2020 AIME I Problems/Problem 9

Revision as of 12:10, 13 March 2020 by Vvluo (talk | contribs) (Solution)


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 $m.$


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 $\boxed{77}$.


See Also

2020 AIME I (ProblemsAnswer KeyResources)
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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