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Difference between revisions of "2020 AIME I Problems/Problem 9"
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The final probability is simply <math>\frac{1330 \cdot 220}{190^3}</math>. Simplification gives <math>\frac{77}{1805}</math>, and therefore the answer is <math>\boxed{077}</math>.
 
The final probability is simply <math>\frac{1330 \cdot 220}{190^3}</math>. Simplification gives <math>\frac{77}{1805}</math>, and therefore the answer is <math>\boxed{077}</math>.
 +
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==Solution 3==
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 +
Similar to before, we calculate that there are <math>190^3</math> ways to choose 3 factors with replacement. Then, we figure out the number of triplets {a,b,c} and {d,f,g}, where a, b, and c represent powers of 2 and d, f, and g represent powers of 5, such that the triplets are in non-descending order. The maximum power of 2 is 18, and the maximum power of 5 is 9. Using the Hockey Stick identity, we figure out that there are <math>\dbinom{12}{3}</math> ways to choose d, f and g, and <math>\dbinom{21}{3}</math> ways to choose a, b, and c. Therefore, the probability of choosing 3 factors which satisfy the conditions is <cmath>\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.</cmath> This simplifies to 77/1805, therefore m = <math>\boxed{077}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 11:04, 29 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 $0\le b_1<b_2+1<b_3+2\le 20$. Therefore, if we pick $3$ integers from $0$ to $20$, they will correspond to a unique solution, forming a 1-1 correspondence between the numbers $b_1$, $b_2+1$, and $b_3+2$. 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 $\boxed{077}$.

-molocyxu

Solution 2

Same as before, say the factors have powers of $b$ and $c$. $b_1, b_2, b_3$ can either be all distinct, all equal, or two of the three are equal. As well, we must have $b_1 \leq b_2 \leq b_3$. If they are all distinct, the number of cases is simply ${19 \choose 3}$. If they are all equal, there are only $19$ cases for the general value. If we have a pair equal, then we have $2 \cdot {19\choose 2}$. We need to multiply by $2$ because if we have two values $b_i < b_j$, we can have either $(b_i, b_i, b_j)$ or $(b_i, b_j, b_j)$.

\[{19 \choose 3} + 2 \cdot {19 \choose 2} + 19 = 1330\]

Likewise for $c$, we get

\[{10 \choose 3} + 2 \cdot {10 \choose 2} + 10 = 220\]

The final probability is simply $\frac{1330 \cdot 220}{190^3}$. Simplification gives $\frac{77}{1805}$, and therefore the answer is $\boxed{077}$.

Solution 3

Similar to before, we calculate that there are $190^3$ ways to choose 3 factors with replacement. Then, we figure out the number of triplets {a,b,c} and {d,f,g}, where a, b, and c represent powers of 2 and d, f, and g represent powers of 5, such that the triplets are in non-descending order. The maximum power of 2 is 18, and the maximum power of 5 is 9. Using the Hockey Stick identity, we figure out that there are $\dbinom{12}{3}$ ways to choose d, f and g, and $\dbinom{21}{3}$ ways to choose a, b, and c. Therefore, the probability of choosing 3 factors which satisfy the conditions is \[\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.\] This simplifies to 77/1805, therefore m = $\boxed{077}$.

See Also

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