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 4 primes (Posted on 2015-05-06)
There is a prime quadruple i.e. four consecutive primes such that:

a. Each one of them consists of distinct digits.
b. For each one of them the sum of the digits' cubes is a prime number.
c. In the new sequence of 4 primes none has repeated digits.

I believe there is a poor chance of multiple solutions but you are welcome to explore the issue after finding "my" quadruple (all are 3 digit primes).

 Submitted by Ady TZIDON No Rating Solution: (Hide) I had in mind only one solution, i.e. 821, 823, 827, and 829. Charlie exploration found:821 823 827 829 521 547 863 1249 (s.o.d. of digits' cubes) 2063 2069 2081 2083 251 953 521 547 2069 2081 2083 2087 953 521 547 863 2081 2083 2087 2089 521 547 863 1249 and last , but not the least: 850613 850631 850637 850673 881 881 1223 1223 The four-digit cases are really one set of six successive primes: 2063 2069 2081 2083 2087 2089 251 953 521 547 863 1249 The last case does not fit comply with puzzle's requirement only distinct digits in each of the digit cube totals.

 Subject Author Date re: computer solution Ady TZIDON 2016-03-03 09:53:53 computer solution Charlie 2015-05-06 15:40:46 the finding Dej Mar 2015-05-06 15:15:32

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