Let n ( 1) be a positive integer. The larges | vedclass

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(C) The sum of the geometric series is given by $S = 1 + n + n^2 + \dots + n^{127} = \frac{n^{128} - 1}{n - 1}$.We are given that $(n^m + 1)$ divides

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$64$

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(C) The sum of the geometric series is given by $S = 1 + n + n^2 + \dots + n^{127} = \frac{n^{128} - 1}{n - 1}$.We are given that $(n^m + 1)$ divides $S$,so $\frac{n^{128} - 1}{(n - 1)(n^m + 1)}$ must be an integer.We know that $n^{128} - 1 = (n^{64} - 1)(n^{64} + 1)$.Substituting this into the expression,we get $\frac{(n^{64} - 1)(n^{64} + 1)}{(n - 1)(n^m + 1)}$.For this to be an integer for all $n > 1$,we compare the terms.If $m = 64$,the expression becomes $\frac{(n^{64} - 1)(n^{64} + 1)}{(n - 1)(n^{64} + 1)} = \frac{n^{64} - 1}{n - 1} = 1 + n + n^2 + \dots + n^{63}$,which is always an integer.Thus,the largest integer $m$ is $64$.

Let $n (> 1)$ be a positive integer. The largest integer $m$ such that $(n^m + 1)$ divides $(1 + n + n^2 + \dots + n^{127})$ is:

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