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

Question

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

Bunuel · Accepted Answer

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.

Bunuel · Answer

Similar questions to practice: if-n-is-a-positive-integer-and-n-2-is-divisible-by-96-then-127364.html if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-90523.html n-is-a-positive-integer-and-k-is-the-product-of-all-integer-104272.html if-n-and-y-are-positive-integers-and-450y-n-92562.html if-m-and-n-are-positive-integer-and-1800m-n3-what-is-108985.html if-x-and-y-are-positive-integers-and-180x-y-100413.html if-x-is-a-positive-integer-and-x-2-is-divisible-by-32-then-88388.html if-5400mn-k-4-where-m-n-and-k-are-positive-integers-109284.html if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-129929.html Hope it helps.

xcusemeplz2009 · Answer

Imo C

32=2^5
nearest squre =2^6=8^2
hence x=8 and the lagest int dividing 8 is 8

prasadrg · Answer

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.

prasadrg · Answer

Thank you and it makes sense.

samsmalldog · Answer

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.

parul1591 · Answer

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 ?

anujbansal · Answer

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.

Pretz · Answer

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.

EMPOWERgmatRichC · Answer

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":     = (12)(12)(12)

GMAT assassins aren't born, they're made,
Rich

Holzbesen · Answer

Bunuel  
If (F) = 128 was one of the possible answers, would that have been the correct choice?
Im trying to understand the concept behind this kind of exercise.

So 128 = 2^7
Because x^2 we have to get a perfect cube (with highest possible number).
If k = 2^7 we get (32 + 128) 2^12 or 4^3 * 4^3.

Is this logic correct?

Mandojay · Answer

maybe a dumb question: The largest positive integer that must divide x, means for lowest value of x which satisfies the given statement in the stem -- why?

Why is it not the largest value of x. For example, if x = 100 why cant x be 100 which will divide 100 and is the largest positive integer. What am I missing?

yashikaaggarwal · Answer

We don't need to get perfect cube but the perfect pair of square. 
Like in Question, X^2 is divisible by 32
The minimum value of x^2 can be 32 but 32 is not a perfect squre. 2^5 need another 2 to make it perfect square. So the least value of X^2 will be 64 {2^6 = (2^3)^2}

And if you say X^2 is divisible by 128 = 2^7
We need perfect square (every even power is one) so multiply 128 with 2
Giving 256 = (2^4)^2
So the maximum number by which x can be divided is 2^4 = 16.
Hope you get it.

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yashikaaggarwal · Answer

We can say that the largest divisor of a number is its least value (the least value of a prime number:2 and it's largest divisor is 2 only, but we can't say the maximum value part for such number which can change. The number has to be fixed to know it's divisor)

In question the least value Of X^2 is 64 (because 32,16,8,4,2 are either not divisible by 32 or is not a perfect square) 
That's why the least value of X is 8. Now 8 is divisible by 4 factors (1, 2,4&8) where 8 is maximum. 
It's not the largest value of X for which we want the divisor because the largest value of x^2 can extend to infinity which won't have a single value.

In your example 100 is divisible by 100 but the 100 must be because of some constraint. Let say x^2 is divisible by 1000 so x^2 has to be a multiple of 1000, we also know x^2 is a perfect square while 1000 is not that's why we will multiply 1000 with 10
Giving 10,000 as perfect square = X^2
And X = 100
Whose maximum divisor is 100.

Posted from my mobile device

gmatpro15 · Answer

The largest positive integer that must divide , means for lowest value of which satisfies the given statement in the stem.
x where is an integer (as is positive).

--> , as is an integer , also must be an integer. The lowest value of , for which is an integer is when --> -->

Posted from my mobile device

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

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