Question

Solve this guyzzzxx​

Answer 1

Given : Code for Lock & Condition

To find : Code of Lock

Solution:

14673  

Four  Digit Correct  & one rightly Placed

02637

Three Digit Correct and all rightly place

637 are common Digits  so atleast two has to be Correct Digits out of 6 , 3 , 7

69308  - one Digit correct & wrongly placed

Has two Digits 6 & 3   from  6 , 3 & 7

So one has to be correct Digit  out of 6 &  3  and one has to be wrong

Hence 1 & 4 are Correct Digit from 1st  ( as only one incorrect Digit is there and that would be 6 or 3)

( 9 , 0  & 8  Can not be Correct Digit  as only correct Digit 6 or  3 )

0 is not correct Digit & one incorrect Digit in 6 & 3  Hence

X2XX7  are Correct Digits  from 2nd

56073  -  Two Digit Correct but Wrongly Placed

7 is correct  & one out of 6 & 3 is correct

Hence  ( 5 & 0 can not be correct Digit)

Hence in total  ( 0 , 5 , 8 , 9 can not be correct Digit )

59418  -   Two Digit Correct one Rightly placed

Here 4 & 1 are Correct Digit  

Correct Digit

2 , 7  ,  4   , 1     & one out of  6 & 3

Lets Assume  6  is the 5th Correct Digit

Then 2nd Code

X26X7

3rd & 4th Code Satisfied

as Per 5th Code

42617  is the code

now Check 1st Code

   - it satisfy all conditions

Hence   14673  is the Code

Now Assume  3  is the 5th Correct Digit  ( not 6)

Then 2nd Code

X2X37

3rd Code Satisfied

4th Code Satisfied

As per 5th Code

12437

Now check 1st Code  it Satisfy all conditions

Hence Code can be

42617

12437

Verification :

42617    

14673  has 4 correct Digits 1  , 4 , 6 , 7  & 6 is correctly placed

02637  has 3 Correct Digits 2 , 6 & 7 Correctly placed

69308  has  1 Correct Digit 6 wrongly placed

56073 has 2 Correct Digits  6 & 7 wrongly placed

59418   has  2 Correct Digits 4 & 1  , 1 is Correctly Placed

12437

14673  has 4 correct Digits 1  , 3, 4 , 6 ,   & 1 is correctly placed

02637  has 3 Correct Digits 2 ,3 & 7 Correctly placed

69308  has  1 Correct Digit 3 wrongly placed

56073 has 2 Correct Digits  3 & 7  wrongly placed

59418   has  2 Correct Digits 4 & 1  , 4 is Correctly Placed

Hence Two possible Codes:

42617

12437

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Answer 2

First point:-It is said that,in first and second number one digit is rightly placed and as we are seeing that all the digits exchange their position in second number except 6 so 6 will beone of the digit at 100th place

Second point:-from the first condition we can say that from 1,4,7,3 three digits are right

From second condition we can say,from 0,2,3,7 two digits are right

Now,from third condition we can say that 9,3,0,8 will not be the digits in required number

Comparing these conditions we can say 2,7,6,4,1 will be the digits

As we get that 6 will be at 100th place

From second condition which is given we can say that 7 will be at one's place and 7 will be at thousandth place

From first given condition we can write that 1 will not be 10000th place so it will be at remaining tenth's place

Hence the required no. is 42617