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The remainder when $2^{2003}$ is divided by $17$ is ?

$\begin{array}{1 1} 1 \\ 2 \\ 3 \\ 8 \end{array}$

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1 Answer

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$2^{2003}=2^{2000}.2^3=\large(2^4)^{500}$$\times 8$
$=8.(17-1)^{500}$
$=8.\big(^{500}C_017^{500}-^{500}C_1.17^{499}+..........-^{500}C_{499}.17+1\big)$
$=17 k+8$
$\therefore$ The remainder is $8$
answered Oct 16, 2013 by rvidyagovindarajan_1
 
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