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If by "this" you mean the 83-digit prime in the question, then no. There's no indication that there are no other 83-digit bitruncatable primes - that's just the
by dmurray 8y ago
If by "this" you mean the 83-digit prime in the question, then no. There's no indication that there are no other 83-digit bitruncatable primes - that's just the only one the author had found.
If you mean the sole 97-digit solution given in the comments, the implication that all the bitruncatable primes have been found and enumerated by your method, and the assertion that there are no such 99-digit primes - then yes, that looks pretty legit, though I'd want to see an independent verification before accepting it as a proof, since computer programs have bugs.
- jemfinch 8y agoI've written some simple code and verified that commenter's numbers through 19 digits. Edit: parallelized the code and proved through 19 digits. That's where I'm stopping due to the uint64 space limitations.
- dmurray 8y agoIt looks exactly like a Project Euler question, I wouldn't be too surprised if it's already been solved there in the forums.
- jwilk 8y ago> I'd want to see an independent verification I wrote a solver for this problem: https://github.com/jwilk/bitruncatable-primes https://github.com/jwilk/bitruncatable-primes I haven't run it yet, because I don't have a powerful-enough machine at hand. It needs ~6 GB of RAM and ~20 CPU core-hours.
- jwilk 8y agoAfter fixing exorbitant memory use, I have now independenty verified that 7228828176786792552781668926755667258635743361825711373791931117197999133917737137399993737111177 (97 digits) is indeed the biggest bi-truncatable prime. This was done under assumption that zero digits are not allowed. If they are allowed, bigger bi-truncatable primes exists, such as: 90072457733413689120801410250233316614403998951220231333193991731791997911317971131797197339199333933 (101 digits)