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

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

ManipalManipal 2018

Solution:

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 $1$ .