GMATinsight wrote:

Find the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600?

A) 639

B) 739

C) 735

D) 651

E) 589

Firstly, yet to see a Q testing sum of factors!!

Also the choices don't seem to be as per GMAT, where they are always in ascending or descending order

But if you want to know the method..

if the number is \(p^a*t^b\) sum is \(p^0+p^1+....p^a)(t^0+t^1+......+t^b\)...

Here the two numbers are

1) 96=2^5*3...

Sum of even divisors= total - sum of odd factors..

Total =\((2^0+2^1+2^2+2^3+2^4+2^5)(3^0+3^1)=63*4=252\)

Sum of odd factors=\(3^0+3^1=4\)..

Sum of even divisors=252-4=248

2) odd divisors of 3600..

3600=\(2^4*3^2*5^2\)..

Odd divisors sum = \((3^0+3^1+3^2)(5^0+5^1+5^2)=13*31=403\)..

Sum = 248+403=651..

D

Small doubt..

while calculating for 96.. you took even divisors = total - odd

But while calculating for 3600... you directly took odd divisors.