Question

If $n$ is an even number, then the digit. In the unit place of $2^{2n+1}+1$ will be ??

Answer 1

$2^{2^1}221+1=5$ ends in $5$.

$Ifn>1$, then $4\mid 2^n$ and so $15=(2^4–1)\mid (2^{2}n-1)$.

Hence $2^{2}n+1\equiv 2$(mod $5$), and since $2^{2}n+1$is odd, we must have $2^{2}n+1\equiv 7$( mod $10$).

Therefore $2^{2}n+1$ ends in $7$ when $n>1$, and in $5$ when $n=1$.