#### [SOLVED] Do powers of 256 all end by 6 and if so, how to prove it?

By Mai Kar

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.

#### 5 comments #### @Wolfgang Kais 2019-02-10 14:09:47

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 $$6$$. #### @Henno Brandsma 2019-02-10 12:24:28

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 $$256 \equiv 6 \pmod{10}$$. #### @don bright 2019-02-10 16:31:48

that is fascinating... can it be explained for people like me with a bit less experience with modular arithmetic? #### @Henno Brandsma 2019-02-10 17:37:17

@donbright the last digit of a number $n$ is just its class modulo $10$. And if you want to answer questions about arithmetic modulo $10$ it makes sense to work in the ring (structure having $+,0,1,-,\times$ with the usual rules) of all integers modulo $10$. You'll need a little bit of elementary algebra. #### @b00n heT 2019-02-10 12:17:06

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. #### @drhab 2019-02-10 12:14:27

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. #### @Mai Kar 2019-02-10 12:45:10

Thank you, my maths are rusty I did not thought about induction, I managed to solve this using it ! I am still marking user's answer as accepted since it provides a quick explanation to this phenomenon. (And my upvotes are not displayed yet since I am new here) #### @drhab 2019-02-10 12:49:30

You are welcome. #### @user 2019-02-10 12:14:11

$$(10x+6)(10y+6)=100xy+60 (x+y)+3\color {red}6.$$ #### @Bill Dubuque 2019-02-10 18:19:04

i.e. RHS $= 10(10xy\!+\!6(x\!+\!y)\!+\!3) + 6 = 10n+6\,$ so it has units digit $= 6.\$ By induction it follows that the product of any number of naturals all ending with $6$ also ends with $6\ \$ #### @Wolfgang Kais 2019-02-13 09:51:57

@BillDubuque The other question doesn't have an "answer", it has a "question". I'd say so.