Question

The sum of the divisors of $2^5\cdot 3^7\cdot 5^3\cdot 7^2$ is

• A. $2^6\cdot 3^8\cdot 5^4\cdot 7^3$
• B. $2^6\cdot 3^8\cdot 5^4\cdot 7^3-2\cdot 3\cdot 5\cdot 7$
• C. $2^6\cdot 3^8\cdot 5^4\cdot 7^3-1$
• D. none of these

Answer 1

Answer: (D)

Any divisor of $2^5.3^7\cdot 5^3\cdot 7^2$ is of the form $2^a{3}^b{5}^c{7}^d$
where $0\le a\le 5$, $0\le 7$, $0\le c\le 3$ and $0\le d\le 2$.
Thus the sum of the.divisors of $2^5.3^7.5^3.7^2$ is $(1+2+.....2^5)$ $(1+\mathrm{3.....}+3^7)$ $(1+5+5^2+5^3)$
$=\left(\frac{2^6-1}{2-1}\right)\left(\frac{3^8-1}{3-1}\right)\left(\frac{5^4-1}{5-1}\right)\left(\frac{7^3-1}{7-1}\right)$
$=\frac{\left(2^6-1\right)\left(3^8-1\right)\left(5^4-1\right)\left(7^3-1\right)}{2\cdot 4\cdot 6}$