Question

If $(2^{32}+1)$ is divisible by a certain number then which of the following is also divisible by that number

• A. $(2^{16}-1)$
• B. $2^{16}+1$
• C. $2^{96}+1$
• D. None of these

Answer 1

Answer: (C)

$2^{96}+1$

Let $2^{32}=x$
Then,$(2^{32}+1)=(x+1)$
Let $(x+1)$ be completely divisible by the natural number N. Then,
$(2^{96}+1)=[{\left(2^{32}\right)}^3+1]=(x^3+1)=(x+1)(x2-x+1)$, which is completely divisible by N, since (x + 1) is divisible by N.