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

Solution: A, C

### Explanation

It is given that n

^{2}is a multiple of both 24 and 108. So the least such n^{2}will be nothing but the LCM of 24 and 108.24 – 2

^{3}*3108 – 2

^{2}*3^{3}Thus the LCM is 2

^{3}*3^{3}However, according to this question, since n should be an integer, n

Hence, the least possible multiple of n

^{2}should be a perfect square.Hence, the least possible multiple of n

^{2}is actually not the LCM but 2^{4}*3^{4}and for which n happens to be 2^{2}*3^{2}= 36.Now since S is a set of all such integers n, S = {36, 72, 108, ……………… }.

We have to find out all the integers among the given choices that are divisors of each of the elements in S.

Because 36 is the least element in the set, 36 and all its factors would be divisors of 36.

And if these numbers are divisors of 36, then they are divisors of all multiples of 36 as well.

The factors of 36 are {36, 18, 12, 9, 6, 4, 3, 2, 1}.

Only 12 and 36 are among the choices.

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