Last digit of $x^{n}$ in Base $12$

Question

I want to show that the last digit of $x^{n}$ in base $12$ when $n=2, 3, \ldots$ is

$0$ when $x=6$

$4$ when $x=A$, $(A=10)$

and also the last digit is $x$ when $x=3,5,7,8,11$ and $n$ is odd.

Here are my thoughts... When $x=6$, I need to prove that $6^n=0 \mod 12$, $n \geq 2$.

When $x=10$, I need to prove that $10^n=4 \mod 12$, $n \geq 2$.

I am not sure where to start on the last part.

Answer 1

You are right about $x=6$. And of course that $n\geqslant 2\implies12\mid6^n$.

On the other hand, $10^2\equiv4\mod12$ and it follows by induction that$$n\geqslant2\implies10^n\equiv4\mod12.$$

Now, $3\equiv3\mod12$ and $3^2\equiv-3\mod12$. It follows from this that, if $n$ is odd, then $3^n\equiv3\mod12$.

In the case of $5$, it's almost the same thing: $5\equiv5\mod12$ and $5^2\equiv1\mod12$. You can exactly the same thing with the number $7$.

Can you take it from here?

Answer 2

$6^2=36=3\cdot 12$ so $6^2=0$ mod $12$ but $6^n=6^{(n-2)}6^2=6^{(n-2)}3\cdot 12$ and so $ 6^n= 0$ mod $12$

In the same way $10^2=100=8\cdot 12 +4$ and so $ 10^2= 4$ mod $12$ and so $10^n=4$ mod $12$