Question

The statement $P(n)=9^n-8^n$, when divided $8$ , always leaves the remainder

• A. 1
• B. 2
• C. 3
• D. 7

Answer 1

Answer: (A)

Given, $P(n)=9^n-8^n$
$\therefore P(1)=9-8=1$
$\therefore P(1)-1=0$,
which is divisible by $8$
Now, $P(2)=9^2-8^2=17$.
When divided by $8$ ,
it leaves the remainder $1$ .