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

m (→‎Solution)
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== Solution ==
 
== Solution ==
 +
 +
<asy>
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size(12cm);
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for (int x = 1; x < 18; ++x) {
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    draw((x, 0) -- (x, 9), dotted);
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}
 +
for (int y = 1; y < 9; ++y) {
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    draw((0, y) -- (18, y), dotted);
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}
 +
 +
draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle);
 +
 +
pair b1, b2, b3;
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pair c1, c2, c3;
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pair a1, a2, a3;
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b1 = (3, 0); b2 = (12, 0); b3 = (16, 0);
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c1 = (0, 2); c2 = (0, 4); c3 = (0, 8);
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a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3;
 +
 +
draw(b1 -- a1 -- c1);
 +
draw(b2 -- a2 -- c2);
 +
draw(b3 -- a3 -- c3);
 +
 +
dot(a1); dot(a2); dot(a3);
 +
label("$a_1$", a1, NE);
 +
label("$a_2$", a2, NE);
 +
label("$a_3$", a3, NE);
 +
label("$b_1$", b1, S);
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label("$b_2$", b2, S);
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label("$b_3$", b3, S);
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label("$c_1$", c1, W);
 +
label("$c_2$", c2, W);
 +
label("$c_3$", c3, W);
 +
 +
</asy>
  
 
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>.

Revision as of 13:10, 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

[asy] size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); c2 = (0, 4); c3 = (0, 8); a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3; draw(b1 -- a1 -- c1); draw(b2 -- a2 -- c2); draw(b3 -- a3 -- c3); dot(a1); dot(a2); dot(a3); label("$a_1$", a1, NE); label("$a_2$", a2, NE); label("$a_3$", a3, NE); label("$b_1$", b1, S); label("$b_2$", b2, S); label("$b_3$", b3, S); label("$c_1$", c1, W); label("$c_2$", c2, W); label("$c_3$", c3, W); [/asy]

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

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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