You are a lightning rod conductor for great puzzles!

This is a very slippery problem, it’s difficult to keep the logic in your head; too much recursion!

The autobiographical number problem allows numbers to have up to 10 digits, they can have less.  I think the smallest one is 1210.

There are a couple of identities which help.

First, the sum of the digits must equal the number of digits.  In 6210001000 we have 6+2+1+1 = 10.  In 1210 we have 1+2+1 = 4.

Second, the sum of the digits times the number they are counting must equal the number of digits.  In 6210001000 this is 6×0 + 2×1 + 1×2 +1×6 = 10.  In 1210 this is 1×0 + 2×1 + 1×2 = 4.

I should explain why the second identity works.  If the third digit (which counts 2s) is equal to 3 then there are three digits with value 2.  Each of these is counting two other digits.  So we have accounted for 3×2 = 6 digits.

Given that each digit is at least zero this is very restrictive.  It doesn’t completely describe the problem but I wonder if it allows any solutions which aren’t autobiographical numbers?

Your solution is the only one with ten digits.  Lets say the first digit is A and that there are ten digits.

If A is five or less then there are at least four other non-zero digits and they sum to at least five .  The smallest value for the second identity is given by A2111 which makes 2+2+3+4=11.  So A > 5.

If A is seven or more then there are at most two other non-zero digits and they sum to three or less.  One of these is the digit which counts A, it will be 1 otherwise the first identity will be at least 2A >= 14.  The digit which counts 1 will be at least 1.  It can’t be 1 as it would then need to be 2.  If it is anything other than 1 then we need another non-zero digit to count it.  So A <7.

So A=6 and it’s easy to see that 6210001000 is the only possibility.

We can find shorter numbers by removing zeros to give 521001000, 42101000 and 3211000.

I have also found 1210 and 2020.  I suspect this is it but I haven’t worked it all through.  That would be one solution for each value of A.

http://mathforum.org/wagon/spring99/p884.html

I’ve only successfully completed the first part.  I found the number 6210001000 .  I started with big ones assuming that little ones are harder as that’s part two of the question.  I suspect, though I’m not sure, that this is the largest qualifying number.

Who can find the smallest one?  And even though they didn’t ask, it would be cool to figure out an algorithm to generate these.  What I did to explore feels like it could be formalized.