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Q. When $2^{1505}$ is divided by $9$, the remainder is

KEAMKEAM 2012Binomial Theorem

Solution:

$2^{1505} =\left(2^{6}\right)^{25} \times 2^{5} $
$=(64)^{25} \times 32$
$=(1+63)^{25} \times 32 $
$=\left[{ }^{25} C_{0}(63)^{0}+{ }^{25} C_{1}(63)^{1}+{ }^{25} C_{2}(63)^{2}\right.$
$\left.\ldots .+{ }^{25} C_{25}(63)^{25}\right] 32 $
$=\left[1+(63)+{ }^{25} C_{2}(63)^{2}+\ldots+(63)^{25}\right] 32$
$=32+63\left[1+{ }^{25} C_{2}(63)+\ldots+(63)^{24}\right] 32 $
The remainder when $2^{1505}$ is divided by
9 $=$ The remainder when 32 divided by 9
$=5$