A person forgets his 4-digit ATM pin code.

Question

A person forgets his 4-digit ATM pin code. But he remembers that in the code all the 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. Then the maximum number of tr...

Answer: 72

Accepted Answer

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.

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