# Proof that $6^n$ always has a last digit of $6$

Without being proficient in math at all, I have figured out, by looking at series of numbers, that $6$ in the $n$-th power always seems to end with the digit $6$.

Anyone here willing to link me to a proof?

I've been searching google, without luck, probably because I used the wrong keywords.

• @Stijn: your comment seems rather obnoxious. The OP did not say that he has "proved" anything; he made an observation which he thinks "seems to always" hold. And he's right... Sep 6, 2011 at 0:47
• @Stijn: Ragnar said "seems", so s/he knows it's not a proof. :) Sep 6, 2011 at 1:19
• @J.M. A p.c. "s/he" in combination with a person called Ragnar is making me chuckle...
– t.b.
Sep 6, 2011 at 1:27
• @Theo: I've been "victimized" by ladies using "manly" names on the Internet, so I'm covering myself just in case. :D Sep 6, 2011 at 1:34
• My comment wasn't meant to be obnoxious. I'll delete it if it comes across as such. Sep 6, 2011 at 8:40

## 6 Answers

We can prove it using mathematical induction.

Claim: $6^n\equiv 6\bmod 10$ for all $n\in\mathbb{N}$ (the symbol $\mathbb{N}$ denotes the natural numbers, and $\bmod 10$ means we are using modular arithmetic with a modulus of 10).

Base case (i.e., showing it's true for $n=1$): $$6^1\equiv 6\bmod 10\qquad\checkmark$$

Induction step (i.e., showing that, if it is true for $n=k$, then it is true for $n=k+1$):

$$6^k\equiv 6\bmod 10\implies 6^{k+1}\equiv 6^k\cdot 6\equiv6\cdot 6\equiv 36\equiv 6\bmod 10\qquad\qquad\checkmark$$

If you multiply any two integers whose last digit is 6, you get an integer whose last digit is 6: $$\begin{array} {} & {} & {} & \bullet & \bullet & \bullet & \bullet & \bullet & 6 \\ \times & {} & {} &\bullet & \bullet & \bullet & \bullet & \bullet & 6 \\ \hline {} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & 6 \end{array}$$ (Get 36, and carry the "3", etc.)

To put it another way, if the last digit is 6, then the number is $(10\times\text{something}) + 6$. So \begin{align} & \Big((10\times\text{something}) + 6\Big) \times \Big((10\times\text{something}) + 6\Big) \\ = {} & \Big((10\times\text{something})\times (10\times\text{something})\Big) \\ & {} + \Big((10\times\text{something})\times 6\Big) + \Big((10\times\text{something})\times 6\Big) + 36 \\ = {} & \Big(10\times \text{something}\Big) +36 \\ = {} & \Big(10\times \text{something} \Big) + 6. \end{align}

HINT $\rm\ \ 6-1\ |\ 6^k-1,\$ so $\rm\:\ 2,5\ |\ 6^n-6\ \Rightarrow\ 10\ |\ 6^n - 6\:,\$ i.e. $\rm\ 6^n\ =\ 6 + 10\ k\:$ for $\rm\:k\in\mathbb Z\:.$

Alternatively: $\rm\ mod\ 10:\ \ 6^n\equiv 6\$ since it is $\rm\ 0^n \equiv 0\pmod 2,\ \ 1^n \equiv 1\pmod 5$

Similarly odd $\rm\:b\: \Rightarrow\: (b+1)^n\equiv b+1\pmod{2\:b}\:,\:$ so $\rm\:(b+1)^n\:$ has last digit $\rm\:b+1\:$ in radix $\rm\:2\:b\:.$

NOTE how modular arithmetic reduces the induction to the trivial inductions $\rm\ 0^n = 0,\ 1^n = 1\:.$ This is a prototypical example of the sort of simplification afforded by reducing arithmetical problems to their counterparts in the simpler arithmetical rings of integers $\rm\:(mod\ m)\:.\:$

• In some contexts, this would be a good way to answer this question. But given the way the question was phrased on this occasion, I wouldn't have considered it probable that this is one of those. Sep 6, 2011 at 1:34
• @Mic Surely the OP can grok the first proof. The rest requires only basic knowledge of modular arithmetic - which it seems is known to the OP given the accepted answer. Even if was not known to the OP, it is known to many other readers. The site is for all to learn - not just OP's. So I disagree. Sep 6, 2011 at 1:58
• Thanks for the answer, I understand it. Actually, I stumbled upon seeing the series, trying to prove $5 | 6^k-1$. My professor has since pointed me in the right direction, using mathematical induction. Sep 6, 2011 at 18:20
• @Ragnar Thanks for the feedback. Should anything I write be unclear, please feel free to ask further questions. I am always happy to elaborate. Sep 6, 2011 at 18:35

This follows from the more general result that the product of two numbers ending with digit 6 also ends with digit 6. This can be proved in an elementary way: $$(10x+6)\cdot(10y+6) = 100xy + 60x +60y + 36 = 10(10xy+6x +6y +3) + 6 = 10z+6$$

Of course, avoiding all these letters is what congruences are all about.

• On the other hand, for the purposes of this question, I prefer this to the mod solutions. :) Sep 6, 2011 at 1:21
• BTW, the same holds for numbers ending with 1 or 5, by the same reasoning.
– lhf
Sep 6, 2011 at 1:51

$6 \times 6 \equiv 6 \pmod{10}$.

Or, more elementarily put, think back to the pen-and-paper multiplication algorithm. When you multiply something by 6, the only part of the original number that can affect the last digit of the result is the last digit of the original. If you start with something that ends in 6, you get 36 for the last position, write 6 down and carry the 3. But no matter what happens after the carry, it cannot affect the final 6 that you've already produced.

I also think that this is true, because $6\times 6=36$, meaning that any ending $6$ digit results in an ending $6$ digit through multiplication.

• You can't just say that one thing is true because it works in one case. If you say: $4^2 = 2^4$ then you could say $x^y = y^x$, which is false. For a better examaple, look for Borwein integrals and you'll get a basic example. Nov 8, 2022 at 19:21