Hi Sarah,

this is wrong isn't it? it says there's one used zero but you used two zeros. Maybe you mean <2,1,2,0,0>? This is how i did it:

We'll start from back, i.e. from box "4". What digit can be in this box? 4? surely not, we'd have to put 4 into at least three other boxes and this would consequently require a huge amount of different digits that must be used, you can see it is not satisfiable. The same applies for putting 3,2,1 into the box "4" - verify this is true. So we must have 0 in the box "4".

Now the box "3", you can eliminate putting 4 into it 'cause the 0 in the box "4" says so. Again try subsequently putting 3,2,1 into it and you'll find quickly, similarly as for the box "4", that this is not possible - verify it. So in the box "3" must be 0. We have < , , ,0,0> so far.

Now for the box "2", you can eliminate 4,3. Trying 2 forces <2,1,2,0,0> so we have a solution. Putting 1 in the box "2" we have

< , ,1,0,0> so we must use digit 2 once, we see we cannot put it into the box "1" because we'd have to put 1 into "0", but there are two used zeros already. So we must put the 2 into "0" which leads to <2, ,1,0,0> which is easily verified to be unsatisfiable. It only remains to try to put 0 into "2", so we have < , ,0,0,0> so far. It is now easily verified that this is not satisfiable because we have at least three used zeros but we cannot put 3 or 4 into the box "0".

It is very likely there's some trick to do it faster which I'm missing, but this is certainly faster then going through all possibilities bruteforce.