Find the sum of all odd and even divisors of 3600. $6300=2^2.3^2.5^2.7$
The divisors are of the form $2^a.3^b.5^c.7^d$ Now sum of it's all divisors (odd and even) given by $(1+2+2^2)(1+3+3^2)(1+5+5^2)(1+7)$--------------(i) Since $3,5,7$ are odd the value of a divisor be odd or even solely depends on power of $2$ The value of sum of odd integers are from (i) $(1)(1+3+3^2)(1+5+5^2)(1+7)$=$3224$ But it is WRONG Also for summation of even integers from (i) $(2+2^2)(1+3+3^2)(1+5+5^2)(1+7)$=$19344$ It's also WRONG What am I doing wrong again and again?
The sum off odd divisors should be $403$ and of even integers should be $12090$
Your prime factorisation of 3600 is wrong. It should rather be : 3600=2^4*3^2*5^2
Hence the sum of all odd divisors =(1)(1+3+3^2)(1+5+5^2)=403
The sum of all even divisors=(2+2^2+2^3+2^4)(1+3+3^2)(1+5+5^2)=12090
The sum of all divisors =(1+2+2^2+2^3+2^4)(1+3+3^2)(1+5+5^2)=12493=12090+403 =Sum of odd divisors +sum of even divisors