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?
[Math] the last digit of $38^{2011}$
modular arithmetictotient-function
Best Answer
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