Problem - 4285
Let $n$ be an odd positive integer, and $$N=6^n + \binom{n}{1}\cdot 6^{n-1}+\cdots + \binom{n}{n-1}\cdot 6-1$$
Find the remainder when $N$ is being divided by $8$.
The answer is $\boxed{5}$ because $$N=(6+1)^n-2=(8-1)^n -2 \equiv (-1)^n -2\equiv -3\equiv 5\pmod{8}$$