Last three digits of $6^{2002}$ Find the last three digits of $6^{2002}$. I did some work and figured out that the last two digits is 36. Can anyone help me with the hundredth digit? By the way, I used modular arithmetic and the recursion method for the tens digit, but it fell short when I attempted to do the hundreds digit. Thank you in advance!
How I figured out the last two digits:
I used the formula $\frac {1}{10^k} [n-a_k (n)]$. The last digit is obviously 6. I obtained the tens digit this way: $\frac {1}{10}(6^{2002}-6)=\frac {6}{10}(6^{2001}-1)=\frac {3}{5} (6^{2001}-1)=\frac {3}{5}(6-1)(6^{2000}+6^{1999}+...+6^2+6+1)=3(6^{2000}+6^{1999}+...+6^2+6+1)\equiv3(6+6+...+6+6+1)=3(2000\cdot6+1)\equiv3 (mod 10)$
Therefore, the last two digits of $6^{2002}$ are $36$.

I used the formula $\frac {1}{10^k} [n-a_k (n)]$. The last two digits is 36. I obtained the hundreds digit this way: $\frac {1}{100}(6^{2002}-36)=\frac {9}{25}(6^{2000}-1)=\frac {9}{25} (6^{2000}-1)=\frac {9}{25}((6^5)^{400}-1)=\frac {9}{25}((7776)^{400}-1)=\frac {9}{25}(7776-1)(7776^{399}+7776^{398}+...+7776^2+77766+1)\equiv9\cdot311\cdot(6+6+...+6+6+1)\equiv9\cdot1\cdot(399\cdot6+1)\equiv9\cdot1\cdot(9\cdot6+1)\equiv9\cdot1\cdot5\equiv5 (mod 10)$
Therefore, the last three digits of $6^{2002}$ are $536$.
