Question

Assertion :The sum of divisors of $n=2^{10}{3}^2{5}^3{7}^2{11}^2{13}^3$ is $\frac{1}{5760}(2^{11}-1)(3^3-1)(5^4-1)(7^3-1)({11}^3-1)({13}^4-1)$ Reason: The number of divisor of $m={p_1}^{{\alpha }_1}{p_2}^{{\alpha }_2}...{p_r}^{{\alpha }_r}$ where $p_1,p_2,...,p_r$ are distinct primes and ${\alpha }_1,{\alpha }_2,...,{\alpha }_r$ are natural number is $({\alpha }_1+1)({\alpha }_2+1)...({\alpha }_r+1)$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Sum of the divisors of $n$

$=(1+2+...+2^{10})(1+3+3^2)(1+5+5^2+5^3)...(1+13+{13}^2+{13}^3)$

$=(2^{11}-1)\left(\frac{3^3-1}{3-1}\right)\left(\frac{5^4-1}{5-1}\right)\left(\frac{7^3-1}{7-1}\right)\left(\frac{{11}^3-1}{11-1}\right)\left(\frac{{13}^4-1}{13-1}\right)$

$=\frac{1}{5760}(2^{11}-1)(3^3-1)(5^4-1)(7^3-1)({11}^3-1)({13}^4-1)$

$A$ divisors of $m$ is of the form ${p_1}^{{\beta }_1}{p_2}^{{\beta }_2}...{p_r}^{{\beta }^r}$ where

$0\le {\beta }_i\le {\alpha }_i$ for $i=1,2,...,r.$

That is, ${\beta }_i$ can take ${\alpha }_i+1$ values.

Thus, the number of divisors of $m$ is $({\alpha }_1+1)({\alpha }_2+1)...({\alpha }_r+1)$