Question

For all $n\in N$, when $7^{2n}+3$ is divided by $8$, the remainder is

• A. $1$
• B. $4$
• C. $7$
• D. $2$

Answer 1

Answer: (B)

$4$

Let $P\left(n\right)=7^{2n}+3$
Now $7^{2n}=(8-1{)}^{2n}{=}^{2n}{C}_0{8}^{2n}{-}^{2n}{C}_1{8}^{2n-1}+........{-}^{2n}{C}_{2n-1}8{+}^{2n}{C}_{2n}1$
$=8{(}^{2n}{C}_0{8}^{2n-1}{-}^{2n}{C}_1{8}^{2n-2}+........{-}^{2n}{C}_{2n-1})+1[{\because }^n{C}_0{=}^n{C}_n=1]$
Thus $P\left(n\right)=8k+1+3=8k+4$
Where, $k{=}^{2n}{C}_0{8}^{2n-1}{-}^{2n}{C}_1{8}^{2n-2}+........{-}^{2n}{C}_{2n-1}=+$ Integer
Hence $P\left(n\right)$ will leave $4$ as remainder after dividing it by $8$