Hello, markjonathan!

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?