Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
$n$ being any odd number greater than $1, n^{65}-n$ is always divisible by
(a)5
(b)13
(c)24
(d)None of these
Answer
Answer (as printed): C
Explanation
$n^{65}-n=n\left(n^{64}-1\right)=n\left(n^{32}-1\right)\left(n^{32}+1\right)$ $=n\left(n^{16}-1\right)\left(n^{16}+1\right)\left(n^{32}+1\right)$ $=n\left(n^{8}-1\right)\left(n^{8}+1\right)\left(n^{16}+1\right)\left(n^{32}+1\right)$ $=n\left(n^{4}-1\right)\left(n^{4}+1\right)\left(n^{8}+1\right)\left(n^{16}+1\right)\left(n^{32}+1\right)$ $=n\left(n^{2}-1\right)\left(n^{2}+1\right)\left(n^{4}+1\right)\left(n^{8}+1\right)\left(n^{16}+1\right)\left(n^{32}+1\right)$ $=(n-1) n(n+1)\left(n^{2}+1\right)\left(n^{4}+1\right)\left(n^{8}+1\right)$ $\left(n^{16}+1\right)\left(n^{32}+1\right)$. Clearly, $(n-1), n$ and $(n+1)$ are three consecutive numbers and they have to be multiples of 2, 3 and 4 as $n$ is odd. Thus, the given number is definitely a multiple of 24 .
Explanation as extracted from the printed page; notation may be imperfect.