Suppose n=2^a3^b 2sigma(n) = (2^(a+1)-1)(3^(b+1)-1) But sigma(n) = 3n, therefore => (2^(a+1)-1)(3^(b+1)-1) = 2^(a+1).3^(b+1) By hcf arguments, therefore, only possible if: 2^(a+1) = 3^(b+1) - 1, and simultaneously, 3^(b+1) = 2^(a+1) - 1. Which latter clearly not possible.