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In trying to prove that $11^{10^3}\equiv 1 \pmod {10^4}$. I have tried to rewrite the expression so that I can use Fermats little theorem using exponentional rules. Also I know that $a^p \equiv a \pmod p$. But I just get stuck in a circular argument going back to the original expression proving nothing. I think this should be fairly easy to prove, I just cant figure it out.

I also plugged this into wolframalpha and I think in general that $11^{10^a} \equiv 1 \pmod {10^{a+1}}$.

marowl
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    Do you mean $11^{103}$? Or maybe $11^{10^3}$? Or what?? – Angina Seng Dec 17 '19 at 21:04
  • Sorry, now the typos are fixed. Yes I mean $11^{10^3}$. – marowl Dec 17 '19 at 21:08
  • Use $x=y\pmod{n}$ for $x=y\pmod{n}$. – Shaun Dec 17 '19 at 21:08
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    First, can you prove $11^{10}\equiv1\pmod{10^2}$? – Angina Seng Dec 17 '19 at 21:09
  • Yes and now I see why it works. – marowl Dec 17 '19 at 21:20
  • It is highly likely that a very similar question has been handled already. The close to 200 hundred questions lumped here and being linked to this mother thread cover a lot of techniques. Admittedly this congruence is a special case better handled by binomial formula as in the answers. That technique has also been used many times on the site already, but in my opinion high rep answerers would be better placed to find a good duplicate. – Jyrki Lahtonen Dec 17 '19 at 21:42
  • I will not use that mother thread as a duplicate target this time due to the special feature. Anyway, take a look Lord Shark the Unknown's second comment (third above this). I recommend induction on $a$. – Jyrki Lahtonen Dec 17 '19 at 21:44

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Maybe consider $(1+10)^{10^3} = \sum_{k=0}^{10^3} \binom{10^3}{k} 10^k = \sum_{k=0}^{3} \binom{10^3}{k} 10^k \mod 10^4$?

It is straightforward to verify by computation that $10^4 \mid \binom{10^3}{k} 10^k$ for $k=1,2,3$.

copper.hat
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  • @MaximilianJanisch: Sorry, forgot the mod part – copper.hat Dec 17 '19 at 21:15
  • Please, why the downvote? – copper.hat Dec 17 '19 at 21:33
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    This time I didn't downvote, but I might have. For we have hundreds of similar threads, and the high rep users arguably should know to search for one. My "favorite" duplicate target is not optimal this time, so I didn't dv, but my general sentiment against duplicates has not changed. This time I was busy trying to help the asker typeset the question properly, and started thinking about finding a duplicate later. May be others share my sentiments. Go figure? – Jyrki Lahtonen Dec 17 '19 at 21:48
  • @JyrkiLahtonen: Thanks for comment. I am not against downvotes, just without knowing why serves little purpose. I have had limited success in the past searching for answers. – copper.hat Dec 17 '19 at 21:51
  • I really wish there were a better way of handling these different ideas on how the site is to be used. The community is too large for a round table discussion, and passions run high as people with diametrically opposite points of view think they are right. Well, that's for meta. – Jyrki Lahtonen Dec 17 '19 at 21:58
  • @JyrkiLahtonen: I can appreciate the frustration, albeit I am a source of such. Generally my goal is to assist the OP, sometimes there is some personal enjoyment if the answer seems to be slightly different than expected... – copper.hat Dec 17 '19 at 22:00
  • When I wrote 'searching for answers' I meant 'searching for duplicates'. – copper.hat Dec 17 '19 at 22:14
  • @JyrkiLahtonen: I have taken your comment to heart, and am spending more time looking for duplicates before answering. However, I must confess that my success rate for locating suitable answers (that I am sure must exist) is pretty low. I have looked at the list of generalisations, etc. – copper.hat Dec 19 '19 at 19:03
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    Appreciated! Searching is not always easy. Have you tried Approach0? You can copy/paste a TeX-snippet there. It often works well, but is no wonderdrug. – Jyrki Lahtonen Dec 19 '19 at 19:32
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    People in this chat room can give more tips. I don't frequent it, but if you comment there some of the regulars can probably help. – Jyrki Lahtonen Dec 19 '19 at 19:36
  • @JyrkiLahtonen: Thanks. Interesting search engine! – copper.hat Dec 19 '19 at 20:50
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A generalized version of copper.hat's answer: \begin{align} 11^{10^a}\pmod{10^{a+1}}&= (10+1)^{10^a}\pmod{ 10^{a+1}} = \left[\sum_{i=0}^{10^a}\binom{10^a}{i}10^i\right] \pmod{ 10^{a+1}} \\ &= \left[\sum_{i=0}^{a}\binom{10^a}{i}10^i\right] \pmod{ 10^{a+1}}\\ & = \left[1+\sum_{i=1}^{a}\frac{10^{i-1}(10^a-1)\ldots (10^a-i)}{i!}10^{a+1}\right] \pmod{ 10^{a+1}}\\ &= 1. \end{align} Here, the last step follows because $\frac{10^{i-1}(10^a-1)\ldots (10^a-i)}{i!}$ is an integer. To prove this, we note that $\binom{10^a}{i}=\frac{10^{a}(10^a-1)\ldots (10^a-i)}{i!}$ is an integer. The factors of $i!$ that are cancelled with the factors of $10^a$ are multiples of $2$ and $5$. Clearly, the largest divisor of $i!$ that cancels with $10^a$ is less than $2^{i-1}5^{i-1}=10^{i-1}$. Therefore, $\frac{10^{i-1}(10^a-1)\ldots (10^a-i)}{i!}$ is also an integer.

Explorer
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  • Whys is $\left[\sum_{i=0}^{10^a}\binom{10^a}{i}10^i\right] \bmod 10^{a+1} = \left[\sum_{i=0}^{a}\binom{10^a}{i}10^i\right] \bmod 10^{a+1}$ – marowl Dec 18 '19 at 12:47
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    $\binom{10^a}{i}10^i$ is a multiple of $10^{a+1}$ for $i>a$, so $\binom{10^a}{i}10^i\bmod 10^{a+1}=0$. – Explorer Dec 18 '19 at 20:38