Question

Sum of proper divisors of the number $1260$ is

Answer 1

$1260=2^2\cdot 3^2\cdot 5\cdot 7$
Sum of divisors $=(2^0+2^1+2^2)(3^0+3^1+3^2)(5^0+5^1)(7^0+7^1)=7\times 13\times 6\times 8=4368$
Sum of proper divisors
$=4368-1-1260=3107$