
\[
2017 \equiv 111x \pmod{31}
\]
Simplifying \( 111 \mod 31 \):
\[
111 \div 31 = 3 \quad \text{remainder} \quad 18 \quad \Rightarrow \quad 111 \equiv 18 \pmod{31}
\]
Thus, the equation becomes:
\[
2017 \equiv 18x \pmod{31}
\]
5. Next, we simplify \( 2017 \mod 31 \):
\[
2017 \div 31 = 65 \quad \text{remainder} \quad 2 \quad \Rightarrow \quad 2017 \equiv 2 \pmod{31}
\]
So, we have:
\[
2 \equiv 18x \pmod{31}
\]
6. To solve for \( x \), we need the multiplicative inverse of 18 modulo 31. We use the Extended Euclidean Algorithm:
\[
31 = 1 \cdot 18 + 13
\]
\[
18 = 1 \cdot 13 + 5
\]
\[
13 = 2 \cdot 5 + 3
\]
\[
5 = 1 \cdot 3 + 2
\]
\[
3 = 1 \cdot 2 + 1
\]
\[
2 = 2 \cdot 1 + 0
\]
Back-substituting to find the inverse:
\[
1 = 3 - 1 \cdot 2
\]
\[
1 = 3 - 1 \cdot (5 - 1 \cdot 3) = 2 \cdot 3 - 1 \cdot 5
\]
\[
1 = 2 \cdot (13 - 2 \cdot 5) - 1 \cdot 5 = 2 \cdot 13 - 5 \cdot 5
\]
\[
1 = 2 \cdot 13 - 5 \cdot (18 - 1 \cdot 13) = 7 \cdot 13 - 5 \cdot 18
\]
\[
1 = 7 \cdot (31 - 1 \cdot 18) - 5 \cdot 18 = 7 \cdot 31 - 12 \cdot 18
\]
Thus, the inverse of 18 modulo 31 is \(-12 \equiv 19 \pmod{31}\).
7. Using the inverse, we solve for \( x \):
\[
2 \cdot 19 \equiv 18x \cdot 19 \pmod{31}
\]
\[
38 \equiv x \pmod{31}
\]
\[
x \equiv 7 \pmod{31}
\]
So, \( x = 7 \) is a solution.
8. Substituting \( x = 7 \) back into the equations for \( m \) and \( n \):
\[
m = 111 \cdot 7 = 777
\]
\[
n = 2017 - 777 = 1240
\]
9. Finally, we find \( n - m \):
\[
n - m = 1240 - 777 = 463
\]
The final answer is \(\boxed{463}\).
|
463
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
In $\triangle{ADE}$ points $B$ and $C$ are on side $AD$ and points $F$ and $G$ are on side $AE$ so that $BG \parallel CF \parallel DE$, as shown. The area of $\triangle{ABG}$ is $36$, the area of trapezoid $CFED$ is $144$, and $AB = CD$. Find the area of trapezoid $BGFC$.
[center][img]https://snag.gy/SIuOLB.jpg[/img][/center]
