Question

Let n(>1) be a positive integer, then the largest integer m such that $(n^m+1)$ divides $(1+n+n^2+....+n^{127})$ is

• A. 32
• B. 63
• C. 64
• D. 127

Answer 1

The correct option is C 64

Since $(n^m+1)$ divides $(1+n+n^2+....+n^{127})$
Therefore $\frac{1+n+n^2+...+n^{127}}{n^m+1}$ is an integer
$\Rightarrow \frac{1-n^{128}}{1-n}\times \frac{1}{n^m+1}$ is an integer
$\Rightarrow \frac{(1-n^{64})(1+n^{64})}{(1-n)(n^m+1)}$
is an integer when largest m=64.