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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.

6 months, 2 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 · 6 months, 2 weeks ago

Can you prove the Bonus ?@Akshat Sharda · 6 months, 2 weeks ago

Nice one @Akshat Sharda · 6 months, 2 weeks ago

@Otto Bretscher , I am still waiting for your reply · 6 months, 1 week ago

I'm afraid that this congruency does not hold in general. As a counterexample, consider $$a=10$$ and $$n=11$$. · 6 months, 1 week ago

a should not be a 2 digit number... · 6 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 $$3$$. · 6 months ago

Thanks for pointing my mistake ! you may also take a look at Another Interesting fact , link is provided above... · 6 months ago

Its true for all values of a such that a is between 1 and 9 · 6 months ago

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... · 6 months, 1 week ago

Comment deleted 6 months ago

Nice property!

I'll try to prove this :) · 6 months, 2 weeks ago

I have already half proven it :)

, lets see if your method is same as mine or different · 6 months, 2 weeks ago

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

How did you proved it? · 6 months, 2 weeks ago

I am trying to disprove it for three. I did it by taking patterns of remainder. · 6 months, 2 weeks ago

It is correct for all prime numbers greater than 7. · 6 months, 2 weeks ago

Comment deleted 6 months ago

Will it be true here $$33 \pmod 2$$ ? No. · 6 months, 2 weeks ago

Yes you are right then :P I forgot to see cases , · 6 months, 2 weeks ago

Comment deleted 6 months ago

Comment deleted 6 months ago

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 .... · 6 months, 2 weeks ago

Aah! Yes... One can prove it by that too! · 6 months, 2 weeks ago

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 ? · 6 months, 2 weeks ago

I also used FLT. · 6 months, 2 weeks ago

ah we both did the same then !!! Lets create a wiki page for this mates @Harsh Shrivastava , @Akshat Sharda · 6 months, 2 weeks ago

Yeah! We can make a nice wiki! When to start? · 6 months, 2 weeks ago

Now ?! · 6 months, 2 weeks ago

Yes! Why not! Who's gonna start it? You?:) · 6 months, 2 weeks ago

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

Sorry guyz. · 6 months, 2 weeks ago

Join us when you have time :) · 6 months, 2 weeks ago

Sure! · 6 months, 2 weeks 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 ...... BTW @Akshat Sharda and @Harsh Shrivastava are you both moderators ? · 6 months, 2 weeks ago

Yes! I'm a moderator.

Well! Can you post other patterns as a note(s)? · 6 months, 2 weeks ago

Of course .... · 6 months, 2 weeks ago

Are you in class 10? · 6 months, 2 weeks ago

Yes! I'm in 10th standard. · 6 months, 2 weeks ago

Please post more of these congruences as notes.

I'll try to prove 'em when i am free :) · 6 months, 2 weeks 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 you @Harsh Shrivastava · 6 months, 2 weeks ago

Are you on slack? · 6 months, 2 weeks ago

yes infinityandbeyond · 6 months, 2 weeks ago

Let's talk there. · 6 months, 2 weeks ago