Make a list of distinct prime numbers, using the decimal digits from 0 to 9 exactly twice each in the list. What is the minimum sum of all the numbers in such a list? What's the minimum product of all the numbers in such a list? (None of the primes may admit leading zeroes).

Bonus Question:

Make a list of distinct prime numbers, using the duodecimal digits from 0 to B exactly twice each in the list. What is the minimum sum of all the numbers in such a list? What's the minimum product of all the numbers in such a list? (None of the primes may admit leading zeroes).

Note: Think of this problem as an extension of Pretty Potent Primes.

(In reply to A possible start! by rod hines)

An examination of the set of numbers to find the lowest sum, I saw that you can switch the 2 and 4 in 23 and 463 to get a lower sum using the primes 43 and 263:

2+43+89+109+263+487+557+601 = 2,151

This sum, though, can also be attained with the following:
2+43+89+109+283+467+557+601, 
2+43+89+109+263+457+587+601, and
2+43+89+157+263+409+587+601
Other combinations may also be possible. I suspect, therefore, there may be a lower sum solution.

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No claim to have found the lowest sum solution, but I did find an even lower sum solution:
2+53+89+157+263+409+487+601 = 2,061

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For the duodecimal digits, I offer the following as a possible solution:
[base 12]:
2+3+67+81+205+3AB+447+69B+801+A95 =>
[base 10]:
2+3+79+97+293+563+631+978+1153+1553 = 5,352

Edited on December 10, 2008, 11:24 pm

Posted by Dej Mar on 2008-12-09 21:46:46