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Let $n$ be a positive integer where $n$ and $10$ are coprime number. Prove that then the last 3 digits of $n^\text{101}$ will be equal to the last 3 digits of $n$.

SOURCE: Bangladesh Math Olympiad

I basically know how to find the last digit of a base with a large exponent. Whatever the exponent in the event of one digit such as $1, 5, 6$ and $9$ will be, the last digit of the number will always result in that specific number which we have used.

So, the last 3 digit of $n$ can be constructed in $^4P_3$ ways with having that 4 digits $1, 5, 6$ and $9$ respectively. But how to determine only that distince value of $n$ and justify that the number and $10$ are coprime number>

The instruction of any kind of reference or any good books related with decimal expansion or number theory will be very helpful for me in the case of a beginner and some basic conception because there is no availability of satisfactory books written on number theory in our country which I have read some. Thanks in advance.

Bill Dubuque
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Anirban Niloy
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    Do you know modular arithmetic? Euler's theorem? – Greg Martin Feb 22 '19 at 05:06
  • @GregMartin To tell about Modular Arithmetic, I have to say that I know few. In the case of Euler's theorem, Absolutely nope. But I heard the name. Because I'm only a 10th grader. Haha. I will be highly gladful to you if you provide me with any reference or books or its conception. – Anirban Niloy Feb 22 '19 at 05:08
  • Everything you want to know about this theme is covered in that umbrella thread (as well as other threads linked to it). Also, if you are serious about participating in a contest, you need to start working on the problems yourself, and showing the effort here. – Jyrki Lahtonen Feb 22 '19 at 06:14
  • @JyrkiLahtonen The contest is too far. Therefore, I have no intention to participate in it. I think my efforts shouldn't be showed as they are so ugly and unacceptable. I ask for clearing the point where I got stuck and understood nothing about how to approach or deny the problem. And from which angle, it looks like as it is a duplicate one? – Anirban Niloy Feb 22 '19 at 06:40
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    I have two problems with that A) You should still check for duplicates. B) IMHO (debatable) the plan not to participate somewhat invalidates the context provided by the question coming from a contest. You should then IMHO provide other context. The reason is that the math underlying contests is not that hot, and has been explained on our site may times over. – Jyrki Lahtonen Feb 22 '19 at 06:45
  • If you prefer, you are welcome to close this as an exact duplicate of this. 3 seconds of your precious time with Approach0. – Jyrki Lahtonen Feb 22 '19 at 06:56
  • @JyrkiLahtonen I think so. I became lost after a nice debate. Yeah, I am bound to close my question as an exact duplicate because at last you gave a precise link where I discovered my question as same as was there. – Anirban Niloy Feb 22 '19 at 07:39
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    Although I will (modestly) add that my solution below will not be found at the earlier question. – Robert Shore Feb 22 '19 at 08:39
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    Have you studied group theory? In particular can you understand RS's answer? If not you shouldn't have changed the number-theory tag to group theory (maybe RS's remark misled you). The problem can be viewed from either standpoint (among others too). – Bill Dubuque Feb 23 '19 at 00:50
  • @BillDubuque Nope. To tell about Robert's solution, I have to say 'absolutely not.' But this can be helpful for other member in this community. Any standard reference or suggested books of group theory for learning? I actually know nothing!!! – Anirban Niloy Feb 23 '19 at 01:27

2 Answers2

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This is a group theory question, not a number theory question. You want to show that any element of the multiplicative group $(\Bbb Z/1000)^*$ has order dividing $100$.

It's fairly easy to see that the order of this group is $400$. Also, both $501$ and $999$ have order $2$ in this group (and therefore $499$ also has order $2$). Any finite abelian group is a direct product of cyclic groups, so if we can find one more element of order $2$, we'll know that every cyclic component must have order dividing $100$ and we'll be done. (Based on what we know so far, it's possible that the group is $\Bbb Z / 2 \times \Bbb Z / 200$.) We also find that $251$ (and therefore $749$) has order $2$ in the group and we are done.

Robert Shore
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  • Perusing the comments shows it is unlikely that the OP knows any group theory. Also you will confuse beginners with opinions like that in your first sentence (this caused the OP to delete the number-theory tag and add the group-theory tag). – Bill Dubuque Feb 23 '19 at 00:58
  • @BillDubuque You have said the speech of my mind. I changed the tag without reason though I don't know specially their literally meaning. A perfect condolence for the beginners and novice!! – Anirban Niloy Feb 23 '19 at 01:31
  • I posted the answer before those comments were up. Nevertheless, you’re correct that I didn’t intend to suggest that the tags were in any way inappropriate. I was simply saying that the group theory solution was, for me, the intuitive approach. – Robert Shore Feb 23 '19 at 08:50
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We need to prove that $n^{101}-n$ is divisible by $1000,$ for which it's enough to prove that $$n^{100}\equiv1\pmod{1000}.$$ Now, since $n^5\equiv n(\mod5),$ we obtain $n^4\equiv1\pmod5$ which gives $$n^{100}-1=(n^4-1)(n^{96}+n^{92}+...+1)=(n^4-1)(\left(n^4\right)^{24}+\left(n^4\right)^{23}+...+1)\equiv0\pmod{125}$$ and since $n$ is an odd number, we obtain: $$n^{100}-1=(n^{50}-1)(n^{50}+1)=(n^{25}-1)(n^{25}+1)(n^{50}+1)\equiv0\pmod8$$ and we are done!

Michael Rozenberg
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