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S be the set of all positive integers n such that n^2
[#permalink]
04 Jan 2016, 16:45

4

Expert Reply

4

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Question Stats:

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

_________________

Indicate all such integers.

A. 12

B. 24

C. 36

D. 72

Practice Questions

Question: 14

Page: 159

Difficulty: hard

Question: 14

Page: 159

Difficulty: hard

Show: :: OA

A,C

_________________

Sandy

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Re: S be the set of all positive integers n such that n^2
[#permalink]
11 Nov 2019, 10:42

6

Expert Reply

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

Answer: A,C

Cheers,

Brent

_________________

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

Answer: A,C

Cheers,

Brent

_________________

Re: S be the set of all positive integers n such that n^2
[#permalink]
07 Nov 2019, 02:12

8

Asmakan wrote:

rapsjade 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]
04 Jan 2016, 18:48

4

Expert Reply

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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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]
29 Mar 2016, 09:38

can you please show the exact repeated steps please?

Re: S be the set of all positive integers n such that n^2
[#permalink]
15 May 2016, 21:09

1

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.

Re: S be the set of all positive integers n such that n^2
[#permalink]
07 Nov 2019, 00:49

rapsjade 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 ?

Re: S be the set of all positive integers n such that n^2
[#permalink]
07 Nov 2019, 03:21

Official explanation.

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Re: S be the set of all positive integers n such that n^2
[#permalink]
07 Nov 2019, 03:56

2

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]
07 Nov 2019, 04:27

1

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]
11 Nov 2019, 01:22

Expert Reply

It is quite easy if you do know how to format a question and the tag usage.

Regards

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Regards

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Re: S be the set of all positive integers n such that n^2
[#permalink]
14 Nov 2019, 09:08

1

I don't understand this one bit why so difficult.

Posted from my mobile device

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

1

Read through all answers provided yet no lead

Posted from my mobile device

Posted from my mobile device

Re: S be the set of all positive integers n such that n^2
[#permalink]
18 Jan 2021, 19:04

1

Consider this approach:

LCM of 24 and 108 is 216.

Since n^2 is a multiple of both of 24 and 108, n^2 should be a multiple of 216.

Now since 216 is not a perfect square, 216 is expanded as 36 x 6,

Now (36 * 6 * x) = n^2

n = sqrt(36 * 6 * n)

We find that for n^2 to be perfect square and therefore have an integer square root, then the value of x should be such that 6*x is a perfect square (because 36 is already a perfect square). In this case the smallest value is 6.

n2 = 36 x 6 x 6

n = 36

Therefore smallest value of n possible is 36 which is a multiple of 12,36.

LCM of 24 and 108 is 216.

Since n^2 is a multiple of both of 24 and 108, n^2 should be a multiple of 216.

Now since 216 is not a perfect square, 216 is expanded as 36 x 6,

Now (36 * 6 * x) = n^2

n = sqrt(36 * 6 * n)

We find that for n^2 to be perfect square and therefore have an integer square root, then the value of x should be such that 6*x is a perfect square (because 36 is already a perfect square). In this case the smallest value is 6.

n2 = 36 x 6 x 6

n = 36

Therefore smallest value of n possible is 36 which is a multiple of 12,36.

gmatclubot

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