60 * n = (some perfect square)

Factorize 60 to its prime factors.

2 | 60
2 | 30
3 | 15
5 | 5

60 = 2^2 * 3 *5

Inorder to make this a perfect square (even powers of prime factors) we need one more '3' and '5' to make it ( 2^2 * 3^2 * 5^2)

So, 60 should be multiplied 15 (3*5) to make it a perfect square. D

60*n = 2*2*3*5*n
for n to be least and still satisfying 60*n as a square of an interger, we just need one more 3 and one more 5.
hence n must 3*5=15
second least number would be = 2*2*3*5 = 60, etc....
Answer D.

Thucydides wrote:Hello fellow GMATers!

I look forward to learning and contributing.

I am having trouble with this question:

If 60*n is the square of an interger, what is the least possible value that n could have?

(A) 6
(B) 9
(C) 12
(D) 15
(E) 60

I have know idea where to start with this one.

60*n is the square of an integer ...

So plug in the values of n from the options given above...


1. 60*6 = 360 - Not a perfect square number...

2. 60*9 = 540 - Not a perfect square number....

3. 60*12 = 720 - Not a perfect square number...

4. 60*15 = 900 - A perfect square number of 30..


So the answer is option D , 15.

Thanks guys. It makes sense now w/ the explanation.