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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
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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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Re: Division & Factor [#permalink]
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24 Dec 2009, 06:21
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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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09 Mar 2014, 02:39
Very interesting question. Here is the similar problem link: ifnisapositiveintegerandn2isdivisibleby72then12992920.html#p1341433I 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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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) GMAT assassins aren't born, they're made, Rich
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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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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.



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