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Difference between revisions of "2020 AIME I Problems/Problem 9"

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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>.
 
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\ge b_2\ge b_3</math>, and <math>c_1\ge c_2\ge 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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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\le b_2\le b_3</math>, and <math>c_1\le c_2\le 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.
  
 
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>.
 
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>.

Revision as of 08:16, 13 March 2020

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. 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

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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