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Answers
Let's assume
(682) → 1st statement
(614) → 2nd statement
(206) → 3rd statement
(738) → 4th statement
(870) → 5th statement
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Δ From the 1st, 3rd statement, we know 2 is that number which is correct and well placed.
So, _ _ 2
Δ From the 3rd, 5th statement, we know 0 is the first digit of the code.
So, 0 _ 2
Δ From the 2nd statement, we know that 4 is that correct number which is the second digit of the code.
So, 042
THE CODE IS 042.
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Question:
A number lock has 3 digit code can you crack it with the following hints
682 => one number is correct and in correct position
614 => one number is correct but in wrong position
206 => two numbers are correct but in wrong position s
738 => nothing is correct
780 => one number is correct but in wrong position
Solution:
First Statement
682 => one number is correct and in correct position
ASSUMPTION => We only know one digit is right but , in wrong place but, don't know which digit.
Possible numbers will be 6,8,2
Second Statement
614 => one number is correct but in wrong position
ASSUMPTION => We removed 6 from our assumption because if we see both statements then 6 is at same place now too which proves previous statement wrong so , we can remove 6 from our assumption.
Possible numbers will be 8,2 and 1,4
Third Statement
206 => two numbers are correct but in wrong position s
ASSUMPTION => We only know two digits are right but , both in wrong place but, don't know which digits and 6 is already out of scope.
Possible numbers will be 8,2 and 1,4 and 2,0
Fourth statement
738 => nothing is correct
ASSUMPTION => All digits are wrong so , we can leave these numbers and also remove 8 from our possible numbers.
Possible numbers will be 2 and 1,4 and 2,0
Fifth statement
780 => one number is correct but in wrong position
ASSUMPTION => We can remove 7 from assumption according to fourth statement.
Possible numbers will be 2 and 1,4 and 2,0 and 0
So , possible numbers which we get from each statement are:-
2
1,4
2,0
0
Final Assumptions from possible numbers for three digit number:-
1. If we see collectively statements first and third and fifth we can come out with result that 2 and 0 are those numbers which will be in our final digit number.
Reason=>
(a) 6,8 are out of scope so , 2 is that number which is right and its place is also right. So , our three digit number will be _ _ 2
(b) In fifth statement , 7,8 are out of scope which leaves us with 0 at wrong place but , it is correct digit and also in third statement it says two digits are correct and we can say by seeing Reason(a) and Reason(b) that those two digits are 2,0 and we already have 2 at our place.
So , the code can't be _ 0 2 and it can be 0 _ 2
2. Now, we have to find last number which is pretty simple it will be 4 because as we see second statement if 1 is the correct digit then in the statement it says its at wrong place. But, we are only left with central place for our final code. 6 is out of scope. 4 is left which is correct digit and is at wrong place. 2 and 0 are already at its place.
The final three digit number will be 042.