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# If n is a positive integer and n^2 is divisible by 96, then

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If n is a positive integer and n^2 is divisible by 96, then  [#permalink]

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10 Feb 2012, 16:42
8
30
00:00

Difficulty:

45% (medium)

Question Stats:

63% (01:39) correct 37% (01:58) wrong based on 921 sessions

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If n is a positive integer and n^2 is divisible by 96, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

This is how I am trying to solve. The prime factorization of 96 is ($$2^5$$)(3), so $$n^2$$ must share those
prime factors.

Lets go straight into answer choices

A --> Is out as 6^6 = 36 is not divisible by 96

B --> 12 = $$2^2$$ * 3. Not divisible by 96 because 12 has $$2^2$$ * 3 as its prime factors. We need $$2^5$$ or more.

C --> 24 . Prime factors are $$2^3$$ * 3. Therefore $$n^2$$ = {$$2^3$$ * 3}$$^2$$ = $$2^6$$ * $$3^2$$. OK its true.

D --> 36 . Prime factors are $$2^4$$ * $$3^4$$. We need $$2^5$$ or more.

E --> 96. Prime factors are $$2^5$$ * 3. So n ^2 is divisible by 96 and also by 48. So this can also be true??

Any thoughts guys?

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Re: largest positive integer that must divide n  [#permalink]

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10 Feb 2012, 16:57
8
23
enigma123 wrote:
If n is a positive integer and n^2 is divisible by 96, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

Given: $$n^2=96*k=2^5*3*k$$, where $$k$$ is some positive integer. $$n=\sqrt{2^5*3*k}=2^2*\sqrt{2*3*k}$$ --> the least value of $$k$$ for which $$n$$ is an integer (hence the least value of $$n$$) is for $$k=2*3$$ --> $$n=2^2*\sqrt{3*2*3*2}=24$$, hence $$n$$ in any case is divisible by 24 (24 is the largest positive integer that MUST divide n evenly).

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if-m-and-n-are-positive-integer-and-1800m-n3-what-is-108985.html
property-of-integers-104272.html
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Hope it helps.
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Re: largest positive integer that must divide n  [#permalink]

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10 Feb 2012, 17:02
enigma123 wrote:
If n is a positive integer and n^2 is divisible by 96, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

This is how I am trying to solve. The prime factorization of 96 is ($$2^5$$)(3), so $$n^2$$ must share those
prime factors.

Lets go straight into answer choices

A --> Is out as 6^6 = 36 is not divisible by 96

B --> 12 = $$2^2$$ * 3. Not divisible by 96 because 12 has $$2^2$$ * 3 as its prime factors. We need $$2^5$$ or more.

C --> 24 . Prime factors are $$2^3$$ * 3. Therefore $$n^2$$ = {$$2^3$$ * 3}$$^2$$ = $$2^6$$ * $$3^2$$. OK its true.

D --> 36 . Prime factors are $$2^4$$ * $$3^4$$. We need $$2^5$$ or more.

E --> 96. Prime factors are $$2^5$$ * 3. So n ^2 is divisible by 96 and also by 48. So this can also be true??

Any thoughts guys?

I think you misinterpreted the question: the correct answer must divide any n, which satisfy the condition that n^2 is divisible by 96. So, we should find the least value of n for which n^2 is a multiple of 96.

Follow the links in my previous post for similar questions to understand the concept better.
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Re: largest positive integer that must divide n  [#permalink]

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19 Feb 2012, 04:32
Bunuel wrote:
enigma123 wrote:
If n is a positive integer and n^2 is divisible by 96, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

Given: $$n^2=96*k=2^5*3*k$$, where $$k$$ is some positive integer. $$n=\sqrt{2^5*3*k}=2^2*\sqrt{2*3*k}$$ --> the least value of $$k$$ for which $$n$$ is an integer (hence the least value of $$n$$) is for $$k=2*3$$ --> $$n=2^2*\sqrt{3*2*3*2}=24$$, hence $$n$$ in any case is divisible by 24 (24 is the largest positive integer that MUST divide n evenly).Answer: C.

