Question

puzzle time challenge its a 5 digit number where... 1st digit denotes how many zeroes are there in the number 2nd digit denotes how many ones are there in the number 3rd digit denotes how many twos are there in the number 4th digit denotes how many threes are there in the number 5th digit denotes how many fours are there in the number can you guess the number?

Answer 1

Number N = A B C D E

A = number of zeros among A, B, C, D, and E.   A ≠ 0. So A can be 1, 2, 3 or 4.
B = number of 1s in the number.       B can be 0, 1, 2, 3 or 4.
C = number of twos.       C can be 0, 1, 2, 3, or 4.
D = number of 3s.     D = 0, 1, 2, 3, or 4.
E = number of fours.  E = 0, 1, 2, 3 or 4

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A can not be 0, as then there is a zero, and hence A must become 1.
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Now let  A = 1.            N = 1 B C D E

  Then B = 1 as there is a one.  But now there are two ones. So B ≠ 1.  Let B=2.  Among C, D and E, we must have one 0 and one 1.  Let E=0. C=1.  Then if D is 0, then A will be 2. If D is 1, then B will be 3.  We try this logical sequence. 

  If D or E is not 0, then the digit 3 or 4 must appear some where.  Then we must have three 0s, or 1s, or 2s in the number.  But then the number becomes some other number.
   Or, we must have four 0s, four 1s, or four 2s some where.  Again, we see clearly that is not possible.

   We don't find a consistent number. Thus, for any choice of C, D and E, we do not find a consistent number.
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Now let A = 2

   The arguments expressed above apply.  There are two zeros among B,C, D and E.  Let E = D = 0.  C is 1 as A = 2.  Since now there is a 1, B = 1.  Now since there are two 1s, B must increase.  Then C or D or E must change.  Thus the recursive effect of change in one digit impacts other digits.
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let A = 3.

   So three of B, C, D, E must be zeros.  D has to be 1 as there is a 3.   Then the remaining have to be B = C = E = 0.   Then B = 1, as D becomes 1.  But then there are two 1s now.  So B becomes 2.  Then C has to become 1.   Thus it is not possible to maintain 3 zeros in the number.
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now let A = 4.

   That means that there are four zeros.  Hence, B = C = D = E = 0.   Now since there is no other digit in the number,  the number remains as it is and the given rules are valid.

   Hence, there is only one solution, that is 4 0  0  0  0

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Alternately, we can start selecting the values for E first.  We find easily that it has to be a 0 only. 

Similarly, we try the values for D.   If it is different from 0, then 3 has to appear in the number.  Then 0 or 1, or 2 must appear three times.  Then the other digits must also appear.  That is not possible.

Thus coming to C, with D and E being 0s.  We have very few possibilities now.  Solve it easily.  The answer is only one number   4 0 0 0 0.

Answer 2

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.