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 $(1+2+..... 2^5)$ $(1+3 .....+3^7)$ $(1+5+5^2 +5^3)$
$=\left(\frac{2^{6}-1}{2-1}\right)\left(\frac{3^{8}-1}{3-1}\right)\left(\frac{5^{4}-1}{5-1}\right)\left(\frac{7^{3}-1}{7-1}\right)$
$=\frac{\left(2^{6}-1\right)\left(3^{8}-1\right)\left(5^{4}-1\right)\left(7^{3}-1\right)}{2 \cdot 4 \cdot 6}$