Which five-digit number has the $n$th digit representing how many occurrences of the digit $(n-1)$ there are in the number, for all $n\le5$?

For example, the fourth digit represents the number of 3s, and the first digit represents the number of 0s.


6 Answers 6


The solution is


This can be found by simply iterating from a starting point. Let's choose 40000 as a start (it's as good as any).

40000 — start with four zeroes (and count them)
40001 — count four zeroes, one four
31001 — count three zeroes, one one, one four
22010 — count two zeroes, two ones, one three
21200 — count two zeroes, one one, two twos
21200 — count two zeroes, one one, two twos

The next row counts the numbers it sees in the previous row. Once we hit two identical numbers back-to-back, we have settled on a solution.

You'll see that 43210, 11111, 22222, 33333, and 44444 are not issues if you make one simple admission.

400001 — allow counting the 5, but don't permanently extend the length

  • $\begingroup$ Just don't start with a permutation of 43210 ;-) All starting points converge to the solution, aside from these permutations and the even more obvious failures 11111, 22222, 33333, 44444. $\endgroup$ Aug 22, 2015 at 12:31

cjm shows in his answer that there is only one correct solution by using a Perl program to brute force the puzzle.

This can also be demonstrated logically:

First digit:

  • Cannot be 0, because then there would be at least 1 zero
  • Cannot be 1, because it would mean that there would be at least a 3, in the best case (1xx10). No matter where you put it, there's no final possible solution.
  • Cannot be 3, because the 4th digit would have to be at least a 1, and all other digits, including the one representing the number of 1s would have to be 0
  • Cannot be a 4, because all other digits, including the one describing the number of 4s would have to be 0s
  • Cannot be a 5, because itself would then have to be a 0

Second digit:

  • Cannot be a 0, because then only one of the last digits can be a 0, so there would have to be, in the best case, at least a 3 in the number, giving us, in the best case, 20310, which is false.
  • Cannot be a 2, 3, 4 or 5, because with more than one 1 we would have less than two 0s.

Third digit:

  • Cannot be a 0, because we already have a 2
  • Cannot be a 1, because we already have one 1 and need only one
  • Cannot be a 3, 4 or 5 because the remaining two digits need to be 0s, so we wouldn't have place to fit them

Fourth and fifth digits:

  • Must be 0s, because we need at least 2.

Therefore the number is 21200.


One answer is:


There is :

2 zeroes, 1 one, 2 twos, 0 three, 0 four


The other answers give a correct solution, but they don't indicate whether it is the only solution.

In fact, 21200 is the only solution.

This is shown by this brute-force Perl solution:

for (10000 .. 44444) {
  say $_ if substr($_, 0, 1) == tr/0//
        and substr($_, 1, 1) == tr/1//
        and substr($_, 2, 1) == tr/2//
        and substr($_, 3, 1) == tr/3//
        and substr($_, 4, 1) == tr/4//;

There are no solutions for 1,2 and 3 digits.

There are two solutions for 4 digits:

1210 and 2020

There is one for 5 digits:


One for 7 digits:


Is there any other possible solution?


And the general math solution (or attempt to build one):



Since the first digit counts the number of 0s and the 2nd digit the number of 1s etc and since the total number digits is 5 the sum of all the digits must be 5. As a counter example if you take a number like 11112 this would imply there are a total of 6 digits which isn't possible. So we need five digits adding up to 5. Possible combinations are:- 11111, 01112, 00122, 00113, 00023 and 00014. If we encode these 6 possibilities according to the rule we get:- 11111 -> 05000 (no zeros, 5 ones, no 2s, 3s or 4s); 01112 -> 13100; 00122 -> 21200; 00113 -> 22100; 00023 -> 30110; 00014 -> 31001; Only one of these is a rearrangement namely 00122 which goes to 21200 which would map onto itself under the rule and so is the solution.


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