The title says it all.I can solve it using Fermat's little theorem.But I cannot use it.

Thanks for any help.


2 Answers 2




Expanding this by binomial we get a number of the form-


[There is a $+1$ term since,last term is $(-1)^{500}$.]

It is easy to derive remainder when $17$ divides $17M+8$.


${\rm mod}\ 17\!:\,\ \color{#c00}{2^{\large 4}\equiv -1}\,\Rightarrow\,2^{\large 3+8N}\!\equiv\, 2^{\large 3}(\color{#c00}{2^{\large 4}})^{\large 2N}\!\equiv\, 2^{\large 3}(\color{#c00}{-1})^{\large 2N}\!\equiv\, 2^{\large 3}$


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