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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 $36$.

KingLogic
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  • One approach is just to compute it in a spreadsheet. Put $6$ in a cell, =mod(6*up,1000) in the cell below, copy down, and look for a repeat. Probably faster than finding the solution any other way except asking Alpha. – Ross Millikan Mar 30 '18 at 05:12
  • You could use that $6^{2002}=2^{2002}\cdot 3^{2002}=2\cdot (10-2)^{667}(10-1)^{1001}$ and expand via binomial theorem. In each binomial theorem expansion, since we are interested only in the last three digits only three terms survive each, but its still frustrating to calculate $\binom{667}{2}$ by hand and the final multiplication. It is still plenty doable though. – JMoravitz Mar 30 '18 at 05:15
  • See https://math.stackexchange.com/questions/789050/last-3-digits-of-3999 and https://math.stackexchange.com/questions/1844558/how-to-find-last-two-digits-of-22016 – lab bhattacharjee Mar 30 '18 at 05:48
  • Agree with Lab, this is an essential duplicate of several other posts. Check out this umbrella thread and others linked to it. – Jyrki Lahtonen Apr 30 '18 at 07:57

6 Answers6

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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 $8$.

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

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

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

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

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

lab bhattacharjee
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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.

Jack D'Aurizio
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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 $$\begin{align} &6^{2002} \\ &\equiv 100\cdot 24^4(101)(1000+1)(2000+1) - 80\cdot 24^4(101)(1000+1) + 16\cdot 24^4(101) \pmod{1000} \\ &\equiv (100-80+16)\cdot 24^4\cdot 101 \pmod{1000} \equiv 36(101)24^4 \pmod{1000} \\ &\equiv (53/2)\cdot 24^5 \pmod{1000} \equiv 53\cdot 312 \pmod{1000} \equiv 600 + 936 \pmod{1000} \\ &\equiv \color{red}{536} \pmod{1000}.\end{align}$$

András Salamon
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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 $536$.

KingLogic
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