MCQ
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$63$
- ✓$64$
- D$127$
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 $m = 64$.
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