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).
Given criteria a and b and that all four of the consecutive primes are 3digit primes, it is fairly easy to find that the prime quadruple is:
821, 823, 827, and 829.8
^{3} + 2
^{3} + 1
^{3} = 521
8
^{3} + 2
^{3} + 3
^{3} = 547
8
^{3} + 2
^{3} + 7
^{3} = 863
8
^{3} + 2
^{3} + 9
^{3} = 1249
Each of 521, 547, 863 and 1249 are prime.

Posted by Dej Mar
on 20150506 15:15:32 