682 - One digit is right and in its place
614 - One digit is right but in the wrong place
206 - Two digits are right, but both are in the wrong place
738 - All digits are wrong
380 - One digit is right, but in the wrong place
Now, keep these numbers in mind. Lots of things are confirmed right?
6 2 - One digit is right and in its place
614 - One digit is right but in the wrong place
206 - Two digits are right, but both are in the wrong place
- All digits are wrong
0 - One digit is right, but in the wrong place
- Statement 1.1. One digit is right (can be 6 or 2)
- Statement 1.2. The right digit is in its place (6 on 1st place or 2 on 3rd position)
If 6
is our number, then 6
is key #1. If 2
is our number then 2
is key #3.
Let's move on.
Pick 6
as correct number, and go back to the rules.
- Statement #1: (6 ) One digit is right and in its place
- Statement #2: (6 ) One digit is right but in the wrong place (Contradiction)
Thus, 6
is eliminated by contradiction consequently giving us 2
as our key #3
(xx2).
At this point, either 1
or 4
is our possible key from statement #2.
- If
1
is the right number, then1
can only key #1, because key #3 is occupied by #2. (1x2) - If
4
is the right number, then4
can be located either in 1st or 2nd position. (4x2, x42)
Let's move on.
0
can only be located in the 1st position, which consequently eliminates 1
in the process.
2
is known as key #3 (xx2)
, thus 0
can only be located at the first position (0x2)
, and
consequently eliminates 1
in the process. Thus 4
can only be located in 2nd position, giving
us a conclustion of 042
as our answer.
- Statement #1 assures
2
in the 3rd position. - Statement #2 assures
4
in the 2nd position. - Statement #3 assures
0
in the first position.
Statement #5 is merely just a supporting statement. We can even eliminate this 5th
statement and we still arrive to same conclusion that the answer is 042
Don't tell your friends the answer. Just keep this for yourself. Don't spoil the fun. :)