Question

It is given that $(2^{32}+1)$ is exactly divisible by a certain number. Which of the following is also definitely divisible by the same number?

• A. $(2^{16}+1)$
• B. $(2^8+1)$
• C. $(2^{16}-1)$
• D. $(2^{96}+1)$

Answer 1

The correct option is D $(2^{96}+1)$

$(2^{96}+1)=(2^{32}{)}^3+(1{)}^3$
$=(2^{32}+1)(\dots \dots )$
The given expression is divisible by $(2^{32}+1)$. Since $(2^{32}+1)$ is a factor of $(2^{96}+1)$ so any factor pf $(2^{32}+1)$ will divide $(2^{96}+1)$.