2020 AIME I Problems/Problem 9

Problem

Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$

Solution

First, prime factorize $20^9$ as $2^{18} \cdot 5^9$ . Denote $a_1$ as $2^{b_1} \cdot 5^{c_1}$ , $a_2$ as $2^{b_2} \cdot 5^{c_2}$ , and $a_3$ as $2^{b_3} \cdot 5^{c_3}$ .

In order for $a_1$ to divide $a_2$ , and for $a_2$ to divide $a_3$ , $b_1\le b_2\le b_3$ , and $c_1\le c_2\le c_3$ . We will consider each case separately. Note that the total amount of possibilities is $190^3$ , as there are $(18+1)(9+1)=190$ choices for each factor.

We notice that if we add $1$ to $b_2$ and $2$ to $b_3$ , then we can reach the stronger inequality $b_1<b_2<b_3$ . Therefore, if we pick $3$ integers from $0$ to $20$ , they will correspond to a unique solution, forming a 1-1 correspondence.This is also equivalent to applying stars and bars on distributing the powers of 2 and 5 through differences. The amount of solutions to this inequality is $\dbinom{21}{3}$ .

The case for $c_1$ , $c_2$ , and $c_3$ proceeds similarly for a result of $\dbinom{12}{3}$ . Therefore, the probability of choosing three such factors is $\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.$ Simplification gives $\frac{77}{1805}$ , and therefore the answer is $\boxed{77}$ .

-molocyxu

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2020 AIME I Problems/Problem 9

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