# A nice property of primes

A while ago, a friend of mine told me that he once knew a number with the following property: "You can prepend any number before that number and will always obtain a prime number."

Of course, he was just kidding: Such a number can't exist, because a number is divisible by 3 if and only if its digit sum is divisible by 3. So, it is no problem to find a number to prepend such the digit sum of the resulting number is divisible by 3. That implies that the resulting number is not a prime.

However, it's possible to prove a similar theorem... We prove the following: For any number \(m\), there is a number with \(m\) digits, such there are infinitely many primes ending with this \(m\)-digit number.

To prove this, let us first consider the case \(m=1\). This means, that we want to show that there is a number \(n \in \left\{ 0,1,2,3,4,5,6,7,8,9 \right\}\) such there are infinitely many primes ending in this number \(n\).

It's a well-known fact that there are infinitely many primes. Thus, due to the pigeonhole principle, there must be a number in \(n \in \left\{ 0,1,2,3,4,5,6,7,8,9 \right\}\) to occur infinitely many often. Thus, such a number exists.

Now, it's no big step to generalize this to more digits: Assume we found our number \(n\) with the desired property. I.e. there are infinitely many primes ending with \(n\). Now, we "cross out" all primes ending *not* in \(n\). So, we are now only considering the infinitely many primes ending in \(n\)

But now, according to the pigeonhole principle, there must be a number \(n' \in \left\{ 0,1,2,3,4,5,6,7,8,9 \right\}\) such that there are infinitely many primes whose next-to-last digit is \(n'\). That means, there are infinitely many primes ending in the digits \(n' n\).

Now, we iterate the procedure: We cross out all primes not ending in \(n' n\). This yields - again - infinitely many primes ending in \(n' n\). So we can argument the same way as before: There must be - due to pigeonhole principle - a number \(n''\) such that infinitely many primes end in \(n'' n' n\).

We can continue this procedure as long as we want - since there are infinitely many primes.

So, this means, there are arbitrarily large numbers \(n_m n_{m-1} n_{m-2} \dots n_2 n_1 n_0\) (concatenated digits \(n_i\)) such that there are infinitely many primes ending with exactly the digits \(n_m n_{m-1} n_{m-2} \dots n_2 n_1 n_0\).

I like this proof because it is - once again - a proof of existence that does not explicitly construct a solution to the problem, but just shows that there is one.