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 $36$.
 A: I did the spreadsheet approach.  Put $6$ in a cell, =mod(6*up,1000) in the cell below, copy down, and look for a repeat. I found a repeat of $25$, so $6^{2002} \equiv 6^{27}\equiv 536 \pmod {1000}$.  You can't use $6^2$ because you need the value to be a multiple of $8$.
A: $3^{400}\equiv 1\mod 1000$ 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 $6^{2002}\equiv 9*504\equiv 536\mod 1000$.
A: $$6=1+5,6^{25n}=(1+5)^{25n}\equiv1\pmod{5^3}$$
$$\implies6^{25n-1}\equiv6^{-1}\equiv21$$
$$6^{25n+2}\equiv6^3(21)\pmod{5^36^3}$$
$$\equiv216\cdot21\pmod{2^35^3}\equiv?$$
A: In order to find the last three digits of $6^{2002}$ it is enough to compute the remainders $\!\!\pmod{8}$ and $\!\!\pmod{125}$, where the former is clearly zero.  About the latter, the binomial theorem grants
$$6^{2002} = (5+1)^{2002} = \sum_{k=0}^{2002}\binom{2002}{k}5^k \equiv \sum_{k=0}^{2}\binom{2002}{k}5^k\equiv 36\pmod{125} $$
hence the last three digits of $6^{2002}$ are $\color{red}{536}$ by the Chinese remainder theorem.
A: 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 $536$.
