2016 AIME II Problems/Problem 8

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Find the number of sets ${a,b,c}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,61$.

Solution

Note that the prime factorization of the product is $3^{2}\cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$. Ignoring overcounting, by stars and bars there are $6$ ways to choose how to distribute the factors of $3$, and $3$ ways to distribute the factors of the other primes, so we have $3^{6} \cdot 6$ ways. However, some sets have $2$ numbers that are the same, namely the ones in the form $1,1,x$ and $3,3,x$, which are each counted $3$ times, and each other set is counted $6$ times, so the desired answer is $\dfrac{729 \cdot 6-6}{6} = \boxed{728}$.

Solution by Shaddoll

See also

2016 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AIME Problems and Solutions
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