Problem Solving - 45

If p is the smallest positive integer such that (p^3)/3920 is also an integer, what is the sum of the digits of p?

(A) 5
(B) 7
(C) 9
(D) 11
(E) 13

Answer: A

m = p*p*p/ 3920
m = p*p*p/ 2*2*2*2*5*7*7
m = p*p*p/ [(4^2)*5*(7^2)]

Hence p must be equal to 4*5*7
thus p = 140
Hence sum of digits = 1+4+0 = 5