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?

• A. $(2^{16}+1)$
• B. $(2^{16}-1)$
• C. $(7\times 2^{23})$
• D. $(2^{96}+1)$

Answer 1

Answer: (D)

$(2^{96}+1)$

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