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.

## Comments

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

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 · 1 year, 5 months ago

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@Akshat Sharda – Chinmay Sangawadekar · 1 year, 5 months ago

Can you prove the Bonus ?Log in to reply

@Akshat Sharda – Chinmay Sangawadekar · 1 year, 5 months ago

Nice oneLog in to reply

@Otto Bretscher , I am still waiting for your reply – Chinmay Sangawadekar · 1 year, 5 months ago

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– Otto Bretscher · 1 year, 5 months ago

I'm afraid that this congruency does not hold in general. As a counterexample, consider \(a=10\) and \(n=11\).Log in to reply

– Chinmay Sangawadekar · 1 year, 5 months ago

a should not be a 2 digit number...Log in to reply

– Otto Bretscher · 1 year, 5 months ago

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 \(3\).Log in to reply

– Chinmay Sangawadekar · 1 year, 5 months ago

Thanks for pointing my mistake ! you may also take a look at Another Interesting fact , link is provided above...Log in to reply

– Ankit Kumar Jain · 1 year, 5 months ago

Its true for all values of a such that a is between 1 and 9Log in to reply

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 · 1 year, 5 months ago

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I'll try to prove this :) – Harsh Shrivastava · 1 year, 5 months ago

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, lets see if your method is same as mine or different – Chinmay Sangawadekar · 1 year, 5 months ago

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How did you proved it? – Harsh Shrivastava · 1 year, 5 months ago

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– Chinmay Sangawadekar · 1 year, 5 months ago

I am trying to disprove it for three. I did it by taking patterns of remainder.Log in to reply

– Akshat Sharda · 1 year, 5 months ago

It is correct for all prime numbers greater than 7.Log in to reply

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– Akshat Sharda · 1 year, 5 months ago

Will it be true here \( 33 \pmod 2\) ? No.Log in to reply

– Chinmay Sangawadekar · 1 year, 5 months ago

Yes you are right then :P I forgot to see cases ,Log in to reply

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BTW i proved it by one other way ! by the fremats little theorem .... – Chinmay Sangawadekar · 1 year, 5 months ago

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– Akshat Sharda · 1 year, 5 months ago

Aah! Yes... One can prove it by that too!Log in to reply

@Harsh Shrivastava did it ? – Chinmay Sangawadekar · 1 year, 5 months ago

Your proof also reflects Fermat's little theorem if noticed clearly , nice job mate , now we have got two good proofs :)))) I wonder howLog in to reply

– Harsh Shrivastava · 1 year, 5 months ago

I also used FLT.Log in to reply

@Harsh Shrivastava , @Akshat Sharda – Chinmay Sangawadekar · 1 year, 5 months ago

ah we both did the same then !!! Lets create a wiki page for this matesLog in to reply

– Akshat Sharda · 1 year, 5 months ago

Yeah! We can make a nice wiki! When to start?Log in to reply

– Chinmay Sangawadekar · 1 year, 5 months ago

Now ?!Log in to reply

– Akshat Sharda · 1 year, 5 months ago

Yes! Why not! Who's gonna start it? You?:)Log in to reply

Sorry guyz. – Harsh Shrivastava · 1 year, 5 months ago

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– Akshat Sharda · 1 year, 5 months ago

Join us when you have time :)Log in to reply

– Harsh Shrivastava · 1 year, 5 months ago

Sure!Log in to reply

@Akshat Sharda and @Harsh Shrivastava are you both moderators ? – Chinmay Sangawadekar · 1 year, 5 months ago

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 ...... BTWLog in to reply

Well! Can you post other patterns as a note(s)? – Akshat Sharda · 1 year, 5 months ago

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– Chinmay Sangawadekar · 1 year, 5 months ago

Of course ....Log in to reply

– Harsh Shrivastava · 1 year, 5 months ago

Are you in class 10?Log in to reply

– Akshat Sharda · 1 year, 5 months ago

Yes! I'm in 10th standard.Log in to reply

I'll try to prove 'em when i am free :) – Harsh Shrivastava · 1 year, 5 months ago

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@Harsh Shrivastava – Chinmay Sangawadekar · 1 year, 5 months ago

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 youLog in to reply

– Akshat Sharda · 1 year, 5 months ago

Are you on slack?Log in to reply

andbeyond – Chinmay Sangawadekar · 1 year, 5 months agoLog in to reply

– Akshat Sharda · 1 year, 5 months ago

Let's talk there.Log in to reply

– Harsh Shrivastava · 1 year, 5 months ago

I am not.Log in to reply