GMATinsight wrote:

If x is a positive integers and x^3 is divisible by 48 then which of the following is the least number of factors that x may have?

A) 1

B) 3

C) 6

D) 12

E) greater than 12

http://www.GMATinsight.comx^3 is divisible by 48

48 = 2*2*2*2*3 = 2^4 * 3

So lets start pairing the factor in group of 3. (2*2*2)*2*3

So x must be having one factor 2 for (2*2*2) , one 2 for 2 and one 3 for 3.

So minimum value of x = 2^2 * 3^1

Least number of factors of x = (2+1)*(1+1) = 6

Answer C

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