Quantitative Aptitude Ques 2322

Question: It is being given that $ (2^{32}+1) $ is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?

Options:

A) $ (2^{16}+1) $

B) $ (2^{16}-1) $

C) $ 7\times 2^{33} $

D) $ (2^{96}+1) $

Show Answer

Answer:

Correct Answer: D

Solution:

  • Let $ 2^{32}=x. $ Then, $ (2^{32}+1)=(x+1) $ Let $ (x+1) $ be completely divisible by the natural number N. Then, $ (2^{96}+1)=[{{(2^{32})}^{3}}+1]=(x^{3}+1) $ $ =(x+1)(x^{2}-x+1) $ which is completely divisible by N, since $ (x+1) $ is divisible by N.
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