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 A Brilliant Member
2 years, 7 months 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 - 2 years, 7 months ago

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Can you prove the Bonus ?@Akshat Sharda

A Brilliant Member - 2 years, 7 months ago

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Nice one @Akshat Sharda

A Brilliant Member - 2 years, 7 months ago

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@Otto Bretscher , I am still waiting for your reply

A Brilliant Member - 2 years, 6 months ago

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

Otto Bretscher - 2 years, 6 months ago

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a should not be a 2 digit number...

A Brilliant Member - 2 years, 6 months ago

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@A Brilliant Member 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 - 2 years, 6 months 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...

A Brilliant Member - 2 years, 6 months ago

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Its true for all values of a such that a is between 1 and 9

Ankit Kumar Jain - 2 years, 6 months 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 - 2 years, 7 months ago

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Comment deleted Mar 13, 2016

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Nice property!

I'll try to prove this :)

Harsh Shrivastava - 2 years, 7 months ago

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I have already half proven it :)

, lets see if your method is same as mine or different

A Brilliant Member - 2 years, 7 months ago

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@A Brilliant Member I think i proved it.I used simple algebraic manipulation and properties of modular arithmetic.

How did you proved it?

Harsh Shrivastava - 2 years, 7 months ago

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@Harsh Shrivastava I am trying to disprove it for three. I did it by taking patterns of remainder.

A Brilliant Member - 2 years, 7 months ago

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@A Brilliant Member It is correct for all prime numbers greater than 7.

Akshat Sharda - 2 years, 7 months ago

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Comment deleted Mar 13, 2016

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@A Brilliant Member Will it be true here \( 33 \pmod 2\) ? No.

Akshat Sharda - 2 years, 7 months ago

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@Akshat Sharda Yes you are right then :P I forgot to see cases ,

A Brilliant Member - 2 years, 7 months ago

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Comment deleted Mar 13, 2016

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Comment deleted Mar 13, 2016

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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 ....

A Brilliant Member - 2 years, 7 months ago

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@A Brilliant Member Aah! Yes... One can prove it by that too!

Akshat Sharda - 2 years, 7 months 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 ?

A Brilliant Member - 2 years, 7 months ago

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@A Brilliant Member I also used FLT.

Harsh Shrivastava - 2 years, 7 months 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

A Brilliant Member - 2 years, 7 months ago

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@A Brilliant Member Yeah! We can make a nice wiki! When to start?

Akshat Sharda - 2 years, 7 months ago

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@Akshat Sharda Now ?!

A Brilliant Member - 2 years, 7 months ago

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@A Brilliant Member Yes! Why not! Who's gonna start it? You?:)

Akshat Sharda - 2 years, 7 months 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 - 2 years, 7 months ago

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@Harsh Shrivastava Join us when you have time :)

Akshat Sharda - 2 years, 7 months ago

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@Akshat Sharda Sure!

Harsh Shrivastava - 2 years, 7 months 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 ?

A Brilliant Member - 2 years, 7 months ago

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@A Brilliant Member Yes! I'm a moderator.

Well! Can you post other patterns as a note(s)?

Akshat Sharda - 2 years, 7 months ago

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@Akshat Sharda Of course ....

A Brilliant Member - 2 years, 7 months ago

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@Akshat Sharda Are you in class 10?

Harsh Shrivastava - 2 years, 7 months ago

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@Harsh Shrivastava Yes! I'm in 10th standard.

Akshat Sharda - 2 years, 7 months ago

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@A Brilliant Member Please post more of these congruences as notes.

I'll try to prove 'em when i am free :)

Harsh Shrivastava - 2 years, 7 months 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

A Brilliant Member - 2 years, 7 months ago

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@A Brilliant Member Are you on slack?

Akshat Sharda - 2 years, 7 months ago

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@Akshat Sharda yes infinityandbeyond

A Brilliant Member - 2 years, 7 months ago

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@A Brilliant Member Let's talk there.

Akshat Sharda - 2 years, 7 months ago

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@A Brilliant Member I am not.

Harsh Shrivastava - 2 years, 7 months ago

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