Problem 1830

Solution:

The answer is $373$.

Let us find all numbers that satisfy the conditions of the problem.

First of all, notice that if we have found all acceptable numbers with $n$ digits, then an acceptable number with $n+1$ digits must necessarily contain one of the numbers with $n$ digits. In particular, if we discover that no number with $k$ digits exists, then no acceptable number with more than $k$ digits will exist.

The acceptable numbers with one digit are the one-digit primes, that is $2, 3, 5$ and $7$. Notice, however, that the digit $2$ and the digit $5$ can only appear at the beginning of an acceptable number (otherwise the number would have a subsequence divisible by $2$ or by $5$ respectively).

The acceptable numbers with two digits are therefore $23, 37, 53, 73$.

Let us now look at the numbers with $3$ digits. None can end in $23$ or in $53$. Then $237$, $537$, $273$ and $573$ are not acceptable since they are multiples of $3$. No number can begin with the same pair of digits, otherwise it would have a subsequence that is a multiple of $11$ (for example in $773$ the subsequence $77$ is a multiple of $11$). What remains to be checked are $373$, which turns out to be prime (and therefore acceptable), and $737$, which is a multiple of $11$.

Let us now see whether there can be numbers with $4$ digits. We only need to check $4$ of them: $2373$, $3373$, $5373$ and $7373$. The first and the third contain a subsequence that is a multiple of $3$. The second contains a subsequence that is a multiple of $11$, and the last turns out to be a multiple of $101$. Therefore no acceptable numbers with $4$ digits exist and, for the reason stated at the beginning, no acceptable numbers with more than $4$ digits will exist either. The sought code is therefore $373$.