I have a ten-digit number. The first digit tells me how many zeros are in the number. The second digit tells me how many ones are in the number. The third digit tells me how many twos are in the number, and so on, until the tenth digit which tells me how many nines are in the number.
What is my number?
Bonus: Are there more solutions or is it a unique solution?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Suppose that the leftmost digit is 9. Then the rightmost digit must be 1. But then there are not enough digits left for 9 0s, a contradiction.
9 Not enough digits for 9 0s 0 0 0 0 0 0 0 0 1
Now suppose that the leftmost digit is 8. Then the second to last digit must be 1. Then, the second from the left digit must be at least 1. But then there are not enough digits left for 8 0s, a contradiction.
Not enough digits for 8 0s 8 1 0 0 0 0 0 0 1 0
Now suppose that the leftmost digit is 7. Then the third to last digit must be 1. Now we can resolve 2s and 1s by placing a 2 in the second digit and a 1 in the third digit. But then there are not enough digits left for 7 0s, a contradiction.
Not enough digits for 7 0s 7 2 1 0 0 0 0 1 0 0
Now suppose that the leftmost digit is 6. Then the fourth to last digit must be 1. Now we can resolve 2s and 1s by placing a 2 in the second digit and a 1 in the third digit. And this works! There are exactly enough digits left to place the 6 0s.
6 2 1 0 0 0 1 0 0 0