By Mai Kar

2019-02-10 12:10:07 8 Comments

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.


@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.

Related Questions

Sponsored Content

2 Answered Questions

How to approach this modulo proof?

1 Answered Questions

1 Answered Questions

[SOLVED] ZFC - How to prove this statement

1 Answered Questions

[SOLVED] Set theory proof with contrapositive

1 Answered Questions

Raising inequality to a power proof

4 Answered Questions

0 Answered Questions

Velleman's How to prove it. Partial order proof.

3 Answered Questions

[SOLVED] Intricate exponential equation

Sponsored Content