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

Thanks for any help.

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2 Answers 2

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Hint:-

$2^{2003}=2^{2000}*2^3=(2^4)^{500}*8=(17-1)^{500}*8$

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

$$(17K+1)*8$$=$$17M+8$$

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

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

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

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