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.

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## 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

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Nice one @Akshat Sharda

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Can you prove the Bonus ?@Akshat Sharda

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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...

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@Otto Bretscher , I am still waiting for your reply

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

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

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a should not be a 2 digit number...

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$a$ 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 $3$.

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