Question 21

Ronny uses a 5-digit key for a combination lock, where 5 digits need to be entered in a fixed sequence. While he remembers that the 5 digits are 9, 8, 7, 5 and 4, he has forgotten the sequence he uses. He also remembers that the sum of the first three digits is a multiple of 3, and so is the sum of the last three digits. Further, the sum of the last four digits is a multiple of 4.
Which of the following is DEFINITELY FALSE?

The 5 digits which are used in the combination lock are 9,8,7,5 and 4.

The sum of first 3 digits and the last 3 digits are divisible by 3. Hence, we can say that the third digit is common in both theΒ combinations.

The possible groups of 3 digits whose sum is divisible by 3 are (9,8,7),(9,8,4),(9,7,5),(9,5,4).Β 

Out of these 4 possible groups, we have to select two groups which can be used as the first and last 3 digits.Β 

The selection should be done in such a way that both the groups should only have one digit in common.Β 

ExampleΒ : If we select (9,8,7) and (9,8,4) and let the middle digit be 9, then we will have 8 in both the first 3 digits and the last 3 digits which is only possible for the 3rd digit which is occupied by 9 in this case. Hence, this case is not possible.Β 

So the only possible combinations for the first and last 3 digits are :

Case 1:Β (9,8,7) and (9,5,4)Β where the 3rd digit will be 9.Β 

Case 2: (9,8,4) and (9,7,5)Β where the 3rd digit will be 9.

Now, it is given that the sum of theΒ last 4 digits is divisible by 4.Β 

The possible cases of the above scenario are :

Case 3:Β (9,8,7,4) where the first digit will be 5.Β 

Case 4:Β (8,7,5,4)where the first digit will be 9 which is not possible as 9 is fixed as the 3rd digit from Case 1 and 2.

Since, the first digit is 5 and the third digit is 9, the second digit will either be 4 or 7 and the last two digits will be either (8,7) or (8,4)

So, the possible combinations of the lock are 54987, 54978, 57984 and 57948.Β Β 

Hence, the option which is definitely false is E as no possible combinations has 8 as its 2nd digit.Β 

$$\therefore\ $$ The required answer is E.Β 

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