Subject: General Aptitude · Chapter: Number System · Exam: 2007 · Marks: · Difficulty:
It is given that $\left(2^{32}+1\right)$ is exactly divisible by a certain number. Which of the following is also definitely divisible by the same number?
(a)$2^{16}+1$
(b)$2^{16}-1$
(c)$7 \times 2^{33}$
(d)$2^{96}+1$
Answer
Answer (as printed): D
Explanation
Let $2^{32}=x$. Then, $\left(2^{32}+1\right)=(x+1)$. Let $(x+1)$ be completely divisible by the natural number $N$. Then, $\left(2^{96}+1\right)=\left[\left(2^{32}\right)^{3}+1\right]=\left(x^{3}+1\right)=(x+1)\left(x^{2}-x+1\right)$, which is completely divisible by $N$ since $(x+1)$ is divisible by $N$.
Explanation as extracted from the printed page; notation may be imperfect.