×

# 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.

1 year, 10 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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

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

- 1 year, 10 months ago

Can you prove the Bonus ?@Akshat Sharda

- 1 year, 10 months ago

Nice one @Akshat Sharda

- 1 year, 10 months ago

- 1 year, 10 months ago

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

- 1 year, 10 months ago

a should not be a 2 digit number...

- 1 year, 10 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 $$3$$.

- 1 year, 10 months ago

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

- 1 year, 10 months ago

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

- 1 year, 10 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...

- 1 year, 10 months ago

Comment deleted Mar 13, 2016

Nice property!

I'll try to prove this :)

- 1 year, 10 months ago

I have already half proven it :)

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

- 1 year, 10 months ago

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

How did you proved it?

- 1 year, 10 months ago

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

- 1 year, 10 months ago

It is correct for all prime numbers greater than 7.

- 1 year, 10 months ago

Comment deleted Mar 13, 2016

Will it be true here $$33 \pmod 2$$ ? No.

- 1 year, 10 months ago

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

- 1 year, 10 months ago

Comment deleted Mar 13, 2016

Comment deleted Mar 13, 2016

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 ....

- 1 year, 10 months ago

Aah! Yes... One can prove it by that too!

- 1 year, 10 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 how @Harsh Shrivastava did it ?

- 1 year, 10 months ago

I also used FLT.

- 1 year, 10 months ago

ah we both did the same then !!! Lets create a wiki page for this mates @Harsh Shrivastava , @Akshat Sharda

- 1 year, 10 months ago

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

- 1 year, 10 months ago

Now ?!

- 1 year, 10 months ago

Yes! Why not! Who's gonna start it? You?:)

- 1 year, 10 months ago

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

Sorry guyz.

- 1 year, 10 months ago

- 1 year, 10 months ago

Sure!

- 1 year, 10 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 ...... BTW @Akshat Sharda and @Harsh Shrivastava are you both moderators ?

- 1 year, 10 months ago

Yes! I'm a moderator.

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

- 1 year, 10 months ago

Of course ....

- 1 year, 10 months ago

Are you in class 10?

- 1 year, 10 months ago

Yes! I'm in 10th standard.

- 1 year, 10 months ago

Please post more of these congruences as notes.

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

- 1 year, 10 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 you @Harsh Shrivastava

- 1 year, 10 months ago

Are you on slack?

- 1 year, 10 months ago

yes infinityandbeyond

- 1 year, 10 months ago

Let's talk there.

- 1 year, 10 months ago

I am not.

- 1 year, 10 months ago