Are there permutation-invariant divisibility tests for primes other than 3?

Question

Every permutation of a number in the form ‘abc’ with a+b+c not divisible by 3 will not be divisible by 3.

This happens because the divisibility test does not depend on the order of the digits.

In a more general case, it seems to me this is not possible, e.g. the test for 11 relies on alternating digit sums.

Other tests based on factorization of radixes near a power of 10 (999 =3^3 x 37 or 1001=7 x 11 x 13) seem to imply a non permutation-invariant formula.

Is this really the case?

With primes less than 50 and integers below 10^6, I found only the tuples 012345 for 11 and 012347 for 37 that remain not divisible for all 720 permutations.

Answer 1

The property you mention is caused by divisibility by nine, and as you might know: $9=10-1$ and ten is the base of the way we write numbers.
If you write numbers using another base, like eight, then the property holds for divisibility by seven: $7 | (142)_8$ and $7 | (124)_8$ and $7 | (421)_8$ ...
If you write numbers in yet another base, like sixteen, then the property holds for divisibility by fifteen, or for the prime numbers $3$ and $5$, ...