A numeric lock has 3 digit key-hint-(6,8,2)one number is correct well placed;(6,1,4)one number is correct but wrong placed;(2,0,6) tow numbers are correct but wrong placed;(7,3,8) nothing is correct;(7,8,0)one number is correct but wrong placed...answer plz dudes...
Answers
Hint 2 : (6,1,4) one number is correct but wrong placed.
Hint 3 : (2,0,6) Two number are correct but wrong placed.
Hint 4: (7,3,8) Nothing is correct.
Hint 5 : (7,8,0) one number is correct but wrong placed.
From Hint 1 & 2 , we come to know that the number 6 is incorrect number.
From Hint 3 , we come to know that 2 & 0 are correct numbers.
From Hint 4 & 1 , we come to know that number 2 is correctly placed.
Hence the number will be _ _ 2
From Hint 3, 4 & 5 we come to know that 0 is wrongly placed in (2,06) & (7,8,0) .
Hence the number will be
(0 _ 2)
And from Hint 2 we come to know that the middle number is 4
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Therefore the 3-digit number is ;
0 4 2
Therefore, the co-de of the given numerical lock is 042.
Decoding the clues given:
On observing the rules given, from the fourth rule, we get that the lock contains no numbers from the list [7, 3, 8]. So, from the first rule 6, 2 remains.
From the second rule, we get that one number among 6, 1, 4 is correct but wrongly placed. But the number 6 is in the same position in rules 1 and 2. So, there is no chance of 6 being among the three numbers of the lock.
So, from the first rule, we get that the number at the last position is 2 as it says that the number is well placed.
_ _ 2
In the third rule, if we eliminate 6, the remaining numbers are 2, 0. They are correct but wrongly placed. As 2 is fixed with the last position, 0 goes for the first position as per the rule.
0 _ 2
Now, from rule 2, we get that one number is correct among 1, 4 but is wrongly placed. If 1 is correct, it would be well placed and that contradicts the rule which says that the number is wrongly placed. So, the correct number that should be in the second position is 4
0 4 2
Therefore, the co-de of the given numerical lock is 042.
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