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This question already has an answer here:

I know this is more of a 'aops' type of question but here we go, I went to this math competition last year and there was this one problem that clearly I didn't solve but it recently came back to my mind and I want to know how to go about such problem:

Find the last 4 digits of the number: $$2^{{10}^{2018}}$$

My intuition is that one should probably use modular arithmetic on this one, the first things that came to my mind when I saw this one where: Chinese remainder Theorem and Binomial sums, I wasn't able to do much unfortunately... I've read through the "How do I compute $a^b$ (mod c) by hand?" question but most of the answers rely on a and c being coprime which in my case $(2,10^4)=2$ is not true, the answers cover a few cases when a and c are not coprime but nothing very similar to my case...

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marked as duplicate by Parcly Taxel, Chinnapparaj R, José Carlos Santos, ArsenBerk, Vladhagen Oct 29 '18 at 16:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It's about computing $2^a$ modulo $10^4$. The sequence $2^a$ will be eventually periodic modulo $10^4$, so really it's a question of determining the period, and seeing how your particular $a=10^{2018}$ relates to that period. $\endgroup$ – Lord Shark the Unknown Oct 28 '18 at 6:59
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    $\begingroup$ The statement of the problem makes me suspect that this is from some form of recent competition. Where is this problem from, exactly? $\endgroup$ – Arthur Oct 28 '18 at 7:09
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    $\begingroup$ @Arthur It's the second phase of an Italian competition, I will roughly translate it as "Group Mathematical contest of Tor Vergata" (which is a university in Rome) if you're interessed here's the link to the website you can find the texts and everything but they're in italian: mat.uniroma2.it/olimpiadi.php $\endgroup$ – Spasoje Durovic Oct 28 '18 at 7:18
  • $\begingroup$ It seems to have been held in March, which is fine. We don't want people to get help with ongoing competitions, but that's not the case here as far as I can see. $\endgroup$ – Arthur Oct 28 '18 at 7:26
  • $\begingroup$ @Arthur Oh okay yeah, the competition was held in March and the solutions are online, but they only give you the result not how it was obtained, that's why I asked: problemisvolti.it/Docu/GaSquadre/GaSquadreTVG18.pdf $\endgroup$ – Spasoje Durovic Oct 28 '18 at 8:30
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You have $10000 = 2^45^4 = 16\cdot 625.$ You need to find $2^{10^{2018}} \pmod{10000}$ and the Chinese Remainder Theorem will do this nicely. First

$$2^{10^{2018}} \equiv 0 \pmod{16}.$$

Second, note that $\phi(625)=500$, so since $500$ divides $10^{2018},$

$$2^{10^{2018}} \equiv 1 \pmod{625}.$$

Then CRT gives $9376$ for the final answer.

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    $\begingroup$ Thanks, it was $10000$ instead of $1000$, I got a response for this one on another forum just a few minutes before yours... $\endgroup$ – Spasoje Durovic Oct 28 '18 at 13:43
  • $\begingroup$ OK, I'll fix it up. $\endgroup$ – B. Goddard Oct 29 '18 at 0:54
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The first thing we can do is to break down $10^4$ to $2^4\cdot5^4$ and use CRT afterwards.

Clearly, $2^{{10}^{2018}}\equiv0\pmod{2^4}$. To calculate $2^{{10}^{2018}}\pmod{5^4}$, we can use Euler's Theorem: $$\phi(5^4)=4\cdot5^3\mid10^{2018} \rightarrow2^{{10}^{2018}}\equiv1\pmod{5^4}$$Therefore, $$2^{{10}^{2018}}\equiv2^0\equiv1\pmod{5^4}$$And finally, use CRT to get the final answer. $$\begin{cases}2^{{10}^{2018}}\equiv0\pmod{2^4}\\2^{{10}^{2018}}\equiv1\pmod{5^4}\end{cases}\implies2^{{10}^{2018}}\equiv\boxed{9376}\pmod{10^4}$$

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