Xavier Aptitude Test 2017 Paper

Question : 59 of 97

Marks: +1, -0

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:

Solution:

(e) N =

\ (11^{p+7}\ ) (7^{q-2}\ )\ (5^{r+1}\ ) (3^{s} )

It N is a perfect cube. Then exponents of 11,7,5 and 3 would be a multiple 3 or 0.
So smallest value ofp + q + r + s = 2 + 2 + 2 + 3=9.

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