Problem - 2817
A prime number is called an absolute prime if every permutation of its digits in base 10 is also a prime number. For example: 2, 3, 5, 7, 11, 13 (31), 17 (71), 37 (73) 79 (97), 113 (131, 311), 199 (919, 991) and 337 (373, 733) are absolute primes. Prove that no absolute prime contains all of the digits 1, 3, 7 and 9 in base 10.
Let $N$ be an absolute prime which contains all of the digits 1, 3, 7 and 9 in base 10. Let $L$ be any number formed from the remaining digits. Consider the following seven permutations of $N$: $10000L+7931$, $10000L+1793$, $10000L+9137$, $10000L+7913$, $10000L + 7193$, $10000L + 1973$ and $10000L + 7139$. They have different remainders when divided by 7. Therefore one of them is a multiple of 7, and is not a prime. Hence $N$ is not an absolute prime.