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
    $127$

Answer

Correct option: C.
$64$
c
(c) 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 $m = 64$.

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