|
1. Let \( AB = CD = x \) and \( BC = y \).
2. Given that the area of \(\triangle ABG\) is 36, we can express this area in terms of the area of \(\triangle ADE\). Since \(BG \parallel CF \parallel DE\), the triangles \(\triangle ABG\) and \(\triangle ADE\) are similar. The ratio of their areas is the square of the ratio of their corresponding sides:
\[
[ABG] = \left(\frac{x}{2x+y}\right)^2 \cdot [ADE]
\]
Given \([ABG] = 36\), we have:
\[
36 = \frac{x^2}{(2x+y)^2} \cdot [ADE]
\]
3. Given that the area of trapezoid \(CFED\) is 144, we can express this area in terms of the area of \(\triangle ADE\). The area of trapezoid \(CFED\) is the difference between the area of \(\triangle ADE\) and the area of \(\triangle ACF\):
\[
[CFED] = [ADE] - [ACF]
\]
Since \(\triangle ACF\) is similar to \(\triangle ADE\), we have:
\[
[ACF] = \left(\frac{x+y}{2x+y}\right)^2 \cdot [ADE]
\]
Therefore,
\[
[CFED] = [ADE] - \left(\frac{x+y}{2x+y}\right)^2 \cdot [ADE] = \left(1 - \left(\frac{x+y}{2x+y}\right)^2\right) \cdot [ADE]
\]
Given \([CFED] = 144\), we have:
\[
144 = \left(1 - \left(\frac{x+y}{2x+y}\right)^2\right) \cdot [ADE]
\]
4. Dividing the equation for \([CFED]\) by the equation for \([ABG]\), we get:
\[
\frac{144}{36} = \frac{\left(1 - \left(\frac{x+y}{2x+y}\right)^2\right) \cdot [ADE]}{\frac{x^2}{(2x+y)^2} \cdot [ADE]}
\]
Simplifying, we get:
\[
4 = \frac{(2x+y)^2 \left(1 - \left(\frac{x+y}{2x+y}\right)^2\right)}{x^2}
\]
\[
4 = \frac{(2x+y)^2 - (x+y)^2}{x^2}
\]
\[
4 = \frac{4x^2 + 4xy + y^2 - x^2 - 2xy - y^2}{x^2}
\]
\[
4 = \frac{3x^2 + 2xy}{x^2}
\]
\[
4 = 3 + \frac{2y}{x}
\]
\[
1 = \frac{2y}{x}
\]
\[
y = \frac{x}{2}
\]
5. Now, we need to find the area of trapezoid \(BGFC\). The area of \(BGFC\) is the difference between the area of \(\triangle ACF\) and the area of \(\triangle ABG\):
\[
[BGFC] = [ACF] - [ABG]
\]
Using the similarity ratios, we have:
\[
[ACF] = \left(\frac{x+y}{2x+y}\right)^2 \cdot [ADE]
\]
\[
[BGFC] = \left(\frac{x+y}{2x+y}\right)^2 \cdot [ADE] - \left(\frac{x}{2x+y}\right)^2 \cdot [ADE]
\]
\[
[BGFC] = \left(\left(\frac{x+y}{2x+y}\right)^2 - \left(\frac{x}{2x+y}\right)^2\right) \cdot [ADE]
\]
\[
[BGFC] = \left(\frac{(x+y)^2 - x^2}{(2x+y)^2}\right) \cdot [ADE]
\]
\[
[BGFC] = \left(\frac{x^2 + 2xy + y^2 - x^2}{(2x+y)^2}\right) \cdot [ADE]
\]
\[
[BGFC] = \left(\frac{2xy + y^2}{(2x+y)^2}\right) \cdot [ADE]
\]
Substituting \(y = \frac{x}{2}\), we get:
\[
[BGFC] = \left(\frac{2x \cdot \frac{x}{2} + \left(\frac{x}{2}\right)^2}{(2x + \frac{x}{2})^2}\right) \cdot [ADE]
\]
\[
[BGFC] = \left(\frac{x^2 + \frac{x^2}{4}}{\left(\frac{5x}{2}\right)^2}\right) \cdot [ADE]
\]
\[
[BGFC] = \left(\frac{\frac{5x^2}{4}}{\frac{25x^2}{4}}\right) \cdot [ADE]
\]
\[
[BGFC] = \left(\frac{5}{25}\right) \cdot [ADE]
\]
\[
[BGFC] = \frac{1}{5} \cdot [ADE]
\]
Since \([ADE] = 36 \cdot \frac{(2x+y)^2}{x^2}\), we have:
\[
[BGFC] = \frac{1}{5} \cdot 180 = 45
\]
The final answer is \(\boxed{45}\)
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the number of rearrangements of the letters in the word MATHMEET that begin and end with the same letter such as TAMEMHET.
|
To find the number of rearrangements of the letters in the word "MATHMEET" that begin and end with the same letter, we can break the problem into three cases based on which letter is at the ends.
1. **Case 1: The word begins and ends with the letter T.**
- The word "MATHMEET" contains 8 letters: M, A, T, H, M, E, E, T.
- If T is at both ends, we are left with the letters M, A, T, H, M, E, E to arrange in the middle 6 positions.
- The number of ways to arrange these 6 letters, considering the repetitions of M and E, is given by:
\[
\frac{6!}{2! \cdot 2!}
\]
where \(6!\) is the factorial of 6, and \(2!\) accounts for the two M's and the two E's.
\[
\frac{6!}{2! \cdot 2!} = \frac{720}{4} = 180
\]
2. **Case 2: The word begins and ends with the letter M.**
- Similarly, if M is at both ends, we are left with the letters A, T, H, M, E, E, T to arrange in the middle 6 positions.
