Question

The sum of the divisors of $2^5{.3}^4{.5}^2$ is

• A. $3^3\cdot 7^1\cdot {11}^2$
• B. $3^2\cdot 7^1\cdot {11}^2\cdot 31$
• C. $3\cdot 7\cdot 11\cdot 31$
• D. $3^2\cdot 7^3\cdot {11}^2\cdot 31$

Answer 1

The correct option is B $3^2\cdot 7^1\cdot {11}^2\cdot 31$

Any divisor of $2^5{.3}^4{.5}^2$ is of form $2^a{.3}^b{.5}^c$ where $0\le a\le 5,0\le b\le 4$ and $0\le c\le 2.$
Hence the sum of the divisors
$=\sum _{a=0}^5\sum _{b=0}^4\sum _{c=0}^2{2}^a{3}^b{5}^c$
$=(1+2+..{.2}^5)(1+3+...+3^4)(1+5+5^2)$
$=\left(\frac{2^6-1}{2-1}\right)\left(\frac{3^5-1}{3-1}\right)\left(\frac{5^3-1}{5-1}\right)$
$=63\times 121\times 31=3^2{.7}^1{.11}^2.31$