> Allow me one nit-pick:
Yeah I noticed this, but was too tired to correct it. =)
Anyway my guess was wrong (9!+8!+7!+...) you got it right.
> (way faster than my brute force approach).
If it's about speed you can limit the $maxlevel, because the longest number can't have more than 7 digits:
- Evidently 0 is excluded!
- Any number this long includes even ciphers. So 5 is excluded!
But any number divisible by 5 and 2 must end with a 0
- An number from the remaining 8 digits would include 9 and 3, but the digit sum would be 40, which is impossible.¹
Even your approach with a brute force loop could compete when only considering 7 digits, cause you don't have the overhead of 1 million function calls.
¹) and therefor a 7 digit number excludes 4 to be divisible by 9.
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