Master Mind

In 6152 +0, 1 and 2 cannot be a + because of 4182 00. 5 cannot be a + or 0 because it is eliminated in 5789 +(Only one in these numbers is included in the answer but 5 is not because if it is there'll be a + in 5314 00). 8 is not included too because of 4182 00. eliminated = 5, 8 accepted = 6 _ _ _ Since 5 is eliminated and 6 is accepted, we only have 2 choices for 6152, either 1 or 2. 7 or 9 in 5789 +. 4 is automatically a number because it is either 1 or 2 in 4182 but we need two numbers. left: "1 or 2," "4", "7 or 9" 2 is eliminated because of 5314 00, if 2 is one of the numbers, there will be only one 0. left: "1", "4", "7 or 9" 1 can only be placed in the fourth position(no + in 4182 and 5314). 9 is eliminated. So 7 is the number in the right position. answer. 6 7 4 1

1. For the third and fourth number we have 5 in the first place, but 4th number has a + with it. So 5 can't be the starting digit (in fact it can't occur anywhere in the required number because if it would then the 4th number would have had a 0 on its right which it doesn't). Removing 5 from all 4 numbers and seeing the first number, observe that 6 1 2 are the only digits left. Since 1 and 2 occur at the exact same position in the 2nd number but it doesn't have a + with it implies that 6 is the first digit of the required number.
2. The 2nd and 4th number have 8 in the third position however 2nd number has no + with it implying that 8 cannot occur anywhere in the required number. So the + for the 4th number means that we either have 7 in 2nd position or 9 in the 4th position.
3. Since we already got 6 from the 1st number and 8 is invalid, it means that either 1 or 2 are present in the required number. Using this for the 2nd number we see that 4 has to be present somewhere in the required number. But the 1st position is already allotted to 6 and the 3rd number has 4 in the last position meaning 4 can occur in either the 2nd or 3rd position. But it has to occur at one of these places.
4. Including 1 as a possible digit rules out 2 or 3 from the required number ( we see this from contradictions in 2nd and 3rd number) and including 2 , includes 3 and rules out 1. The possible location for 1 is the 4th position only. Possible locations for 2 and 3 are 2nd and 3rd or 3rd and 4th only. Since including 2 and 3 leaves violates point number 2. Hence 1 is the included number at position 4. From point 2, we have 7 at 2nd location and the 3rd position is 4 The required number = 6 7 4 1.

Following the last line of clue, only four possible combinations come up: (5---), (-7--), (--8-) or (---9). Now following the first line of clue and the first possible combination, we can have (51--), (5-5-) and (5--2). Now, (5-5-) can be easily eliminated as repeat numbers are not allowed. Similarly, using the other possibilities and the first line of clue, then eliminating the impossible ones, we are left with (67--), (-75-), (6--9) and (--59). Take care that the possibilities that come up should exactly match the + and 0 zero signs, i.e., there should not be any + matches if there is none. Now for the rest two clues we have two places vacant. So we need two digits that are common to both the clues as both have a double 0 sign. The only possible digits are 4 and 1. Now only the first possibility, (67--), goes with that as others bring up mismatches of + and 0 signs. Hence the answer is 6 7 4 1.

6741............. For the first four & last four there is a matching digit in matching position. For the first number 1, 5,2 cannot be the matching position digit since they have the same position for two numbers ( 1 for first & third, 5 for 3rd and 4th, and 2 for 1st and 2nd). So 6 must be the required number for matching position. Now consider the forth number. From here 7 or 9 can be the number for respective matching position ( since 8 has the same position for 2nd & 4th number). Now for 2nd & 3rd number 4 and 1 are the only matching numbers in non matching position because only two numbers are left. Suppose 9 from the last number be the matching number in matching position. Then we can put 4 as the 2nd & 1 as the 3rd digit or vice verse. In both cases the digit 1 will be in a conflicting position with the 2nd or 3rd number. Now if we consider 7 as the matching number in the matching position then 4 & 1 will be placed either in 3rd & 4th position or vice verse. In the second case the position of 1 will conflict with the position of 1 in the 3rd number. But in the 1st case all the conditions provided are getting fulfilled. So the number would be 6741

a + in 1st number and a + in 4th number, while they have no digit in common, show that we have to pick two numbers from 1st and 4th for unknown number and on same positions. After that only two numbers are left to be chosen and 2nd and 3rd number shows there are two numbers from each on different positions and they have 1 and 4 in common so we should pick 1 and 4 and put them on different positions than their original.

6741