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Interesting fact


I have found that ,

\[\Large \underbrace{aaa\ldots }_{n \text{ times}} \equiv a \mod n\]

\[\forall \text{ n} \in \left \{ \text{ Primes} \right \}\] also \(a\) belongs to 1 to 9 and also 0.

and \(n\geq7\)


\(Bonus\) \(for\) \(reader\)

Disprove this for \(n <7\)


\(Note\)

Also see Another Interesting fact

\(\overline {aaa...}\) is a single number.

Note by Chinmay Sangawadekar
8 months, 4 weeks ago

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\[\underbrace{aaa\ldots aa}_{n \text{ times} } \equiv a \pmod n \\ \underbrace{aa \ldots }_{n \text{ times}}-a=\underbrace{aaa\ldots a}_{n-1\text{ times}}0 \equiv 0 \pmod n\]

Now we have to prove that,

\[\underbrace{aaa\ldots aa}_{n-1 \text{ times}}\equiv 0 \pmod n , \quad \forall n \in \text{ Prime≥7} \\ a\cdot \frac{10^{n-1}-1}{9} \pmod n \]

As \(n\) is a prime number,

\[10^{\phi(n)}=10^{n-1}\equiv 1\pmod n \\ \therefore a\cdot \frac{10^{n-1}-1}{9} \equiv a\cdot \frac{1-1}{9}=0 \pmod n\]


\[\underbrace{aaa\ldots aa}_{n \text{ times} } \equiv a \pmod n \\ a\cdot \frac{10^n -1}{9}\pmod n \]

By Fermat's Little Theorem,

\[10^n \equiv 10 \pmod n \\ a\cdot \frac{10^n -1}{9}\equiv a\cdot \frac{10 -1}{9}= a \pmod n\]

• QED Akshat Sharda · 8 months, 4 weeks ago

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@Akshat Sharda Can you prove the Bonus ?@Akshat Sharda Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Akshat Sharda Nice one @Akshat Sharda Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Otto Bretscher , I am still waiting for your reply Chinmay Sangawadekar · 8 months, 2 weeks ago

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@Chinmay Sangawadekar I'm afraid that this congruency does not hold in general. As a counterexample, consider \(a=10\) and \(n=11\). Otto Bretscher · 8 months, 2 weeks ago

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@Otto Bretscher a should not be a 2 digit number... Chinmay Sangawadekar · 8 months, 2 weeks ago

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@Chinmay Sangawadekar You are saying \(a\) is a natural number... so you need to change that in the statement. Essentially you are saying this is true for \(a=1\), we can then multiply with \(a= 2..9\). The case \(a=1\) is trivial as long as the prime is not \(3\). Otto Bretscher · 8 months, 2 weeks ago

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@Otto Bretscher Thanks for pointing my mistake ! you may also take a look at Another Interesting fact , link is provided above... Chinmay Sangawadekar · 8 months, 2 weeks ago

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@Otto Bretscher Its true for all values of a such that a is between 1 and 9 Ankit Kumar Jain · 8 months, 2 weeks ago

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I guess this is true for all primes except 3...To disprove this for 3....We see that aaa will always be divisible by 3 because for divisibility by 3 we need to check the sum of digits . The sum of the digits for the given number will be 3a which is divisible by 3..Hence the given number will always be congruent 0 mod 3... Ankit Kumar Jain · 8 months, 3 weeks ago

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@Chinmay Sangawadekar Nice property!

I'll try to prove this :) Harsh Shrivastava · 8 months, 4 weeks ago

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@Harsh Shrivastava I have already half proven it :)

, lets see if your method is same as mine or different Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Chinmay Sangawadekar I think i proved it.I used simple algebraic manipulation and properties of modular arithmetic.

How did you proved it? Harsh Shrivastava · 8 months, 4 weeks ago

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@Harsh Shrivastava I am trying to disprove it for three. I did it by taking patterns of remainder. Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Chinmay Sangawadekar It is correct for all prime numbers greater than 7. Akshat Sharda · 8 months, 4 weeks ago

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@Chinmay Sangawadekar Will it be true here \( 33 \pmod 2\) ? No. Akshat Sharda · 8 months, 4 weeks ago

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@Akshat Sharda Yes you are right then :P I forgot to see cases , Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Akshat Sharda Obviously it is correct !!!I actually got inspired by your problem .... wait I will mention your name there ,,,,@Akshat Sharda ,

BTW i proved it by one other way ! by the fremats little theorem .... Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Chinmay Sangawadekar Aah! Yes... One can prove it by that too! Akshat Sharda · 8 months, 4 weeks ago

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@Akshat Sharda Your proof also reflects Fermat's little theorem if noticed clearly , nice job mate , now we have got two good proofs :)))) I wonder how @Harsh Shrivastava did it ? Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Chinmay Sangawadekar I also used FLT. Harsh Shrivastava · 8 months, 4 weeks ago

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@Harsh Shrivastava ah we both did the same then !!! Lets create a wiki page for this mates @Harsh Shrivastava , @Akshat Sharda Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Chinmay Sangawadekar Yeah! We can make a nice wiki! When to start? Akshat Sharda · 8 months, 4 weeks ago

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@Akshat Sharda Now ?! Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Chinmay Sangawadekar Yes! Why not! Who's gonna start it? You?:) Akshat Sharda · 8 months, 4 weeks ago

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@Akshat Sharda I am sorry i can't help now 'coz tmmrw is my FIT exam, havta study for that.

Sorry guyz. Harsh Shrivastava · 8 months, 4 weeks ago

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@Harsh Shrivastava Join us when you have time :) Akshat Sharda · 8 months, 4 weeks ago

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@Akshat Sharda Sure! Harsh Shrivastava · 8 months, 4 weeks ago

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@Akshat Sharda Sure , well I have got more such patterns to be proved , we will prove them first and then we could categorize them into a new types of congruences ...... BTW @Akshat Sharda and @Harsh Shrivastava are you both moderators ? Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Chinmay Sangawadekar Yes! I'm a moderator.

Well! Can you post other patterns as a note(s)? Akshat Sharda · 8 months, 4 weeks ago

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@Akshat Sharda Of course .... Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Akshat Sharda Are you in class 10? Harsh Shrivastava · 8 months, 4 weeks ago

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@Harsh Shrivastava Yes! I'm in 10th standard. Akshat Sharda · 8 months, 4 weeks ago

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@Chinmay Sangawadekar Please post more of these congruences as notes.

I'll try to prove 'em when i am free :) Harsh Shrivastava · 8 months, 4 weeks ago

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@Harsh Shrivastava For sure , but mate many of them are proved by me before,...... but still i will post for new and brilliant proofs by brilliant guys like you @Harsh Shrivastava Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Chinmay Sangawadekar Are you on slack? Akshat Sharda · 8 months, 4 weeks ago

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@Akshat Sharda yes infinityandbeyond Chinmay Sangawadekar · 8 months, 4 weeks ago

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@Chinmay Sangawadekar Let's talk there. Akshat Sharda · 8 months, 4 weeks ago

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@Chinmay Sangawadekar I am not. Harsh Shrivastava · 8 months, 4 weeks ago

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