Find the remainder when 2^ 81 is divided by 17 - Vidyadip

Question

Find the remainder when $2^{81}$ is divided by $17$

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Answer

$02^{81} \equiv x(\bmod 17)$
$2^{40} \times 2^{40} \times 2^{41} \equiv x(\bmod 17)$
$\left(2^4\right)^{10} \times\left(2^4\right)^{10} \times 2^1 \equiv x(\bmod 17)$
$(16)^{10} \times(16)^{10} \times 2 \equiv x(\bmod 17)$
$\left(16^5\right)^2 \times\left(16^5\right)^2 \times 2$
$\left(16^5\right) \equiv 16(\bmod 17)$
$\left(16^5\right)^2 \equiv 16^2(\bmod 17)$
$\left(16^5\right)^2 \equiv 256(\bmod 17)$
$=1(\bmod 17) \ldots[\because 255 \text { is divisible by } 17]$
$\left(16^5\right)^2 \times\left(16^5\right)^2 \times 2 \equiv 1 \times 1 \times 2(\bmod 17)$
$\therefore 2^{81} \equiv 2(\bmod 17)$
$\therefore x=2$

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