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 $n\geq7$

Disprove this for $n <7$

**$Note$**

Also see Another Interesting fact

$\overline {aaa...}$ is a single number.

No vote yet

1 vote

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest@Otto Bretscher , I am still waiting for your reply

Log in to reply

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

Log in to reply

a should not be a 2 digit number...

Log in to reply

$a$ 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 $3$.

You are sayingLog in to reply

Log in to reply

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

Log 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...

Log in to reply

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

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

Log in to reply

Can you prove the Bonus ?@Akshat Sharda

Log in to reply

Nice one @Akshat Sharda

Log in to reply