Question

$P\left(n\right)=9^n-8^n$ , when divided by $8$, it always leaves the remainder that is?

• A. $2$
• B. $3$
• C. $1$
• D. $7$

Answer 1

The correct option is C $1$

Given, $P\left(n\right)=9^n-8^n$
$=(8+1{)}^n-8^n$
Applying Binomial expansion gives us
$P\left(n\right)=8^n+{}^n{C}_1{8}^{n-1}+...{}^n{C}_n-8^n$
$P\left(n\right)=8^n+8k+1-8^n$, where $k$ is positive real number.
$=8k+1$
Hence, $\frac{9^n-8^n}{8}$
$=\frac{8k+1}{8}$
$=k+\frac{1}{8}$
Thus the remainder is $1$.