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 3-digit primes, it is fairly easy to find that the prime quadruple is:

821, 823, 827, and 829.


8³ + 2³ + 1³ =  521
8³ + 2³ + 3³ =  547
8³ + 2³ + 7³ =  863
8³ + 2³ + 9³ = 1249
Each of 521, 547, 863 and 1249 are prime.

Posted by Dej Mar on 2015-05-06 15:15:32