Do powers of 256 all end by 6 and if so, how to prove it? [duplicate]
Solution 1:
$$(10x+6)(10y+6)=100xy+60 (x+y)+3\color {red}6. $$
Solution 2:
In $\mathbb{Z}_{10}$ (the integers moduo $10$) we have that $6^2 = 6$, an idempotent, so all its positive powers are $6$ too. And $256 \equiv 6 \pmod{10}$.
Solution 3:
Let $P(n)$ be the statement that $6^n$ ends on $6$.
Then evidently $P(1)$ is true.
If $P(n)$ is true then it is not difficult to prove that also $P(n+1)$ is true (try it yourself).
According to the principle of induction on natural numbers we are allowed to conclude that $P(n)$ is true for every positive integer.