jmc

Prime factorize $48=2^4\cdot 3$. The sum of the the positive factors of 48 is $1+2+2^2+2^3+2^4+3+2\cdot 3+2^2\cdot 3+2^3\cdot 3+2^4\cdot 3=\boxed{124}$. Note that this also could be obtained from multiplying $(2^0+2^1+2^2+2^3+2^4)(3^0+3^1)=(31)(4)$, since expanding the left-hand side of this equation yields the left-hand side of the previous equation.