If N = (11^(p+7)) (7^(q-2)) (5^(r+1)) (3^s) is a perfect cube

• Numbers

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:

1. 5
2. 6
3. 7
4. 8
5. 9

Answer

For N to be perfect cube, the powers must be multiples of 3.

As p, q, r and s are positive integers; and we need to find smallest value of p + q + r + s,

p = 2, q = 2, r = 2 and s = 3

p + q + r + s = 9

The correct option is E.