Find the remainder when the number $9^{100}$ is divided by 8

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Solution

$(9)^{100}$
$= (1 + 8)^{100}$
$= 1 +^{100} C_{1} .8 +^{100} C_{2} 8^{2} + . . .^{100} C_{100} 8^{100}$
$= 1 + 800 + 800 (k)$ where $k$ is a positive integer.
$= 801 + 800 (k)$
Hence
When $9^{100}$ divided by $8$ , we get
$= \frac{801 + 800 k}{8}$
$= \frac{801}{8} + 100 k$
$= \frac{800 + 1}{8} + 100 k$
Hence remainder is $1$ .

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