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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – user88
    Commented Jul 12, 2014 at 12:18
  • 2
    $\begingroup$ This problem has been discussed intensively here: math.stackexchange.com/questions/19061/… $\endgroup$
    – justhalf
    Commented Jul 15, 2014 at 4:17

2 Answers 2


Here we go:

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

  • $\begingroup$ I count 7 1's there. $\endgroup$ Commented Jul 12, 2014 at 7:04
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    $\begingroup$ @Calvin'sHobbies, you mean like this: puzzling.stackexchange.com/questions/1904/… ? Do you know the answer? $\endgroup$
    – klm123
    Commented Jul 12, 2014 at 7:25
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jul 12, 2014 at 7:32
  • 1
    $\begingroup$ 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, ... $\endgroup$ Commented Jul 12, 2014 at 7:44
  • 3
    $\begingroup$ "Are there infinitely many solutions? (I'll bet there are.)" I disagree. I don't think there are any solutions that use a three digit number. Suppose a sentence has the phrase "100 1s"; this implies that at least one other part of the sentence has a a number containing ten or more digits. So "1000000000 2s" (or some other similarly large number) is in the sentence as well. But then another part of the sentence must have a number containing a hundred million digits... I can't fit a formal proof in this comment, but I think you get the idea that solutions only exist for small numbers. $\endgroup$
    – Kevin
    Commented Jul 14, 2014 at 16:40

This statement contains 1 0, 11 1's, 2 2's, 1 3, 1 4, 1 5, 1 6, 1 7, 1 8, and 1 9.

  • $\begingroup$ This one seems "cleaner" than the other answer. $\endgroup$ Commented Jul 14, 2017 at 22:31

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