Can you prove that if \[\frac{10^{n}-1}{9} \] is a prime number then 'n' is also a prime number?

Can you prove that if \[\frac{10^{n}-1}{9} \] is a prime number then 'n' is also a prime number?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestUmm.. Yes, I can :)

Let's prove the contrapositive. In other words, we're going to prove that if \(n\) is not a prime number, then \(\frac{10^9-1}{9}\) will never be a prime number.

Let \(n\) be equal to \(ab\) where \(a\) and \(b\) are two positive integers greater than \(1\).

Now notice that

\[\frac{10^{ab}-1}{9}=\frac{10^a-1}{9}\times (1+10^a+10^{2a}+10^{3a}+\cdots 10^{a(b-1)})\]

which is definitely not a prime number.

A fun fact: primes in the form of \(\frac{10^n-1}{9}\) are called repunit primes. – Mursalin Habib · 3 years, 6 months ago

Log in to reply

– Rahul Saha · 3 years, 6 months ago

This is true for any base.A repunit is of the form \[\dfrac{b^n-1}{b-1}\] in base \(b\).The above result holds in any base that a repunit is a prime if \(n\) is a prime.Log in to reply

Here's a related problem:

How many numbers of the form \( 1010\ldots 0101 \) are prime? – Calvin Lin Staff · 3 years, 6 months ago

Log in to reply

– Sagnik Saha · 3 years, 1 month ago

only \(101\) is a primeLog in to reply

\(101\) is prime.

\(10101\) is a multiple of \(3\). (This holds for all other cases where the number of \(1\)s are a multiple of \(3\))

\(1010101\) is the product of at least two factors, namely \(101\) and \(10001\). This holds for all greater cases where the number of \(1\)s are even.

Haven't thought of when the number of \(1\)s are odd though. – Michael Tong · 3 years, 6 months ago

Log in to reply

– Patrick Corn · 3 years, 6 months ago

When the number of \( 1 \)s is odd, the number is divisible by the repunit you get by taking out all of the zeroes; e.g. \( 10101 \) is divisible by \( 111 \), \( 101010101 \) is divisible by \( 11111 \), and so on. This is because \( 1+10^2 + 10^4 + \cdots + 10^{2n} = (1+10+10^2+\cdots+10^n)(1-10+10^2-\cdots+10^n). \) Note that the last sign in that alternating factor is a \( + \) only if \( n \) is odd.Log in to reply

Log in to reply

Yeah the one way of doing this. I was not sure how to put proof related questions on this forum. But this is good level 4 question otherwise :) – Vikram Pandya · 3 years, 6 months ago

Log in to reply