Question

Let $n(>)$ be a positive integer. Then, largest integer $m$ such that $(n^m+1)$ divides $1+n+n^2+...n^{225}$ is

• A. $128$
• B. $63$
• C. $64$
• D. $32$

Answer 1

The correct option is A $128$

$S=1+n+n^2+....+n^{255}$

$\Rightarrow S=\frac{1(n^{256}-1)}{n-1}=(1+n^{128})\frac{(-1+n^{128})}{n-1}$

$\Rightarrow S=(1+n^{128})(1+n+n^2+....+n^{128})$

Thus the largest value of m for which $1+n+n^2+....+n^{255}$ is divisible by $n^m+1$ is 128.