# Write a true self-reflective statement about the digits from 0 to 9 (or prove it can't be done)

This statement contains 3 0's, 1 1, 4 2's, 1 3, 5 4's 92 5's, 6 6's, 53 7's, 58 8's, and 9 9's.

Clearly that statement is incorrect. It only has 1 occurrence of the character "0", not 3; and there are definitely not 92 instances of "5".

Can you rewrite it to make it true? You still must follow the same format, giving the exact number of times each digit has appeared in the statement.

Specifically your statement must be in the format: This statement contains A 0's, B 1, C 2's, D 3, E 4's F 5's, G 6's, H 7's, I 8's, and J 9's., where A is the number of 0's, B is the number of 1's, etc. The number of 1's (or any digit) includes the 1 needed for B 1's.

If not can you prove that this is impossible?

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Aside from klm's solution, there is also a solution for a self-reflective number - that is, a number that contains the number of occurrences of each digit from 0 to 9 within itself. That number is 6210001000. – Joe Z. Jul 12 '14 at 12:18
This problem has been discussed intensively here: math.stackexchange.com/questions/19061/… – justhalf Jul 15 '14 at 4:17

Kind of. I was actually thinking of doing it in binary or other bases. Like: This statement contains 1011 0's and 101 1's, then of course fixing 1011 (11 base 10) and 101 (5 base 10) to numbers that are correct. Yet your variant is good too. – Helka Homba Jul 12 '14 at 7:32
In fact a lot of questions could be asked relating to this. Are there infinitely many solutions? (I'll bet there are.) Is there an efficient algorithm for generating them without missing any? Can it be solved if we replace digits with letters and write numbers as words? e.g. This statement has one a's, zero b's, ..., four t's, ... – Helka Homba Jul 12 '14 at 7:44