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:
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Answers
Answered by
6
Answer:
The correct option is E.
Explanation:
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
Answered by
0
Answer:
The correct option is B.
Explanation:
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