# Absolute Prime numbers in base 10.

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.

There are 24 permutations when all the four digits are considered to form one number. I am trying to prove that each number is divisible by another number (other than one and itself); so each number is not prime; and hence that no absolute prime contains all of the digits 1,3,7 & 9.

This approach is taking too long; but am I correct? Is there a shorter method? Please advise.

• Are you certain that no digits can be repeated? – Arthur Jun 16 '18 at 20:53
• well, the question says permutations. ... and i get 24 permutations (non-distinct) by using these four digits. ... so I think that no digits can be repeated. ... – Math Tise Jun 16 '18 at 21:13
• An absolute prime with more than $3$ digits , not being a rep-unit (number of the form $\frac{10^n-1}{9}$) , must (if it exists) have at least $6\cdot 10^{175}$ digits – Peter Jun 16 '18 at 21:17
• Thank You @Peter for your answer. ... I request you to please explain your answer and also provide references if possible. ... – Math Tise Jun 16 '18 at 21:20
• en.wikipedia.org/wiki/Permutable_prime – Peter Jun 16 '18 at 21:21

$1379, 1397, 1739, 1793, 1937, 1973$ and $3719$ form a complete residue system modulo $7$. In this case, $1379$ is divisible by $7$.
Similarly, integers of the form $x1379, x1397, x1739, x1793, x1937, x1973$ and $x3719$, where $x$ represents any fixed digits you choose, will also form a complete residue system modulo $7$, so one of those integers will be divisible by $7$.