- The number of ways to arrange these 6 letters, considering the repetitions of T and E, is given by:
\[
\frac{6!}{2! \cdot 2!} = 180
\]
3. **Case 3: The word begins and ends with the letter E.**
- If E is at both ends, we are left with the letters M, A, T, H, M, T to arrange in the middle 6 positions.
- The number of ways to arrange these 6 letters, considering the repetitions of M and T, is given by:
\[
\frac{6!}{2! \cdot 2!} = 180
\]
Adding the number of rearrangements from all three cases, we get:
\[
180 + 180 + 180 = 540
\]
The final answer is \(\boxed{540}\).
|
540
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Let $x$, $y$, and $z$ be real numbers such that
$$12x - 9y^2 = 7$$
$$6y - 9z^2 = -2$$
$$12z - 9x^2 = 4$$
Find $6x^2 + 9y^2 + 12z^2$.
|
1. We start with the given equations:
\[
12x - 9y^2 = 7
\]
\[
6y - 9z^2 = -2
\]
\[
12z - 9x^2 = 4
\]
2. Add all three equations:
\[
(12x - 9y^2) + (6y - 9z^2) + (12z - 9x^2) = 7 - 2 + 4
\]
Simplifying the right-hand side:
\[
12x + 6y + 12z - 9x^2 - 9y^2 - 9z^2 = 9
\]
3. Rearrange the equation:
\[
12x + 6y + 12z = 9x^2 + 9y^2 + 9z^2 + 9
\]
Divide the entire equation by 3:
\[
4x + 2y + 4z = 3(x^2 + y^2 + z^2 + 1)
\]
4. Rearrange to isolate the quadratic terms:
\[
3(x^2 + y^2 + z^2) - 4x - 2y - 4z = -3
\]
5. Complete the square for each variable:
\[
\left(x^2 - \frac{4}{3}x\right) + \left(y^2 - \frac{2}{3}y\right) + \left(z^2 - \frac{4}{3}z\right) = -1
\]
\[
\left(x - \frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 + \left(y - \frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 + \left(z - \frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 = -1
\]
\[
\left(x - \frac{2}{3}\right)^2 + \left(y - \frac{1}{3}\right)^2 + \left(z - \frac{2}{3}\right)^2 = 0
\]
6. Since the sum of squares is zero, each square must be zero:
\[
x - \frac{2}{3} = 0 \implies x = \frac{2}{3}
\]
\[
y - \frac{1}{3} = 0 \implies y = \frac{1}{3}
\]
\[
z - \frac{2}{3} = 0 \implies z = \frac{2}{3}
\]
7. Substitute \(x = \frac{2}{3}\), \(y = \frac{1}{3}\), and \(z = \frac{2}{3}\) into the expression \(6x^2 + 9y^2 + 12z^2\):
\[
6\left(\frac{2}{3}\right)^2 + 9\left(\frac{1}{3}\right)^2 + 12\left(\frac{2}{3}\right)^2
\]
\[
= 6 \cdot \frac{4}{9} + 9 \cdot \frac{1}{9} + 12 \cdot \frac{4}{9}
\]
\[
= \frac{24}{9} + \frac{9}{9} + \frac{48}{9}
\]
\[
= \frac{24 + 9 + 48}{9}
\]
\[
= \frac{81}{9}
\]
\[
= 9
\]
The final answer is \(\boxed{9}\)
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
The set of positive real numbers $x$ that satisfy $2 | x^2 - 9 | \le 9 | x | $ is an interval $[m, M]$. Find $10m + M$.
|
To solve the problem, we need to find the set of positive real numbers \( x \) that satisfy the inequality \( 2 | x^2 - 9 | \le 9 | x | \). We will consider different cases based on the value of \( x \).
1. **Case 1: \( x > 3 \)**
- Here, \( | x^2 - 9 | = x^2 - 9 \) and \( | x | = x \).
- The inequality becomes:
\[
2(x^2 - 9) \le 9x
\]
- Simplifying, we get:
\[
2x^2 - 18 \le 9x
\]
\[
2x^2 - 9x - 18 \le 0
\]
\[
(2x + 3)(x - 6) \le 0
\]
- Solving the quadratic inequality, we find:
\[
x \in \left( -\frac{3}{2}, 6 \right]
\]
- Since \( x > 3 \), we have:
\[
x \in (3, 6]
\]
2. **Case 2: \( 0 < x \le 3 \)**
- Here, \( | x^2 - 9 | = 9 - x^2 \) and \( | x | = x \).