Hope it helps.

hence $$n$$ in any case is divisible by 24 (24 is the largest positive integer that MUST divide n evenly).
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Posts: 51307
Re: largest positive integer that must divide n  [#permalink]

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19 Feb 2012, 11:46
docabuzar wrote:
Bunuel wrote:
enigma123 wrote:
If n is a positive integer and n^2 is divisible by 96, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

Given: $$n^2=96*k=2^5*3*k$$, where $$k$$ is some positive integer. $$n=\sqrt{2^5*3*k}=2^2*\sqrt{2*3*k}$$ --> the least value of $$k$$ for which $$n$$ is an integer (hence the least value of $$n$$) is for $$k=2*3$$ --> $$n=2^2*\sqrt{3*2*3*2}=24$$, hence $$n$$ in any case is divisible by 24 (24 is the largest positive integer that MUST divide n evenly).Answer: C.

Hope it helps.

hence $$n$$ in any case is divisible by 24 (24 is the largest positive integer that MUST divide n evenly).

The question asks about the largest positive integer that must divide n, since the lowest value of n is 24, then the largest positive integer that must divide n is 24 (so even for the lowest value of n it's divisible by 24).

Hope it's clear.
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Re: largest positive integer that must divide n  [#permalink]

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19 Feb 2012, 23:33
[/quote]

hence $$n$$ in any case is divisible by 24 (24 is the largest positive integer that MUST divide n evenly).[/quote]

The question asks about the largest positive integer that must divide n, since the lowest value of n is 24, then the largest positive integer that must divide n is 24 (so even for the lowest value of n it's divisible by 24).

Hope it's clear.[/quote]

Thanks for the explanation.

I m just repeating in my words what I understand that since the lowest value for n is 24, & all values of n are multiples of 24, therefore, 24 is the largest number that MUST divide any value of n.
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Re: largest positive integer that must divide n  [#permalink]

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20 Feb 2012, 00:50
5
1
enigma123 wrote:
If n is a positive integer and $$n^2$$ is divisible by 96, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

This is how I am trying to solve. The prime factorization of 96 is ($$2^5$$)(3), so $$n^2$$ must share those
prime factors.

Lets go straight into answer choices

A --> Is out as 6^6 = 36 is not divisible by 96

B --> 12 = $$2^2$$ * 3. Not divisible by 96 because 12 has $$2^2$$ * 3 as its prime factors. We need $$2^5$$ or more.

C --> 24 . Prime factors are $$2^3$$ * 3. Therefore $$n^2$$ = {$$2^3$$ * 3}$$^2$$ = $$2^6$$ * $$3^2$$. OK its true.

D --> 36 . Prime factors are $$2^4$$ * $$3^4$$. We need $$2^5$$ or more.

E --> 96. Prime factors are $$2^5$$ * 3. So n ^2 is divisible by 96 and also by 48. So this can also be true??

Any thoughts guys?

Let me tell you why your method didn't work.
"The prime factorization of 96 is ($$2^5$$)(3), so $$n^2$$ must share those
prime factors."

You are right. n^2 must share those prime factors. But remember, it is a square so its prime factors have even powers. So n^2 must have 2^6*3^2. We know n^2 has one 3. It must have another 3 since it is a square. Similarly, we know n^2 has 2^5. It must have another 2 to make 2^6 so that the power of 2 is even.
n^2 = 2^6 * 3^2 (at least)
n = 2^3 * 3 = 24 (atleast)
Therefore, n is definitely divisible by 24 at least.
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Re: If n is a positive integer and n^2 is divisible by 96, then  [#permalink]

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26 Jun 2013, 01:24
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Divisibility/Multiples/Factors questions to practice: search.php?search_id=tag&tag_id=354
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Re: If n is a positive integer and n^2 is divisible by 96, then  [#permalink]

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16 Mar 2016, 00:29
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Re: If n is a positive integer and n^2 is divisible by 96, then  [#permalink]

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25 Jan 2017, 08:40
1
2
1) Since n is squared, it should have two identical sets of prime factors
2) Since n^2 is divisible by 96, all prime factors of 96 should be the prime numbers of n^2.
3) The prime factors of 96 are 2*2*2*2*2*3. A squared number that includes all these prime factors should have an additional factor 2 and an additional factor 3.
4) n^2 has to be at least 2^6*3^2, and n has to be 2^3*3=24
5) If n has to be at least 24, then the largest integer it MUST be divisible by is 24.

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Re: If n is a positive integer and n^2 is divisible by 96, then  [#permalink]

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18 Feb 2018, 03:28
It is given that $$n^2$$ is divisible by 96. We can write it as -
$$n^2=96*m$$
$$n^2=2^5*3*m$$

Also it is given that n is an integer, so the RHS of above expression should be a perfect square. So that m should be 2*3, which gives the value of $$n^2$$ as 576 and thus the value of n as 24. So the largest positive integer that can divide n is 24.

