Question

For the number $n=2^5\times 3^7\times 5^3\times 7^2$

• A. the sum of divisors $=\left\{\frac{(2^6-1)(3^8-1)(5^4-1)(7^3-1)}{48}\right\}$
• B. number of divisors$=576$
• C. the sum of divisors $=\left\{\frac{(2^8-1)(3^6-1)(5^3-1)(7^4-1)}{48}\right\}$
• D. number of divisors$=574$

Answer 1

Answer: (A), (B)

number of divisors$=576$
the sum of divisors $=\left\{\frac{(2^6-1)(3^8-1)(5^4-1)(7^3-1)}{48}\right\}$

$n=2^5\times 3^7\times 5^3\times 7^2$

Hence total number of divisors of the number is $(5+1)(7+1)(3+1)(2+1)$

$=576$

Sum of all divisors$=(1+2+2^2+...+2^5)(1+3+3^2+...+3^7)(1+5+...+5^3)(1+7+7^2)$

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

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