Crack the Code ?
Code
A Numeric Lock has a 3 Digit Key
Answers
Answer:
042
Explanation:
From given hints, we have,
1) 6 8 2 - one number is correct and well placed
2) 6 1 4 - one number is correct and but wrongly placed
3) 2 0 6 - two numbers are correct but wrongly placed
4) 7 3 8 - nothing is correct
5) 7 8 0 - one number is correct but wrongly placed
Now lets take the 4th point and highlight the digits that cannot be in the codes in every hint
1) 6 8 2 - one number is correct and well placed
2) 6 1 4 - one number is correct and but wrongly placed
3) 2 0 6 - two numbers are correct but wrongly placed
4) 7 3 8 - nothing is correct
5) 7 8 0 - one number is correct but wrongly placed
Now lets see the first two hints.Consider 6 as the number in the code but according to first hint the code is well placed but according to second it is wrong placed i.e contradicts to first statement so it cannot be the digit.So lets highlight six.
1) 6 8 2 - one number is correct and well placed
2) 6 1 4 - one number is correct and but wrongly placed
3) 2 0 6 - two numbers are correct but wrongly placed
4) 7 3 8 - nothing is correct
5) 7 8 0 - one number is correct but wrongly placed
Now we know from the first hint that 2 goes to one's place ie
_ _ 2
From 3 and 5 we know 0 is the digit in code but cannot be in ones and tens place so it has position in hundreds ie
0 _ 2
Now the left position is tens and seeing statement 2 we know that the digit that lies in tens place in second hint does not have any position in the code i.e 4 is the digit in tens place so the code becomes
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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