If x^2 is divisible by 240 what is the least possible value of integer x?  Choices:
A) 120
B) 30
C) 90
D) 60
E) 12
240 can be written as (2^4)*3*5.   For x^2 to be divisible by 240 it should contain at least 2^4 and 3 and 5 in its factors. We can leave out option E because 12 doesnt have 5 as one of its factor. Now if we check for option B, 30 can be written as 2*3*5, hence 30^ 2 will have 2 as the maximum power of 2, so we can leave out this option too.   Option D is the right answer if we follow the same method as we followed for other two previous options.   60 = (2^2)*3*5; 60^2 = (2^4)*(3^2)*(5^2). So it shows that 60^2 is divisible by 240 and hence the answer.   Answer : 60
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If x^2 is divisible by 240 what is the least possible value of integer x?