If x is a positive integer and x^2 is divisible by 32, then the largest positive integer that must divide x is
(A) 2
(B) 6
(C) 8
(D) 12
(E) 16
OA 8
(A) 2
(B) 6
(C) 8
(D) 12
(E) 16
OA 8
x^2 is divisible by 32.
(Since 32 = 2^5): x^2 is divisible by 2^5
if x = 2^3,(2^3)^2 should be divisible by 2^5 although we have one extra power of 2. but if we take x = 2^2 then (2^2)^2 or 2^4 = 16 will not divide 32.
therefore minimum x can be 8 and 8^2 = 64 and that divides 32.
Ans C
