# Oh dang, I have to change my whole answer!

One day, you enter your math classroom. On the board, there are the words POP QUIZ.

Dang it! Was it about fractions? you knew you were bad with fractions.

You look down.

$\begin{array}{|cc|} \hline \text{You can only use your brain, and paper. } \\ \text{The number of digits in this box that are not 1 is} & \text{_______} \\ \text{The number of digits in this box that are not 2 is} & \text{_______} \\ \text{The number of digits in this box that are not 3 is} & \text{_______} \\ \text{The number of digits in this box that are not 4 is} & \text{_______} \\ \text{The number of digits in this box that are not 5 is} & \text{_______} \\ \text{The number of digits in this box that are not 6 is} & \text{_______} \\ \text{The number of digits in this box that are not 7 is} & \text{_______} \\ \text{The number of digits in this box that are not 8 is} & \text{_______} \\ \text{The number of digits in this box that are not 9 is} & \text{_______} \\ \hline \end{array}$

Hint.

All of the blanks are not mostly the same number.

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People keep underestimating the work involved here. Cool puzzle. – Solocutor Feb 24 at 23:30
@Oray It's asking about the number of digits that aren't x, which is different from the number of x-digit numbers. Your answer would be correct if it was asking you to count the number of x-digit numbers, but it would be worded like "The number of numbers that are not 1-digit long in this box is ____". instead of "The number of digits in this box that are not 1 is ____".. – Paulpro Feb 24 at 23:41
Isn't this the same as Puzzle C here: puzzling.stackexchange.com/questions/23554/… - not that it has an answer there! – abligh Feb 25 at 14:46
There's also some interesting discussion here : puzzling.stackexchange.com/questions/23557/… – dennisdeems Feb 25 at 20:04
If this were a real classroom and a real test, I would write out the word "eight" in every blank. But I don't think that's what you're intending... – matt Feb 25 at 20:26

## 1 Answer

One answer is

1: 25
2: 18
3: 25
4: 24
5: 23
6: 24
7: 26
8: 25
9: 26

To find this, I first noted that each of the blanks must be at least 8, since there are already 8 digits different from the given one. None of the blanks can be 8, though, since that would mean all of the digits in the blanks are the given digit, which would have to be 8, and there can't be 88 digits in the box - at most 27. If a blank were 9, all of the digits would have to be the given digit except, presumably, that 9. But then there would be 24 digits different from 9, which contains two different digits, '2' and '4,' so 9 is ruled out.

Thus, we know every blank is a 2-digit number, so there are indeed 27 digits in total. Thus the average number of digits that aren't a given digit is (9*27 - 27)/9 = 24.

So I started by writing 24 in every blank. I then counted the number of each digit, and wrote what should be in the blanks in that case. Since only '2' and '4' appear in the blanks, the number of non-1 digits is 26, same as 3, 5, 6, 7, 8, and 9, while there are 17 non-2's and 17 non-4's. So I changed the blanks to 26, 17, 26, 17, 26, 26, 26, 26, 26. Then I counted again and corrected. After 11 iterations, I found that I hadn't changed any of the numbers, so I arrived at an answer, which I've written above.

I can't prove that it's unique though. Also, my method may have ended in a cycle, so I was lucky to have found a fixed point.

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Matches the computer's answer – humn Feb 25 at 8:26
Seems to match the computer's algorithm too, after efficiently bypassing some iterations. Thank you for including the number of iterations, Would be interesting to compare with the computer. – humn Feb 25 at 9:06
If you would have not started from 24, but with the "empty string" everywhere, you would have finished in fewer iterations. – Pakk Feb 25 at 13:55

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