QJEEM23Ma6e88: JEE Main 2023 Mathematics - Class 11 JEEM23Ma6e88 Solution: A person forgets his 4-digit ATM pin code. But he remembers…

Let the 4-digit ATM pin code be represented by the digits $abxy$ , where all digits are different, the greatest digit is 7, and the sum of the first two digits is equal to the sum of the last two digits: $a + b = x + y$ .

Since the greatest digit is 7, the possible digits for the pin code are $0, 1, 2, 3, 4, 5, 6,$ and $7$ .

1. Let's denote the sum of the first two digits and the sum of the last two digits as $\lambda$ . Since the largest digit is 7, we can analyze different possible values for $\lambda$ .

We will analyze different possible values for $\lambda$ and the corresponding possible pairs of digits:

- $\lambda = 7$ : The pairs of digits that have a sum of 7 are $(0,7),(7,0),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$ . Since 7 must be present, we consider the pairs with 7: $(0,7),(7,0)$ .

When the pair $(0,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$ . For each of these pairs, there are 2 possible pin codes: $07xy$ and $07yx$ , where $x$ and $y$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.

When the pair $(7,0)$ is chosen, the other two digits can also be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$ . For each of these pairs, there are 2 possible pin codes: $70xy$ and $70yx$ , where $x$ and $y$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.


So, there are 24 possible pin codes for $\lambda = 7$ .



- $\lambda = 8$ : The pairs of digits that have a sum of 8 are $(1,7),(7,1),(2,6),(6,2),(3,5),(5,3)$ . Since 7 must be present, we consider the pairs with 7: $(1,7),(7,1)$ .

When the pair $(1,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$ . For each of these pairs, there are 2 possible pin codes: $17xy$ and $17yx$ , where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

When the pair $(7,1)$ is chosen, the other two digits can also be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$ . For each of these pairs, there are 2 possible pin codes: $71xy$ and $71yx$ , where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

So, there are 16 possible pin codes for $\lambda = 8$ .



- $\lambda = 9$ : The pairs of digits that have a sum of 9 are $(2,7),(7,2),(3,6),(6,3),(4,5),(5,4)$ . Since 7 must be present, we consider the pairs with 7: $(2,7),(7,2)$ .

When the pair $(2,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$ . For each of these pairs, there are 2 possible pin codes: $27xy$ and $27yx$ , where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

When the pair $(7,2)$ is chosen, the other two digits can also be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$ . For each of these pairs, there are 2 possible pin codes: $72xy$ and $72yx$ , where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

So, there are 16 possible pin codes for $\lambda = 9$ .



- $\lambda = 10$ : The pairs of digits that have a sum of 10 are $(3,7),(7,3),(4,6),(6,4)$ . Since 7 must be present, we consider the pairs with 7: $(3,7),(7,3)$ .

When the pair $(3,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(4,6),(6,4)$ . For each of these pairs, there are 2 possible pin codes: $37xy$ and $37yx$ , where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

When the pair $(7,3)$ is chosen, the other two digits can also be any one of the remaining pairs: $(4,6),(6,4)$ . For each of these pairs, there are 2 possible pin codes: $73xy$ and $73yx$ , where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

So, there are 8 possible pin codes for $\lambda = 10$ .



- $\lambda = 11$ : The pairs of digits that have a sum of 11 are $(4,7),(7,4),(5,6),(6,5)$ . Since 7 must be present, we consider the pairs with 7: $(4,7),(7,4)$ .

When the pair $(4,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(5,6),(6,5)$ . For each of these pairs, there are 2 possible pin codes: $47xy$ and $47yx$ , where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

When the pair $(7,4)$ is chosen, the other two digits can also be any one of the remaining pairs: $(5,6),(6,5)$ . For each of these pairs, there are 2 possible pin codes: $74xy$ and $74yx$ , where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

So, there are 8 possible pin codes for $\lambda = 11$ .



- $\lambda = 12$ : The only pair of digits that have a sum of 12 is $(5,7)$ . Since 7 must be present, this case is not possible.

- $\lambda = 13$ : The only pair of digits that have a sum of 13 is $(6,7)$ . Since 7 must be present, this case is not possible.

- $\lambda = 14$ : The only pair of digits that have a sum of 14 is $(7,7)$ . Since all the digits must be different, this case is not possible.

Adding up all the possible pin codes for each value of $\lambda$ , we get:

$16 + 24 + 24 + 16 + 8 = 72$

Therefore, the maximum number of trials necessary to obtain the correct code is 72.