Question

$n$ is an integer. The largest integer $m$, such that $n^m+1$ divides $1+n+n^2+.....n^{127},$ is

• A. $127$
• B. $63$
• C. $64$
• D. $32$

Answer 1

The correct option is C $64$

$1+n+n^2+...n^{127}=\frac{n^{128}-1}{n-1}=\frac{(n^{64}-1)(n^{64}+1)}{(n-1)}(a^n-b^m)$ is divisible by $(a-b)\bigvee m\in z^{+}$(positive Integers)

$(n^{64}-1)$ is divisible by $(n-1)$

$\frac{(n^{64}-1)}{(n-1)}$ is Integer value

$1+n+n^2+...n^{127}$ is divisible by $(n^{64}+1)$

Given $1+n+n^2+...n^{127}$ is divisible by $n^m+1$

Maximum possible value of m $m=64$