Interesting fact

I have found that ,

aaa…⏟n times≡amod  n\Large \underbrace{aaa\ldots }_{n \text{ times}} \equiv a \mod n

∀ n∈{ Primes}\forall \text{ n} \in \left \{ \text{ Primes} \right \} also aa belongs to 1 to 9 and also 0.

and n≥7n\geq7

BonusBonus forfor readerreader

Disprove this for n<7n <7

NoteNote

Also see Another Interesting fact

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

#NumberTheory

Note by A Former Brilliant Member
5 years, 3 months ago

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1 vote

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Comments

aaa…aa⏟n times≡a(modn)aa…⏟n times−a=aaa…a⏟n−1 times0≡0(modn)\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,

aaa…aa⏟n−1 times≡0(modn),∀n∈ Prime≥7a⋅10n−1−19(modn)\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 nn is a prime number,

10ϕ(n)=10n−1≡1(modn)∴a⋅10n−1−19≡a⋅1−19=0(modn)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

aaa…aa⏟n times≡a(modn)a⋅10n−19(modn)\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,

10n≡10(modn)a⋅10n−19≡a⋅10−19=a(modn)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 - 5 years, 3 months ago

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Nice one @Akshat Sharda

A Former Brilliant Member - 5 years, 3 months ago

Can you prove the Bonus ?@Akshat Sharda

A Former Brilliant Member - 5 years, 3 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...

Ankit Kumar Jain - 5 years, 2 months ago

@Otto Bretscher , I am still waiting for your reply

A Former Brilliant Member - 5 years, 2 months ago

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

Otto Bretscher - 5 years, 2 months ago

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

Ankit Kumar Jain - 5 years, 2 months ago

a should not be a 2 digit number...

A Former Brilliant Member - 5 years, 2 months ago

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@A Former Brilliant Member – You are saying aa is a natural number... so you need to change that in the statement. Essentially you are saying this is true for a=1a=1, we can then multiply with a=2..9a= 2..9. The case a=1a=1 is trivial as long as the prime is not 33.

Otto Bretscher - 5 years, 2 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 Former Brilliant Member - 5 years, 2 months ago
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