Question

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

Accepted Answer

E. For $N$ to be a perfect cube, exponents must be multiples of 3. 1) $p + 7$: Smallest positive integer $p \ge 1$ giving a multiple of 3 is $p = 2$ (so $p+7 = 9$). 2) $q - 2$: Smallest positive integer $q \ge 1$ giving a multiple of 3 (or 0) is $q = 2$ (so $q-2 = 0$). 3) $r + 1$: Smallest positive integer $r \ge 1$ giving a multiple of 3 is $r = 2$ (so $r+1 = 3$). 4) $s$: Smallest positive integer $s \ge 1$ giving a multiple of 3 is $s = 3$. Sum $p + q + r + s = 2 + 2 + 2 + 3 = 9$.

Smallest Exponent Sum for Perfect Cube

Explanation

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