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
645 two numbers are
correct but in wrong
Podition
206 two numbers are
correct but in wrong
position
738 nothing is correct
780 two numbers are
correct but in wrong
position

Answer 1

ANSWER :

Dude is the number lock code 608

Answer 2

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.