# 1985 AHSME Problems/Problem 12

## 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?

$\mathrm{(A)\ } p^8q^8r^8 \qquad \mathrm{(B) }\left(pq^2r^2\right)^3 \qquad \mathrm{(C) } \left(p^2q^2r^2\right)^3 \qquad \mathrm{(D) } \left(pqr^2\right)^3 \qquad \mathrm{(E) \ }4p^3q^3r^3$

## Solution

A number of the form $p^aq^br^c$ will be a perfect cube precisely when $a$, $b$, and $c$ are multiples of 3. (Clearly, since we are looking for the smallest possible perfect cube, we can assume that it has no other prime factors.) Furthermore, for it to be a multiple of $pq^2r^4$, we must have $a \geq 1$, $b \geq 2$, and $c \geq 4$. The smallest multiple of $3$ that is at least $1$ is $3$, the smallest that is at least $2$ is again $3$, and the smallest that is at least $4$ is $6$. Hence the smallest possible perfect cube with $n$ as a divisor is $p^3q^3r^6 = \boxed{\text{(D)} \left(pqr^2\right)^3}$.