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Re: Division & Factor [#permalink]
24 Dec 2009, 06:21

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prasadrg wrote:

Thank you for solving. However I am confused why not x is divisible by16.

if x is 16 then x^2 is 256 which is divisible by 32. Appreciate your help.

The largest positive integer that must divide x, means for lowest value of x which satisfies the given statement in the stem.

Given: 32k=x^2, where k is an integer \geq1 (as x is positive).

32k=x^2 --> x=4\sqrt{2k}, as x is an integer \sqrt{2k}, also must be an integer. The lowest value of k, for which \sqrt{2k} is an integer is when k=2 --> \sqrt{2k}=\sqrt{4}=2 --> x=4\sqrt{2k}=4*2=8

Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]
29 Oct 2013, 16:09

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Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]
05 Jul 2014, 06:15

Expert's post

parul1591 wrote:

Hello Bunuel

"The largest positive integer that must divide x, means for lowest value of x which satisfies the given statement in the stem."

Why are we looking for the smallest value of x ?

Because we need the largest positive integer that MUST divide x. So, we should find the least value of x for which x^2 is divisible by 32, and if that x is divisible by some number then so will be every other x's (MUST condition will be satisfied). The least positive x for which x^2 is divisible by 32 is 8 (8^2 = 64, which is divisible by 32). So, even if x = 8 it is divisible by 8 but not divisible by 6, 12 or 16.

Check similar problems in my post above. _________________