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If x is a positive integer and x^2 is divisible by 32, then

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

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24 Dec 2009, 02:59
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Jun 2013, 00:46, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Division & Factor [#permalink]

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24 Dec 2009, 03:18
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Imo C

32=2^5
nearest squre =2^6=8^2
hence x=8 and the lagest int dividing 8 is 8
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Re: Division & Factor [#permalink]

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24 Dec 2009, 05:46
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.
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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 $$k=2$$ --> $$\sqrt{2k}=\sqrt{4}=2$$ --> $$x=4\sqrt{2k}=4*2=8$$

Hope it's helps.
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Re: Division & Factor [#permalink]

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24 Dec 2009, 08:25
Thank you and it makes sense.
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Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]

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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]

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09 Mar 2014, 02:39
Very interesting question.

Here is the similar problem link:
if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-129929-20.html#p1341433

I got one of the questions wrong and then got another right, I think some problem has been discussed in the above link on the first page.

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

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09 Mar 2014, 02:50
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Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]

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02 Jul 2014, 22:11
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 ?
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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, 06:15
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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.
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Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]

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06 Jul 2014, 10:34
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Simply put x = 8 and get the answer. If you put a number greater than 8, then you would have less options to eliminate and more to check.
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Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]

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07 Apr 2015, 20:44
Hi Bunnel,

What if the question is as below:

If P is a positive integer and p^3 is divisible by 144, then the largest positive integer that must divide p is

How do we calculate using the "k" method when the cubeth root is asked for.
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Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]

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08 Apr 2015, 22:22
Hi Pretz,

In the original question, we're essentially looking for the smallest multiple of 32 that is a perfect square....

32(2) = 64......which is 8^2.

Using prime factorization, we know that 32K = (2^5)K

By making K = 2, we have (2^5)(2) = 2^6

We can then break 2^6 into 2 equal "pieces": (2^3)(2^3) which equals (8)(8)

To answer your question, we're going to use a similar approach:

We're looking for the smallest multiple of 144 that is a perfect cube....

144K = (2^4)(3^2)K

By making K = 12, we have (2^4)(3^2)(2^2)(3) = (2^6)(3^3)

We can break this down into 3 equal "pieces": [(2^2)(3)][(2^2)(3)][(2^2)(3)] = (12)(12)(12)

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

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26 Oct 2015, 10:43
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?
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Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]

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14 Mar 2016, 00:40
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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Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]

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14 Mar 2016, 04:20
prasadrg wrote:
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

x^2 is a square number and divisible by 32. The first square number divisible by 32 is 64 = 2^6 or 8^2
Therefore, 8 will certainly divide x.
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Re: If x is a positive integer and x^2 is divisible by 32, then [#permalink]

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16 Mar 2016, 00:45
Mathivanan Palraj wrote:
prasadrg wrote:
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

x^2 is a square number and divisible by 32. The first square number divisible by 32 is 64 = 2^6 or 8^2
Therefore, 8 will certainly divide x.

I don't get this approach
what if x isnt 32 its 37488594
i think the correct approach is taking the prime factors into account
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Mock Test -2 (Evensand Odds Basic Quiz) --->http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1768023

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

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16 Mar 2016, 00:55
prasadrg wrote:
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

The first square number, which is divisible by 32 is 64. It implies x is 8.
Re: If x is a positive integer and x^2 is divisible by 32, then   [#permalink] 16 Mar 2016, 00:55
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