The Math Help Topic

Hey mat, I have a game for you (and for others as well, but I guess you'll be the only one interested) ! It's about auto-referent numbers (couldn't find anything in English about those, so I don't even know how they're called).

Let's consider numbers made of 4 digits, like 1230. The referent of a 4 digits number is a 4 digits number as well. To find it, you count the zeros in it : that number will be the first digit of the referent. Then the number of 1, the number of 2, and the number of 3. So the referent of 1230 is 1111.
Here are some examples (the referent of a number is below it) :

1111 - 3000 - 3213 - 4444
0400 - 3001 - 0112 - 0000

An auto-referent is obviously a number equal to its referent.

Can you find all the auto-referent numbers made of 8 digits or less ? :P
 
Interesting. Can't think of a clever algorithm just yet, but I'll look into it.
1210 is the lowest auto-referent, right?
 
They suck though.
Anyways I can prove that sum of digits must be equal to the length of the number, and that the digits must be less than (length/2). It also seems that consecutive digits can't be the same, but I'm yet to prove that. If that's so (gonna look into it soon) it really shouldn't leave many numbers to check. If it isn't, I'll look for other properties.
 
How is this obviously not a pen&paper problem then? They all can be found manually since there aren't too many of them rite?
I just don't like coding at all anymore. It's annoying.
 
I'm not sure you can find them all manually. I'll give it a try though. But if you want some pen&paper problem, I'll come up with some good algebra. :P
 
Oh no, there are just 5, my mistake. There's a 9 and a 10 digits ones.
Well it didn't seem too hard, I'll try it too !
 
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