What is the last digit of $38^{2011}$?

Question

I would like to find the last digit of $38^{2011}$.

We have by Euler's therem :
$19^4\equiv 1\pmod{10}$ and $2^4\equiv 6\pmod{10}\implies 38^4\equiv 6\pmod{10}$. What can I do after this?

Answer 1

No need to use big guns here ... just focus on the last digit as you keep multiplying:

8,4,2,6,8,...

OK, so it cycles with a period of 4

Since 2011 divided by 4 leaves 3, we should be looking at the third entry of this cycle, i.e. it's a 2

Answer 2

$19^4\equiv 1\pmod{10}$ & $2^4\equiv 6\pmod{10}\implies 38^4\equiv 6(\mod{10})$. Notice that $6^n\equiv 6\pmod{10}\ \forall n\in \mathbb{N}$. So $38^{4n}\equiv 6\pmod{10}.$ Now you should do it.