If a^2 is divisible by 240 what is the least possible value of integer a?	Choose the correct choice from the following choices.
A) 120
B) 30
C) 90
D) 60
E) 12
240 can be written as (2^4)*3*5.   For a^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 a^2 is divisible by 240 what is the least possible value of integer a?