The sum of the divisors of 2^5 times 3^4 time | vedclass

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(B) Any divisor of $2^5 	imes 3^4 	imes 5^2$ is of the form $2^a 	imes 3^b 	imes 5^c$ where $0 \le a \le 5$,$0 \le b \le 4$,and $0 \le c \le 2$. T

Vedclass · Accepted Answer

$3^2 	imes 7^1 	imes 11^2 	imes 31$

Vedclass · Answer

(B) Any divisor of $2^5 	imes 3^4 	imes 5^2$ is of the form $2^a 	imes 3^b 	imes 5^c$ where $0 \le a \le 5$,$0 \le b \le 4$,and $0 \le c \le 2$. The sum of the divisors is given by the product of the geometric series for each prime factor:Sum $= (1 + 2 + 2^2 + 2^3 + 2^4 + 2^5) 	imes (1 + 3 + 3^2 + 3^3 + 3^4) 	imes (1 + 5 + 5^2)$Using the formula for the sum of a geometric series $\frac{r^n - 1}{r - 1}$:Sum $= \left( \frac{2^6 - 1}{2 - 1} ight) 	imes \left( \frac{3^5 - 1}{3 - 1} ight) 	imes \left( \frac{5^3 - 1}{5 - 1} ight)$Sum $= (64 - 1) 	imes \left( \frac{243 - 1}{2} ight) 	imes \left( \frac{125 - 1}{4} ight)$Sum $= 63 	imes \frac{242}{2} 	imes \frac{124}{4}$Sum $= 63 	imes 121 	imes 31$Prime factorization of $63 = 3^2 	imes 7$. So,Sum $= 3^2 	imes 7^1 	imes 11^2 	imes 31$.

The sum of the divisors of $2^5 \times 3^4 \times 5^2$ is

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