Question

Let $n>1$ be the positive integer. The largest positive integer $m$, such that $n^m+1$ divides $1+n+n^2.....n^{125}$ is

• A. $60$
• B. $62$
• C. $63$
• D. $64$

Answer 1

The correct option is C $63$

We have,

$1+n+n^2.....n^{125}$

$=\frac{n^{126}-1}{n-1}$

$=\frac{(n^{63}{)}^2-1}{n-1}$

$=\frac{(n^{63}-1)(n^{63}+1)}{n-1}$

This will be divisible by $n^m+1$ for largest $m$ when $m=63$.