A number theory problem by Sumukh Bansal

Sum of the factors is using prime factorization:

756 = $2^2$ x $3^3$ x 7. Therefore, sum of all positive factors of 756 is calculated as follows:

( $2^0$ + $2^1$ + $2^2$ )( $3^0$ + $3^1$ + $3^2$ + $3^3$ )( $7^0$ + $7^1$ )=(1+2+4)(1+3+9+27)(1+7)=2240

By using sum of a geometric series ( $2^0$ + $2^1$ + $2^2$ )( $3^0$ + $3^1$ + $3^2$ + $3^3$ )( $7^0$ + $7^1$ )= ( $\frac{2^3-1}{2-1}$ )( $\frac{3^4-1}{3-1}$ )( $\frac{7^2-1}{7-1}$ )=( $\frac{7}{1}$ )( $\frac{80}{2}$ )( $\frac{48}{6}$ )=2240

(or)

Factors of 756 are 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 27, 28, 36, 42, 54, 63, 84, 108, 126, 189, 252, 378, 756.

Therefore, sum of all positive factors of 756 is 1+2+3+4+6+7+9+12+14+18+21+27+28+36+42+54+63+84+108+126+189+252+378+756=2240