Problems & Puzzles: Puzzles

Puzzle 793. Entry 1399 from Claudio Meller's site The current puzzle is the same that Claudio Meller posted in the entry 1399 of his always interesting site. Here we ask for prime numbers P such that if composed by K distinct digits, then every one of them appears at least once contiguous to each other. Examples: For K=3, then P=1021 is a solution because: a) there are 3 distinct digits {0, 1, 2} and b) in P, 0 is contiguous to 1 & 2; 1 is contiguous to 0 & 2 and 2 is contiguous to 0 & 1. For K=4, then P=10307137 is a solution because: a) there are 4 distinct digits {0, 1, 3, 7} b) in P, 0 is contiguous at least once to 1, 3 &7; 1 is contiguous to 0, 3 & 7; 3 is contiguous at least once to 0, 1 & 7 and 7 is contiguous to 1, 1 & 3. Carlos Rivera claims also that 1021 and 10307137 are the smallest prime numbers solutions for K=3 & 4, respectively. Q. Send the smallest prime solutions for K for 5 to 10?

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Contribution came from Emmanuel Vantieghem

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Emmanuel wrote:

Here is my final answer :

  K           p 

  5 10213053251

  6 1021302341504253451

  7 1021304150624354652361

  8 10213041506170234253456275637467

  9 1021304150617082342536273845647687581

 10 102031405123416071809152453627382938465749676859789

 

With the methods I used I cannot find smaller examples.  But I still believe that there might be smaller solutions.

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