Question
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
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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$\left| {1 - {{\log }_{\frac{1}{6}}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{\frac{1}{6}}}x + {{\log }_{\frac{1}{2}}}x} \right|$ is $\left[ {\frac{a}{b},a} \right],a,b, \in N,$ then the value of $(a + b)$ is
$4 x+3 y=69$
$4 y-3 x=17 \text { and }$
$x+7 y=61$
Then $(\alpha-\beta)^2+\alpha+\beta$ is equal to $.........$.