How many 5-digit numbers that consist only of digits 1, 2 and 3 are divisible by 6?
There is no neat formula for this problem.
I had to use brute-force Listing.
To be divisible by 2, the number must be even; it must end with 2.
To be divisible by 3, the digit-sum of the number must be a multiple of 3.
I found only five cases (and their permutations).
Did I miss any?