Find the remainders of the number $2^{2001}$ when divided by $3, 5, 7, 9, 11, 13, 15, 17$.

#
$2^{3} \equiv-1(\bmod 9)$, so, $2^{2001}=\left(2^{3}\right)^{667} \equiv-1(\bmod 9)$.

$2^{4} \equiv 1(\bmod 15)$, so, $2^{2001}=2 \cdot\left(2^{4}\right)^{500} \equiv 2(\bmod 15)$.

$2^{3} \equiv 1(\bmod 7)$, so, $2^{2001}=\left(2^{3}\right)^{667} \equiv 1(\bmod 7)$.

$2^{5} \equiv-1(\bmod 11)$, so, $2^{2001}=2 \cdot\left(2^{5}\right)^{400} \equiv 2(\bmod 11)$.
$2^{6} \equiv-1(\bmod 13)$, so, $2^{2001}=8 \cdot\left(2^{6}\right)^{333} \equiv 5(\bmod 13)$.

$2^{4} \equiv-1(\bmod 17)$, so, $2^{2001}=2 \cdot\left(2^{4}\right)^{500} \equiv 2(\bmod 17)$.

## Answer

$2,2,1,8,2,5,2,2$.