Note that m can also have the value of $$2^3*3$$ to make the RHS a perfect square again, which gives the value of $$n^2$$ as 2304 and thus the value of n as 48, which is also an option. But 48 will only be able to divide n when n = 48, it won't be able to divide n when n =24, which is also a feasible value of n. Thus 24 is the correct option.
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Re: If n is a positive integer and n^2 is divisible by 96, then  [#permalink]

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14 Apr 2018, 08:33
+1 for option C.
n^2=96k=16*6*k
k has to contain 6.
n is a multiple of 4*6 or 24.
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If n is a positive integer and n^2 is divisible by 96, then  [#permalink]

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11 May 2018, 10:41
Bunuel wrote:
enigma123 wrote:
If n is a positive integer and n^2 is divisible by 96, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

Given: $$n^2=96*k=2^5*3*k$$, where $$k$$ is some positive integer. $$n=\sqrt{2^5*3*k}=2^2*\sqrt{2*3*k}$$ --> the least value of $$k$$ for which $$n$$ is an integer (hence the least value of $$n$$) is for $$k=2*3$$ --> $$n=2^2*\sqrt{3*2*3*2}=24$$, hence $$n$$ in any case is divisible by 24 (24 is the largest positive integer that MUST divide n evenly).

Similar questions to practice:
http://gmatclub.com/forum/properties-of ... 90523.html
http://gmatclub.com/forum/if-m-and-n-ar ... 08985.html
http://gmatclub.com/forum/property-of-i ... 04272.html
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http://gmatclub.com/forum/division-factor-88388.html
http://gmatclub.com/forum/if-5400mn-k4- ... 09284.html

Hope it helps.

Bunuel hello, i have two questions

how from this $$n=\sqrt{2^5*3*k}$$= how you got this i mean $$2^2$$ outside of radical sign wheras only one 2 is left under radical sign, but we have five 2s 2^2*$$\sqrt{2*3*k}$$

here- $$n=2^2*\sqrt{3*2*3*2}=24$$ --- but 3*2*3*2 = 36 how can 3*2*3*2 be equal 24
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Re: If n is a positive integer and n^2 is divisible by 96, then  [#permalink]

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11 May 2018, 12:07
1
dave13 wrote:
Bunuel wrote:
enigma123 wrote:
If n is a positive integer and n^2 is divisible by 96, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

Given: $$n^2=96*k=2^5*3*k$$, where $$k$$ is some positive integer. $$n=\sqrt{2^5*3*k}=2^2*\sqrt{2*3*k}$$ --> the least value of $$k$$ for which $$n$$ is an integer (hence the least value of $$n$$) is for $$k=2*3$$ --> $$n=2^2*\sqrt{3*2*3*2}=24$$, hence $$n$$ in any case is divisible by 24 (24 is the largest positive integer that MUST divide n evenly).

Similar questions to practice:
http://gmatclub.com/forum/properties-of ... 90523.html
http://gmatclub.com/forum/if-m-and-n-ar ... 08985.html
http://gmatclub.com/forum/property-of-i ... 04272.html
http://gmatclub.com/forum/if-x-and-y-ar ... 00413.html
http://gmatclub.com/forum/number-properties-92562.html
http://gmatclub.com/forum/can-someone-a ... 92066.html
http://gmatclub.com/forum/og-quantitative-91750.html
http://gmatclub.com/forum/division-factor-88388.html
http://gmatclub.com/forum/if-5400mn-k4- ... 09284.html

Hope it helps.

Bunuel hello, i have two questions

how from this $$n=\sqrt{2^5*3*k}$$= how you got this i mean $$2^2$$ outside of radical sign wheras only one 2 is left under radical sign, but we have five 2s 2^2*$$\sqrt{2*3*k}$$

here- $$n=2^2*\sqrt{3*2*3*2}=24$$ --- but 3*2*3*2 = 36 how can 3*2*3*2 be equal 24

1. $$n=\sqrt{2^5*3*k}=\sqrt{(2^4*2)*3*k}=\sqrt{((2^2)^2*2)*3*k}=2^2*\sqrt{2*3*k}$$.

2. $$n=2^2*\sqrt{3*2*3*2}=4*\sqrt{36}=4*6=24$$.

Hope it helps.
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Re: If n is a positive integer and n^2 is divisible by 96, then &nbs [#permalink] 11 May 2018, 12:07
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