### 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.