MHB Is it possible to form six prime numbers using three distinct digits?

[Ackbach]

Here is this week's POTW, which marks my first anniversary of being the University POTW Director:

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Given three distinct digits, six numbers can be formed such that each of the given digits appears exactly once in any of them; e.g., using 1, 2, and 5, you can form 125, 152, 215, 251, 512, and 521. Is it possible to choose the three digits in such a way that all of the six numbers so formed are prime?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

We had excellent participation this week, with kiwi, johng, and petek all providing correct solutions. Congratulations to you!

petek's solution follows below:

Spoiler

None of the digits can equal 0, 2, 4, 6, 8 or 5, since a permutation ending in one of those digits will be divisible either by 2 or 5. Therefore, the three digits must occur in the set {1, 3, 7, 9}. The four possibilities of three digits are {1, 3, 7}, {1, 3, 9}, {1, 7, 9} or {3, 7, 9}. However, each of these sets of three digits contains at least one permutation that's composite:

371 = 7 x 53
931 = 7^2 x 19
791 = 7 x 113
973 = 7 x 139

(Some other permutations also are composite.)