Last three digits of $6^{2002}$

Question

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 $36$.

Answer 1

It is possible to do this without the Chinese Remainder Theorem, just using the binomial expansion. Note that $$\begin{align}2^{500k} &= (30+2)^{100k} = 2^{100k} \pmod{1000},\\ 2^{10k} &= (1000+24)^k \equiv 24^k \pmod{1000},\\ 24^{5k} &= 2^{15k}3^{5k} = (32000+768)^k 243^k \equiv (768\cdot 243)^k \pmod{1000} \\ &\equiv (100+600+580+344)^k \pmod{1000} \equiv 624^k \pmod{1000},\end{align}$$ and $$\begin{align} 624^{4k} &= (600+24)^{4k} \equiv 4k\cdot 600\cdot 24^{4k-1} + 24^{4k} \pmod{1000} \\ &\equiv (100k + 1)\cdot 24^{4k} \pmod{1000}.\end{align}$$ In particular, $$\begin{align} 2^{4000} &\equiv 2^{10\cdot 80} \pmod{1000} \equiv 24^{5\cdot 16} \pmod{1000} \\ &\equiv 624^{16} \pmod{1000} \equiv (401)24^{16} \pmod{1000} \\ &\equiv 24(401)624^3 \pmod{1000} \equiv (8020+1604)624^3 \pmod{1000} \\ &\equiv 624^4 \pmod{1000} \equiv (101)24^4 \pmod{1000}.\end{align}$$

We then have $$\begin{align} 6^{2002} &= (10-4)^{2002} \\ &= \binom{2002}{2} 100\cdot 4^{2000} - 2002\cdot 10\cdot 4^{2001} + 4^{2002} \pmod{1000},\end{align}$$ so $$\begin{align} &6^{2002} \\ &\equiv 100\cdot 24^4(101)(1000+1)(2000+1) - 80\cdot 24^4(101)(1000+1) + 16\cdot 24^4(101) \pmod{1000} \\ &\equiv (100-80+16)\cdot 24^4\cdot 101 \pmod{1000} \equiv 36(101)24^4 \pmod{1000} \\ &\equiv (53/2)\cdot 24^5 \pmod{1000} \equiv 53\cdot 312 \pmod{1000} \equiv 600 + 936 \pmod{1000} \\ &\equiv \color{red}{536} \pmod{1000}.\end{align}$$

Answer 2

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 $8$.

Answer 3

$3^{400}\equiv 1\mod 1000$ 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 $6^{2002}\equiv 9*504\equiv 536\mod 1000$.

Answer 4

$$6=1+5,6^{25n}=(1+5)^{25n}\equiv1\pmod{5^3}$$

$$\implies6^{25n-1}\equiv6^{-1}\equiv21$$

$$6^{25n+2}\equiv6^3(21)\pmod{5^36^3}$$

$$\equiv216\cdot21\pmod{2^35^3}\equiv?$$

Answer 5

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.

Answer 6

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 $536$.