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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]

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29 Oct 2013, 17:09

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

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05 Jul 2014, 07:15

Expert's post

1

This post was BOOKMARKED

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. _________________

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

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26 Oct 2015, 11:43

Bunuel wrote:

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 \(k=2\) --> \(\sqrt{2k}=\sqrt{4}=2\) --> \(x=4\sqrt{2k}=4*2=8\)

Hope it's helps.

so if 18 was on the list would that also satisfy the statement?

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

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14 Mar 2016, 01:40

Excellent Question... here the same old rule applies that x and x^n have the same number of prime factors.. hence 2 is the prime factor of x. now x^2/32=integer so x=8 is the least value hence 8 can be the largest integer that divides 8 hence C _________________

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