Question

Let "k" be the number of factors (excluding $1$ and the expression itself) of the number 38808 and "m" be the sum of its divisors. Let "h" be the remainder of m/k. Find the sum of the digits of "h" ?

• A. $1$
• B. $0$
• C. $2$
• D. $5$

Answer 1

The correct option is A $1$

Part 1: The number $38808=2^3{.3}^2{.7}^2.11$
Hence the total number of divisors (excluding $1$ and itself i.e. 38808)
$=(3+1)(2+1)(2+1)(1+1)-2=70$
Part 2: The sum of these divisors
$=(2^0+2^1+2^2+2^3)(3^0+3^1+3^2)(7^0+7^1+7^2)({11}^0+{11}^1)=\left(15\right)\left(13\right)\left(57\right)\left(12\right)-1-38808=133308-1-38808=94571$
remainder is 1