• # question_answer 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 [IIT 1995] A) 32 B) 63 C) 64 D) 127

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 $m=64$.