How to show that $6^n$ always ends with a $6$ when $n\geq 1$ and $n\in\mathbb{N}$

Question

Is there a proof that for where $n$ is a natural number $$6^n$$ will end with a $6$?

I can understand conceptually that $6\cdot 6$ ends with $6$ and then multiplying that by $6$ will still end with $6$, but is there an actual proof?

Answer 1

Let's prove by induction that $6^n$ ends with $6$ for every $n\in\mathbb{N}$.

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First, show that this is true for $n=1$:

$6^1=6$

Second, assume that this is true for $n$:

$6^n=10k+6$

Third, prove that this is true for $n+1$:

$6^{n+1}=6\cdot\color{red}{6^n}=6\cdot(\color{red}{10k+6})=60k+36=10(6k+3)+6$

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Please note that the assumption is used only in the part marked red.

Answer 2

As pretty much everyone is indicating, what you want to show is probably easiest shown using induction. Are you familiar with this proof technique? I'll try to make the proof as clear as possible--simply comment if a step doesn't make sense.

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Claim: For all $n\geq 1$, where $n\in\mathbb{N}$, the number $6^n$ ends with a $6$. Let $S(n)$ denote this statement where we have $$ S(n) : 6^n = 10m+6, m\in\mathbb{Z}. $$ Now let's prove that $S(n)$ holds for all $n\geq 1$ by using induction.

Base step ($n=1$): $S(1)$ says that $6^1=6$ ends with a $6$ and this is clearly true.

Induction step: Fix some $k\geq 1$ and assume that $S(k)$ is true where $$ S(k) : 6^k = 10\ell+6, \ell\in\mathbb{Z}. $$ It remains to show that $S(k+1)$ follows where $$ S(k+1) : 6^{k+1} = 10\eta+6, \eta\in\mathbb{Z}. $$ Starting with the left-hand side of $S(k+1)$, \begin{align} 6^{k+1} &= 6\cdot 6^k\tag{by definition}\\[0.5em] &= 6(10\ell+6)\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &= 60\ell+36\tag{expand}\\[0.5em] &= 10(6\ell+3)+6\tag{rearrange}\\[0.5em] &= 10\eta+6, \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.

Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$

Answer 3

Say $6^n=10k+6$, then $6^{n+1}=(10k+6)\times 6=60k+36=10(6k+3)+6$, hence, it always ends with "$6$", as shown inductively.

By the way, the question may be formulated as: Prove $6^n \equiv 6$ (mod $10$).

Answer 4

We'll show if a number $n$ ends with a six then $6\cdot n$ ends with 6 too. If $n$ ends wit a six then $n=k\cdot10 + 6$ so $6\cdot n = 6k 10+ 36=(6k+3)\cdot10+6$. This means $6\cdot n$ ends with a six.

Answer 5

We know $6$ ends in a $6$. If we can prove $6^n-6$ is divisible by $10$ for $n\gt1$ we are done. $6^n-6=6(6^{n-1}-1)=6(6-1)(\sum_{k=0}^{n-2}6^k)$, which equals $30\sum_{k=0}^{n-2}6^k$ for $n\gt1$, and means we are done.

Answer 6

Assuming you mean $n$ is strictly a positive integer, this is something you could prove inductively. The case of $n=1$ is easy, $6^1 = 6 \space \checkmark$.

Now suppose $6^n$ has a $6$ in the one's column for all $n \in \{1,2,\ldots k\}$. Consider $6^{k+1} = 6^k \cdot 6$. By the induction hypothesis we know $6^k$ has a $6$ in the one's column. Then we can write $6^k = M+6$ for some number $M$ where $M$ has a $0$ in the one's column. Further, $(M+6)6 = 6M+36 = (6M+30)+6$. Note that $6M+30$ will still have a zero in the one's column, since $6M$ and $30$ have zeroes in the one's column. So $$(6M+30)+6 = 6^k6 = 6^{k+1}$$ will have a six in the one's column.

Answer 7

The product of two integers with unit digits $ = \color{#c00}6\,$ also has unit digit $= \color{#c00}6\,$ since

$\qquad (\color{#c00}6 + 10j)(\color{#c00}6+10k)\, =\, 36 + 10(\color{}{\cdots})\, =\, \color{#c00}6 + 10(3 + \cdots)$

So, by induction, the same holds for a product of $\,n\,$ such integers, e.g. for $\,6^n$.

Answer 8

Surprised that after so many answers, there still doesn't seem to be a Chinese Remainder Theorem based approach. So here's one, just for completeness.

We want to show that $6^n \equiv 6 \pmod{10} \ \forall \ n \in \mathbb{Z}^+$

Since $10 = 5\cdot 2$, express this as the pair of simultaneous linear congruences:

$6^n \equiv x \pmod 5 \implies x \equiv 1 \pmod 5$

and

$6^n \equiv x \pmod 2 \implies x \equiv 0 \pmod 2$

From the second congruence, $x = 2k, k \in \mathbb{Z}$.

Putting that back into the first congruence, we get:

$2k \equiv 1 \pmod 5 \implies k \equiv 2^{-1} \equiv 3 \pmod 5$.

That gives $k = 5t + 3, t \in \mathbb{Z}$.

Hence $x = 2(5t + 3) = 10t + 6 \equiv 6 \pmod{10}$.

So $6^n \equiv 6 \pmod{10}$, as required.