Difference between revisions of "2020 AIME I Problems/Problem 9"
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== Solution == | == Solution == | ||
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+ | First, prime factorize <math>20^9</math> as <math>2^{18} \cdot 5^9</math>. Denote <math>a_1</math> as <math>2^{b_1} \cdot 5^{c_1}</math>, <math>a_2</math> as <math>2^{b_2} \cdot 5^{c_2}</math>, and <math>a_3</math> as <math>2^{b_3} \cdot 5^{c_3}</math>. | ||
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+ | In order for <math>a_1</math> to divide <math>a_2</math>, and for <math>a_2</math> to divide <math>a_3</math>, <math>b_1<=b_2<=b_3</math>, and <math>c_1<=c_2<=c_3</math>. We will consider each case separately. Note that the total amount of possibilities is <math>190^3</math>, as there are <math>(18+1)(9+1)=190</math> choices for each factor. | ||
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+ | We notice that if we add <math>1</math> to <math>b_2</math> and <math>2</math> to <math>b_3</math>, then we can reach the stronger inequality <math>b_1<b_2<b_3</math>. Therefore, if we pick <math>3</math> integers from <math>0</math> to <math>20</math>, they will correspond to a unique solution, forming a 1-1 correspondence. The amount of solutions to this inequality is <math>\dbinom{21}{3}</math>. | ||
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+ | The case for <math>c_1</math>,<math>c_2</math>, and <math>c_3</math> proceeds similarly for a result of <math>\dbinom{12}{3}</math>. Therefore, the probability of choosing three such factors is <cmath>\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.</cmath> Simplification gives <math>\frac{77}{1805}</math>, and therefore the answer is <math>\boxed{77}</math>. | ||
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+ | -molocyxu | ||
==See Also== | ==See Also== |
Revision as of 16:13, 12 March 2020
Note: Please do not post problems here until after the AIME.
Problem
Solution
First, prime factorize as . Denote as , as , and as .
In order for to divide , and for to divide , , and . We will consider each case separately. Note that the total amount of possibilities is , as there are choices for each factor.
We notice that if we add to and to , then we can reach the stronger inequality . Therefore, if we pick integers from to , they will correspond to a unique solution, forming a 1-1 correspondence. The amount of solutions to this inequality is .
The case for ,, and proceeds similarly for a result of . Therefore, the probability of choosing three such factors is Simplification gives , and therefore the answer is .
-molocyxu
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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