Permutations And Combinations Question 322

Question: If $ n={2^{p-1}}(2^{p}-1) $ , where $ 2^{p}-1 $ is a prime, then the sum of the divisors of n is equal to

Options:

A) n

B) 2n

C) pn

D) $ p^{n} $

Show Answer

Answer:

Correct Answer: B

Solution:

  • [b] If $ N=p_1^{{\alpha_1}}P_2^{{\alpha_2}} $ than the sum of the divisors of N is $ ( \frac{p_1^{{\alpha_1}+1}-1}{p_1-1} )( \frac{p_2^{{\alpha_2}+1}-1}{p_2-1} ) $