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XAT 2017 — QA & DI Question 15

FactorsEasy

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

Answer & solution

  • A

    5

  • B

    6

  • C

    7

  • D

    8

  • 9

Solution

In order for N to be a perfect cube, every prime number should be a power of multiple of 3 i.e., 0, 3 , 9 …

Also, we need to consider minimum possible values of p, q, r and s. (p, q, r and s > 0)

For 3s to be a perfect cube, minimum value of s = 3.

For 5r+1 to be a perfect cube, minimum value of r = 2.

For 7q-2 to be a perfect cube, minimum value of q = 2.

For 11p+7 to be a perfect cube, minimum value of p = 2. 

So, smallest value of p + q + r + s = 2 + 2 + 2 + 3 = 9

Hence, option (e).

XAT 2017 QA & DI Q15: 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 — Solution | TheCATExam