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Bob selects a prime number, and then every second, he adds the digit $1$ or $3$ to the right end of this number (after the unit digit), such that the new number is also a prime, can Bob continue indefinitely?

I calculated: The prime No is $3k+1$, or $3k+2$; So if $3k+2$, you cant add any digit $1$, which will be divided by $3$. of if $3k+1$ you can only add 1 time digit 1, and all others are $3$. So this question becomes add $3$ always. $ p1=$prime; $p2=10P1+3$ is prime or not? p3=10p2+3 p4=10P3+3? obviously will meet composite no, How to prove?

Red Five
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jun ma
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    What have you tried? – Aphelli Aug 11 '24 at 07:53
  • Welcome to MSE. In this forum you're expected to show some work of your own in order to get help. – jjagmath Aug 11 '24 at 09:52
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    Connected. See as well Wikipedia page about (right)truncable primes here. – Jean Marie Aug 11 '24 at 10:32
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    Sad that this interesting question is presented utterly contextless. Pleas provide any trials (for examples search for a long valid sequence) , then this can become a really good question. – Peter Aug 11 '24 at 10:33
  • @Peter, there's a limit to how good a question can become, when the answer has been known for decades and is widely available in books & on websites. – Gerry Myerson Aug 11 '24 at 13:05
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    @Gerry Myerson "when the answer has been known for decades" maybe by you, surely because you have a good knowledge of the domain, but not by me for example (it has taken me some time before catching the good keywords to find the sites I have indicated above)... and a fortiori not by a student... – Jean Marie Aug 11 '24 at 15:03
  • @GerryMyerson This makes it a duplicate and gives a close-reason , this does however not prevent a question to be a good question. Also , I was not aware of this either. I heard from right truncatable primes , but this is not quite what is asked here. Here we do not start with a single digit , but with an arbitary prime. – Peter Aug 11 '24 at 16:33

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