How do I find the last digits of large numbers using modular arithmetic? I know I have to consider things base 10 and use the multiplicative and additive properties of modular arithmetic. But I still am not sure how to do this.
For example: can someone help me with showing the methods of getting the last digit to these numbers? 
1) $(3^5)^7$
2) $(7^5)^3$
3) $(11^{10})^6$
4) $(8^5)^4$
Thanks! 
 A: As $3$, for instance, is prime to $10$, we can apply Euler's theorem: $3^4\equiv 1\mod 10$, hence
$$\bigl(3^5\bigr)^7=3^{35}\equiv3^{35\bmod4}\mod 10=3^3\equiv 7\mod 10.$$
As to $8$, which is not coprime to $10$, its powers modulo $10$ follow this pattern:
$$\begin{array}{c|cccccc}
n&1&2&3&4&5&\dots\\
\hline
8^n&8&4&2&6&8&\dots
\end{array}$$
so $8^n\equiv 8^{n\bmod4}\mod 10$ if we agree to represent integers modulo $4$ by a number in $\{1,2,3,4\}$ instead of $\{0,1,2,3\}$. Thus $$\bigl(8^{5}\bigr)^4=8^{20}\equiv8^4\equiv 6.$$
A: To find the last digit, $d_0$, of a number $N$, you want: $d_0 = N \mod 10$.  (I.e., $d_0$ is the remainder from dividing $N$ by 10.)
Now, to find the next-to-last digit of $N$, realize that it is the last digit of
$$
{1 \over 10}(N - d_0)
$$
A: I'lll explain 4) as an example. So we have $(8^5)^4$ (mod10). So the goal is to either reduce 8 modulo 10, or arrange the powers to make the number more manageable. Since 8 is congruent to 8 mod 10, we will work with the exponents. So $(8^5)^4=(8^4)^5=(8^{2})^{2x5}=16^{2x5}=6^{2x5}$mod 10. Then $6^{2x5}=36^5=6^5$ mod 10. Finally, $6^5=46,656=6$ mod 10. And there is your last digit. 
