Question

Let $n(>1)$ be a positive integer. Then largest integer $m$ such that $\left(n^m+1\right)$ divides $1+n+n^2+\dots +n^{255}$ is

• A. 128
• B. 63
• C. 64
• D. 32

Answer 1

Answer: (A)

We have, $S=1+n+n^2+\dots +n^{255}$
$\Rightarrow S=\frac{1\left(n^{256}-1\right)}{n-1}=\left(n^{128}+1\right)\frac{n^{128}-1}{n-1}$
$\therefore S=\left(n^{128}+1\right)\left(1+n+n^2+\dots +n^{127}\right)$
Thus, the largest value of $m$ for which $1+n+n^2$
$+\dots +n^{255}$ is divisible by $n^m+1$ is $128$.