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?

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

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 $p^3q^3r^6=(pqr^2)^3, \boxed{\text{D}}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png