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 $36$.

$3^{400}\equiv 1\mod 1000$ so $3^{2002}\equiv 9\mod 1000$.
$2^{100}\equiv 1\mod 125$ so $2^{2002}\equiv 4\mod 125$.
And $2^{2002}\equiv 0\mod 8$.  So by Chinese Remainder  theorem $2^{2002}\equiv 504 \mod 1000$.
So $6^{2002}\equiv 9*504\equiv 536\mod 1000$.
