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4 primes (Posted on 2015-05-06) Difficulty: 3 of 5
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    
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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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Hints/Tipsre: computer solutionAdy TZIDON2016-03-03 09:53:53
Solutioncomputer solutionCharlie2015-05-06 15:40:46
Solutionthe findingDej Mar2015-05-06 15:15:32
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