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If x is a positive integer and x^2

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by srcc25anu » Mon Apr 22, 2013 5:03 pm
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
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by Anju@Gurome » Mon Apr 22, 2013 9:25 pm
guerrero wrote:If x is a positive integer and x^2 is divisible by 32, then the largest positive integer that must divide x is
Let us assume, x² = 32k = (2�)*k, where k is some positive integer.
We can see that, minimum possible value of k such that (2�)*k becomes a square of an integer is 2.
Hence, minimum possible value of x² is 2�.
So, minimum possible value of x is 2³ = 8.

The correct answer is C.

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