First slide

Combinations

Question

the sum of the divisors of $2^{5} {.3}^{7} {.5}^{3} {.7}^{2}$ is

Very difficult

A

$2^{6} . 3^{8} . 5^{4} . 7^{3}$
B

$2^{6} {.3}^{8} . 5^{4} {.7}^{3} - 2.3 5.7$
C

$2^{6} . 3^{8} . 5^{4} . 7^{3} - 1$
D

none of these

Solution

Any divisor of $2^{5} . 3^{7} . 5^{3} . 7^{2}$ is of the form $2^{a} 3^{b} 5^{c} 7^{d},$ where $0 \leq b \leq 7, 0 \leq c \leq 3$ and $0 \leq d$ $\leq 2.$
Thus, the sum of the divisors of $2^{5} . 3^{7} . 5^{3} . 7^{2}$ is

            $(1 + 2 + ... + 2^{5}) (1 + 3 + ... + 3^{7})$ $(1 + 5 + 5^{2} + 5^{3}) (1 + 7 + 7^{2})$

           $= (\frac{2^{6} - 1}{2 - 1}) (\frac{3^{8} - 1}{3 - 1}) (\frac{5^{4} - 1}{5 - 1}) (\frac{7^{3} - 1}{7 - 1})$

         $= \frac{(2^{6} - 1) (3^{8} - 1) (5^{4} - 1) (7^{3} - 1)}{48}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App