- The inequality becomes:
\[
2(9 - x^2) \le 9x
\]
- Simplifying, we get:
\[
18 - 2x^2 \le 9x
\]
\[
2x^2 + 9x - 18 \ge 0
\]
\[
(2x - 3)(x + 6) \ge 0
\]
- Solving the quadratic inequality, we find:
\[
x \in \left[ \frac{3}{2}, 3 \right]
\]
3. **Case 3: \( x \le -3 \)**
- Since we are only interested in positive \( x \), we can ignore this case.
4. **Case 4: \( -3 < x \le 0 \)**
- Since we are only interested in positive \( x \), we can ignore this case.
Combining the results from the relevant cases, we find that the set of positive \( x \) that satisfy the inequality is:
\[
x \in \left[ \frac{3}{2}, 6 \right]
\]
Thus, the interval \([m, M]\) is \(\left[ \frac{3}{2}, 6 \right]\).
To find \( 10m + M \):
\[
m = \frac{3}{2}, \quad M = 6
\]
\[
10m + M = 10 \left( \frac{3}{2} \right) + 6 = 15 + 6 = 21
\]
The final answer is \( \boxed{ 21 } \)
|
21
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
Find the sum of all values of $a + b$, where $(a, b)$ is an ordered pair of positive integers and $a^2+\sqrt{2017-b^2}$ is a perfect square.
|
1. We start with the equation \(a^2 + \sqrt{2017 - b^2} = k^2\) where \(k\) is an integer. For this to be true, \(\sqrt{2017 - b^2}\) must also be an integer. Let \(\sqrt{2017 - b^2} = m\) where \(m\) is an integer. Therefore, we have:
\[
2017 - b^2 = m^2
\]
This can be rewritten as:
\[
2017 = b^2 + m^2
\]
This implies that \(b^2 + m^2 = 2017\).
2. We need to find pairs \((b, m)\) such that both \(b\) and \(m\) are integers. We test values of \(b\) to find corresponding \(m\) values that satisfy the equation.
3. Testing \(b = 44\):
\[
2017 - 44^2 = 2017 - 1936 = 81
\]
\[
\sqrt{81} = 9
\]
So, one pair is \((b, m) = (44, 9)\).
4. Testing \(b = 9\):
\[
2017 - 9^2 = 2017 - 81 = 1936
\]
\[
\sqrt{1936} = 44
\]
So, another pair is \((b, m) = (9, 44)\).
5. Now, we have two pairs \((b, m)\): \((44, 9)\) and \((9, 44)\). We need to find corresponding \(a\) values such that \(a^2 + m = k^2\).
6. For \((b, m) = (44, 9)\):
\[
a^2 + 9 = k^2
\]
This can be rewritten as:
\[
k^2 - a^2 = 9
\]
\[
(k - a)(k + a) = 9
\]
The factor pairs of 9 are \((1, 9)\) and \((3, 3)\). Solving these:
- For \((1, 9)\):
\[
k - a = 1 \quad \text{and} \quad k + a = 9
\]
Adding these equations:
\[
2k = 10 \implies k = 5
\]
Subtracting these equations:
\[
2a = 8 \implies a = 4
\]
So, one solution is \((a, b) = (4, 44)\).
7. For \((b, m) = (9, 44)\):
\[
a^2 + 44 = k^2
\]
This can be rewritten as:
\[
k^2 - a^2 = 44
\]
\[
(k - a)(k + a) = 44
\]
The factor pairs of 44 are \((1, 44)\), \((2, 22)\), and \((4, 11)\). Solving these:
- For \((2, 22)\):
\[
k - a = 2 \quad \text{and} \quad k + a = 22
\]
Adding these equations:
\[
2k = 24 \implies k = 12
\]
Subtracting these equations:
\[
2a = 20 \implies a = 10
\]
So, another solution is \((a, b) = (10, 9)\).
8. The solutions \((a, b)\) are \((4, 44)\) and \((10, 9)\). The sums \(a + b\) are:
\[
4 + 44 = 48
\]
\[
10 + 9 = 19
\]
Therefore, the total sum is:
\[
48 + 19 = 67
\]
The final answer is \(\boxed{67}\)
|
67
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
aops_forum
| false
|
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