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
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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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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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can you please show the exact repeated steps please?

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 ?

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

you can write it down
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Well said. Fortunately, There is no Easy way to write something long here. Make it easy after-all.
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It is quite easy if you do know how to format a question and the tag usage.

Regards

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

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Read through all answers provided yet no lead

Posted from my mobile device

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 1236.

Therefore answer is A,C

S is the set of values of all positive values which satisfy the conditions. consider n value as 216,then square of 216 is multiple of 24 and 108,so if n is 216 then the remaining two options should also be the correct answers.

Am I doing anything wrong here?

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S is the set of values of all positive values which satisfy the conditions.

We need to consider all the values for which the answers will be A & C only.
You can verify by taking \(n = 108\)


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