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TAGS: Retired Moderator Joined: 07 Jun 2014
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S be the set of all positive integers n such that n^2 [#permalink]
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Expert's post 00:00

Question Stats: 48% (02:15) correct 51% (02:14) wrong based on 94 sessions
Let S be the set of all positive integers n such that $$n^2$$ is a multiple of both 24 and 108. Which of the following integers are divisors of every integer n in S ?

Indicate all such integers.

A. 12
B. 24
C. 36
D. 72

Practice Questions
Question: 14
Page: 159
Difficulty: hard

[Reveal] Spoiler: OA
A,C

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Sandy
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Try our free Online GRE Test Retired Moderator Joined: 07 Jun 2014
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Re: S be the set of all positive integers n such that n^2 [#permalink]
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Expert's post
Now here we are looking for a Set S of numbers n such that $$n^2$$ is multiple of 24 and 108.

Lets take one such case Least Common Multiple of 108 and 24 is 216.

$$n^2$$ = 216
n = 6$$\sqrt{6}$$.

This is not an integer however if we multiply 216 by any number it is still divisible by 24 and 108. So we multiply by 6 and repat the above steps to find n. So n =36.

Now n= 36 is one of the number in set S.

Hence all the options that divide n =36 are correct. Hence option A and C.
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Re: S be the set of all positive integers n such that n^2 [#permalink] Manager Joined: 23 Jan 2016
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Re: S be the set of all positive integers n such that n^2 [#permalink]
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LCM of 108 and 24 is 216 = 2^3*3^3. As n^2 is a square, the least value n^2 can take is 2^4*3^4 = 6*216 = 36 * 36, so n should be at least 36 or multiples of 36. So any factor of 36 should always divide n. Hence A and C are correct.
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Re: S be the set of all positive integers n such that n^2 [#permalink]
LCM of 108 and 24 is 216 = 2^3*3^3. As n^2 is a square, the least value n^2 can take is 2^4*3^4 = 6*216 = 36 * 36, so n should be at least 36 or multiples of 36. So any factor of 36 should always divide n. Hence A and C are correct.

Though I understand what you wrote but I need clearer explanation and a step by step pls. Now, after we find out LCM 216 and it is not a square of any number what do we do ? VP Joined: 20 Apr 2016
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Re: S be the set of all positive integers n such that n^2 [#permalink]
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Asmakan wrote:
LCM of 108 and 24 is 216 = 2^3*3^3. As n^2 is a square, the least value n^2 can take is 2^4*3^4 = 6*216 = 36 * 36, so n should be at least 36 or multiples of 36. So any factor of 36 should always divide n. Hence A and C are correct.

Though I understand what you wrote but I need clearer explanation and a step by step pls. Now, after we find out LCM 216 and it is not a square of any number what do we do ?

don't get confused with LCM and GCF, use ur own logic

Back to ques:

$$24 = 2^3 * 3$$

$$108 = 2^2 * 3^3$$

For $$2$$, we have maximum value = $$2^3$$

For $$3$$, we have maximum value =$$3^3$$

We need the value of n in such a way that squaring n will contains both maximum value of $$2$$(i.e $$2^3$$) and$$3 (i.e 3^3)$$

But, squaring a number won't raise the power to $$3$$

so the minimum value should be $$2^2$$ and squaring this will be = $$2^4$$

Similarly for $$3$$, the minimum value =$$3^2$$and squaring will be = $$3^4$$

therefore, the minimum value = $$2^2 * 3^2 = 36$$. This is divisible by $$36$$ and $$12$$
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Re: S be the set of all positive integers n such that n^2 [#permalink] VP Joined: 20 Apr 2016
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Re: S be the set of all positive integers n such that n^2 [#permalink]
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huda wrote:
Official explanation.

PLZ, don't attach screenshot of official explanation. This is not entertained in this forum.

you can write it down
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Re: S be the set of all positive integers n such that n^2 [#permalink]
pranab01 wrote:
huda wrote:
Official explanation.

PLZ, don't attach screenshot of official explanation. This is not entertained in this forum.

you can write it down

Well said. Fortunately, There is no Easy way to write something long here. Make it easy after-all.
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Re: S be the set of all positive integers n such that n^2 [#permalink]
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Re: S be the set of all positive integers n such that n^2 [#permalink]
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Expert's post
GIVEN: n² is a multiple of 24
24 = (2)(2)(2)(3)
So, n² must have at least three 2's and one 3 in its prime factorization.
What does this tell us about n?
Since n² = (n)(n), we can conclude that n must have at least two 2's and one 3 in its prime factorization.

GIVEN: n² is a multiple of 108
108 = (2)(2)(3)(3)(3)
So, n² must have at least two 2's and three 3's in its prime factorization.
What does this tell us about n?
Since n² = (n)(n), we can conclude that n must have at least one 2 and two 3's in its prime factorization.

So, when we combine both pieces of information, we can see that n must have at least two 2's and two 3's in its prime factorization.
In other words, n = (2)(2)(3)(3)(k), where k is some positive integer

Now let's check the answer choices...
A. 12 = (2)(2)(3)
Since n = (2)(2)(3)(3)(k), we can see that n IS divisible by 12

B. 24 = (2)(2)(2)(3)
Since n = (2)(2)(3)(3)(k), we can see that n need NOT be divisible by 24

C. 36 = (2)(2)(3)(3)
Since n = (2)(2)(3)(3)(k), we can see that n IS divisible by 36

D. 72 = (2)(2)(2)(2)(3)
Since n = (2)(2)(3)(3)(k), we can see that n need NOT be divisible by 72

Cheers,
Brent
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Re: S be the set of all positive integers n such that n^2 [#permalink]
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I don't understand this one bit why so difficult.

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Re: S be the set of all positive integers n such that n^2 [#permalink]

Posted from my mobile device  Re: S be the set of all positive integers n such that n^2   [#permalink] 14 Nov 2019, 09:09
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