There is a ten digit number where the first digit is also the number of zeros in the number, the second digit is the number of ones and so forth until the last number is the number of nines in that number. What is this number and is it unique?
As there are ten digits in the number, and the digits themselves reflect the number of digits in the number, the sum of the digits in the number must add up exactly to ten. Therefore, the largest number that can exist in the last (ninth) digit is a one, because anything more would mean that there are two nines in the number, which already adds up to eighteen which is more than ten. This same argument can be applied to every digit in the number to obtain the following maximum values for each digit:
9953221111
These theoretical maxima can be reduced by taking into account the digit as well as the digits that it represents. For example, in the ones position (second digit), any number greater than four is not possible. If a value of five or higher was used, then there would be at least five ones in the number, without considering that number itself, which means that the sum of the digits is already over ten. This argument can be applied to most of the lower order digits to reduce the maximum values for each digit:
9533211111
There are still practical limits to some of these values, and one can reduce the range of values by taking into consideration the positions of each of the digits that each number represents. For example, if the third digit was a three, then that would mean that there are three twos in the number. The lowest order positions that these twos can take are the first (zeros), second (ones), and fourth (threes) position.
2232______
This means that there are at least two more ones, and another three in the number, which puts us over the ten limit. Using this argument for the other digits, the practical maxima become:
9321111111
Now we can look at each position individually. Consider the last digit in the number.
_________x
We know that this last digit must be a one or zero. If the number was one, then there must be a nine in the number. The only possible location is at the front (the zeros),
9________1
which would mean that there are nine zeros. As there are only eight spots left in the number, this is impossible. Therefore, the last digit is a zero. Continuing onto the eight's digit,
________x0
we know that there are only two possible values it can take. If this number was one, then the only possible location for the eight is in the first spot, which would suggest that there are eight zeros. As there are seven spots left, we can satisfy the eight zeros,
8000000010
but now there is also a one in the number, and as there are no more spots available, the one cannot be accounted for. Therefore, the second last digit must be a zero. We can now look at the seven's digit.
_______x00
Again, there are only two possibilities. If this number was one, then the only possible location for the seven is in the zeros spot.
7______100
Again, the only logical places to put the extra five zeros are in all of the positions except for the ones position (since we already have a one).
7_00000100
Unfortunately, this condition cannot be satisfied because a one in the one's position gives us two ones, while a two in the ones position is false. Thus, the sevens digit must be a zero and we can continue onto the six's digit.
______x000
Again, there are only two possibilities. If this number was one, then the only possible location for the six is in the zeros spot.
6_____1000
Before the zeros are placed, consider the number of ones. A one in the ones position was a one, then it will be false, therefore the digit must be two or more. A value of three is impossible since 6+3+1*3 > 10, therefore the ones digit must be a two. Furthermore, to satisfy the condition that three of the remaining four positions are zeros, the two's digit must be a one.
This leaves us with one possible answer of:
6210001000
Now we will attempt to eliminate all other possible answers.
If the six's position was a zero, then the zero's digit can be at most a five.
5_____0000
The five's digit then must be a one and the one's digit must be at least a two (since a one's in the one's position means that there are two ones in the number!).
52___10000
At this point, we have a problem, since we can only have one more zero in the number, the remaining digits must be at least ones. Since the sum of digits is already eight, the most that the remaining digits can be are also ones! However, this means that there will be three ones in the number, and there are no threes in the number, which means that this is not the correct number. Therefore, the five's digit must be a zero, while the zero's digit can be at most a four.
4____00000
It is obvious already there is a contradiction with this number. The zeros digit maximum is already less than the number of zeros in the digit, therefore, we cannot add any more zeros to the number! At this point, we can safely conclude that there will be no more solutions to the problem, which means that the solution that's been found already is unique.
In conclusion, there is indeed a solution to the problem. The solution is 6210001000 and this number is also unique.
This problem can be restated for a smaller number of digits, I have not done an exhaustive search for such numbers, but I have a list of the ones that I have found based on the number of digits in the number:
