Question

Prove that the number of divisors of 8400 is 58 only (excluding the number itself and one). Also find the sum of the divisors.

Answer 1

$8400=4\times 100\times 21=18\times 25\times 3\times 7$
$=3^1{.2}^4{.5}^2{.7}^1=3^a{2}^b{5}^c{7}^d$
a varies from 0 to 1, b from 0 to 4, c from 0 to 2 and d again from 0 to 1.
Thus the total number of divisors is
= $2\times 5\times 3\times 2=60$
We have to exclude the divisor 1 and the divisor 8400. i.e. the number itself. Hence required number is $60-2=58$.
Sum of the divisors
Any divisor is of the form $3^a{2}^b{5}^c{7}^d$
Sum = ${\sum }_0^1{3}^a{\sum }_0^4{2}^b{\sum }_0^2{5}^c{\sum }_0^1{7}^d$
$=(1+3)(1+2+2^2+2^3+2^4)(1+5+5^2)(1+7)$
$=4\times 31\times 31\times 8$