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 \(1\).

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.

\(10101\) 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.

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.

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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.

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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.

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Here's a related problem:

How many numbers of the form \( 1010\ldots 0101 \) are prime?

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only \(101\) is a prime

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Some initial thoughts:

\(101\) 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.

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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.

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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 :)

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