Question

If \(N=(11^{\,p+7})(7^{\,q-2})(5^{\,r+1})(3^{\,s})\) is a perfect cube, where \(p,q,r,s\) are positive integers, then the smallest value of \(p+q+r+s\) is:

• A. 5
• B. 6
• C. 7
• D. 8
• E. 9

Hint:

For “perfect power” questions, set each prime’s exponent to the nearest \emph{nonnegative} multiple of the target power (here, \(3\)). Zero exponents are allowed and often minimize the sum.

Answer 1

Answer: (E)

For \(N\) to be a perfect cube, every prime’s exponent must be a multiple of \(3\).
\(\bullet\) For \(11^{\,p+7}\): \(p+7 \equiv 0 \pmod{3}\Rightarrow p\equiv -7\equiv 2 \pmod{3}\). Smallest positive \(p=2\).
\(\bullet\) For \(7^{\,q-2}\): \(q-2 \equiv 0 \pmod{3}\Rightarrow q\equiv 2 \pmod{3}\). Smallest positive \(q=2\) (gives exponent \(0\), which is a multiple of \(3\)).
\(\bullet\) For \(5^{\,r+1}\): \(r+1 \equiv 0 \pmod{3}\Rightarrow r\equiv 2 \pmod{3}\). Smallest positive \(r=2\).
\(\bullet\) For \(3^{\,s}\): \(s\equiv 0 \pmod{3}\). Smallest positive \(s=3\).
Hence the minimum sum is \[ p+q+r+s=2+2+2+3=9. \]