Primes in random strings of numbers? Given any number with random digits, can you always arrange sequential primes in such a way where that number can be formed?
Example:
$31017643$ is broken into $3,101,7,643$ given all of those are prime.
Of course we can eliminate every number that ends in 0, 2, 4, 5, 6, and 8.
I wrote some code to test this, and it gets out of hand very quickly.
Has someone 200 years ago already solved this? How would I go about proving or disproving this idea?
 A: The question was already answered in the comments, but I decided to post a kind of an extended answer.
First of all, I don't think we should exclude numbers that end in 2, because e.g. 32 can be broken into 3 and 2, both of which are primes. The same goes for numbers ending in 5.
Even without this suggestion, here are all the numbers below 100 that end in 1, 3, 7, or 9, and that cannot be represented as a concatenation of primes in base 10: 9, 21, 39, 49, 51, 63, 69, 81, 87, 91, 93, 99. Here is a C++ program that computes them by simultaneously maintaining a list of prime numbers and a bitmask telling which numbers are prime.
Using the same approach (and switching to 64-bit integers; here's the full code: it won't run online due to time limit restrictions - it ran for 16 minutes on my i9 machine), I computed all numbers less than one billion that cannot be broken into primes (including those ending in 0, 2, 4, 5, 6, 8): there happen to be 711970152 of them (about 71%).
So, if we assume that numbers ending in 0, 2, 4, 5, 6, 8 cannot be broken into primes (as I've said above, this isn't entirely true for 2), among the remaining 40% of all numbers about 3/4 can be broken into primes.
If instead we only exclude number that definitely cannot be broken into primes, i.e. numbers ending in 0, 4, 6, or 8, that constitute 40% of all numbers, among the remaining 60% about half (slightly less than a half, to be precise) of these numbers can be broken into primes, and about half cannot be.
