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I believe I am supposed to use a mod in this problem. I am unsure exactly how to do so. Help would be much appreciated and thanks in advance!

thesmallprint
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all3a
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    Modulo what number, do you suppose? – saulspatz Apr 19 '18 at 18:46
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    $13^{1201}=(10+3)^{1201}=\underbrace{10^{1201}+\binom{1201}{1}10^{1200}\cdot 3 +\dots}_{\text{multiple of 1000}}+\binom{1201}{1199}10^2\cdot 3^{1199}+\binom{1201}{1200}10\cdot 3^{1200}+3^{1201}$ – JMoravitz Apr 19 '18 at 18:49
  • See also : https://math.stackexchange.com/questions/789050/last-3-digits-of-3999 and https://math.stackexchange.com/questions/679842/find-the-last-two-digits-of-345 – lab bhattacharjee Apr 27 '18 at 11:29

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This corresponds to finding $13^{1201}\bmod 1000$ since the last three digits are in between $0$ and $999$.

Now, since $13$ and $1000$ are co-prime; Euler's theorem says $$13^{\varphi(1000)}\equiv 1\bmod 1000,$$ where $\varphi(x)$ is Euler's totient function evaluated at $x$.

Now $\varphi(1000)=400$ so $13^{400}\equiv 1$, that is, $13^{1200}\equiv 1$ and so $$13^{1201}\equiv 13.$$

thesmallprint
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