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
Tiny URL Link: https://tinyurl.com/nice-problem-042