Given some digits $d_1, \dots, d_k$ is there a calculation that I can perform to determine if at least 1 permutation of $d_1, \dots, d_k$ is divisible by 11?
I am interested in all bases (radices) but a solution for base 10 should be sufficient for me to figure out the generalisation.
Generalised question: Given a string of $k$ digits in base $b$, how can I test if at least 1 permutation of $d_1, \dots, d_k$ is divisible by $b+1$ (without computing them all)
Thanks in advance for any help!
To clarify:
I want to categorise a string $S$ of k digits into 1 of 2 sets.
Set1: S has no permutation $S'$ such that $S'$ divides by 11
Set2: S has at least 1 permutation $S'$ s.t. $S'$ divides by 11
An acceptable solution would be a finite list of criteria that applies to any combination of base and string length (b,k) i.e. If $S$ satisfies any of the following:
- Criteria 1
- Criteria 2
- etc
The usual divisibility by 11 test is to calculate the sum of the digits in the even positions minus the sum of the digits in the odd positions. If that is divisible by 11 then the original number is. From now on, I will say "even digit" to mean one in an even position, not necessarily an even value. Similarly "odd digit". I am counting from the right but it should work either way.
So, it is obvious that any permutation of the even digits or any permutation of the odd digits will not affect the divisibility by 11. You could define a canonical form in which the even digits are in ascending order and so are the odd ones.
Also, any permutation can be expressed as a product of two element swaps. From the above, any swaps of two even or two odd digits can be ignored. For another two element swap, the effect on the test sum should be easy to calculate. A swap of an odd digit and an even digit might affect the divisibility by 11.
Edit, due to comments, I have reworded the above to clarify the meaning of "even digit" as one in an even position and not necessarily an even value.