Do powers of 256 all end by 6 and if so, how to prove it? I computed the 10 first powers of 256 and I noticed that they all end by 6.
256^1 = 256
256^2 = 65536
256^3 = 16777216
256^4 = 4294967296
256^5 = 1099511627776
256^6 = 281474976710656
256^7 = 72057594037927936
256^8 = 18446744073709551616
256^9 = 4722366482869645213696
256^10 = 1208925819614629174706176

My intuition is that it is the same for all powers of 256 but I can't figure out how to prove it, any suggestion?
My attempt was to show that $\forall n$ there is a $k$ such that $2^{8n} = k*10 + 6$, $k$ and $n$ being non null integers. I tried to decompose the right member in powers of $2$ but that leaves me stuck.
 A: 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.
A: $$(10x+6)(10y+6)=100xy+60 (x+y)+3\color {red}6. $$
A: Using the binomial theorem
$$256^n{\pmod {10}}=(250+6)^n{\pmod {10}}=6^n{\pmod {10}}.$$
Now note that $$6^2{\pmod {10}}=36{\pmod {10}}=6{\pmod {10}}$$ and thus $6^n{\pmod {10}}=6{\pmod {10}}$, completing the proof.
A: We can prove this by complete induction:
For $n = 1$, we have $256^n = 256^1 = 256$, which ends with a $6$. 
Now, let $256^n$ end with a $6$, so there is an integer $k \ge 0$ such that $256^n = 10k+6$ and we have
$$256^{n+1} = 256(10k+6) = 2560k+1536 = 10(256k+153)+6$$
Therefore, $256^{n+1} = 10k'+6$ with $k'=256k+153$, and so also $256^{n+1}$ ends with a $6$.
A: 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 $256 \equiv 6 \pmod{10}$.
