# 1985 AHSME Problems/Problem 12

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## Problem

Let $p, q$ and $r$ be distinct prime numbers, where $1$ is not considered a prime. Which of the following is the smallest positive perfect cube having $n=pq^2r^4$ as a divisor? $\mathrm{(A)\ } p^8q^8r^8 \qquad \mathrm{(B) \ }(pq^2r^2)^3 \qquad \mathrm{(C) \ } (p^2q^2r^2)^3 \qquad \mathrm{(D) \ } (pqr^2)^3 \qquad \mathrm{(E) \ }4p^3q^3r^3$

## Solution

For a number of the form $p^aq^br^c$ to be a perfect cube and a multiple of $pq^2r^4$, $a, b,$ and $c$ must all be multiples of $3$, $a\ge1$, $b\ge2$, and $c\ge4$. The smallest multiple of $3$ greater than $1$ is $3$, the smallest multiple of $3$ greater than $2$ is $3$, and the smallest multiple of $3$ greater than $4$ is $6$. Therefore, the smallest such $p^aq^br^c$ is $p^3q^3r^6=(pqr^2)^3, \boxed{\text{D}}$.

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