Dtyjlui

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.

$$(10x+6)(10y+6)=100xy+60 (x+y)+3color {red}6. $$

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 $256 equiv 6 pmod{10}$.

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.

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.

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 $6$.