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 n2 is a multiple of both 24 and 108. So the
least such n2 will be nothing but the LCM of 24 and 108.
24 – 23*3
108 – 22*33
Thus the LCM is 23*33
However, according to this question, since n should be an integer, n2
should be a perfect square. Hence, the least possible multiple of n2
is actually not the LCM but 24*34 and for which n happens
to be 22*32 = 36.
Now since S is a set of all such integers n, S = {36, 72, 108, ……………… }.
We are supposed to find out all the integers among the given choices that
are divisors of each of the elements in S. Since 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.