Find a number

**ABCDEFGHIJ** such that

**A** is the count of how many 0's are in the number,

**B** is the number of 1's, and so on.

(*Taken from http://einstein.et.tudelft.nl/~arlet/puzzles/logic.html*)

(In reply to

Puzzle Solution by K Sengupta)

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

*Edited on ***April 10, 2007, 2:10 pm**