Question Details

There is a numeric lock which has a 3-digit PIN. The PIN contains digits 1 to 7. There is no repetition of digits. The digits in the PIN from left to right are in decreasing order. Any two digits in the PIN differ by at least 2. How many maximum attempts does one need to find out the PIN with certainty?

Options

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Correct Answer :

Option C

Solution :

The correct option is 10.

Let the 3-digit PIN be represented by three digits from left to right as $d_{1}$ , $d_{2}$ , and $d_{3}$ .
According to the problem, the digits must satisfy the following conditions:
1. The digits are selected from the set {1, 2, 3, 4, 5, 6, 7}.
2. There is no repetition of digits.
3. The digits are in decreasing order from left to right, which means:
$d_{1} > d_{2} > d_{3}$
4. Any two digits in the PIN differ by at least 2. Since the digits are in decreasing order, this means:
$d_{1} - d_{2} \geq 2$
and
$d_{2} - d_{3} \geq 2$

To find the maximum number of attempts needed to find the PIN with certainty, we need to find the total number of possible PINs that satisfy all the above conditions. We can list the possible combinations systematically by choosing the first digit $d_{1}$ from the highest possible value (7) downwards:

• If $d_{1} = 7$ :
Since $d_{2} \leq d_{1} - 2$ , the possible values for $d_{2}$ are 5, 4, and 3 (since $d_{2}$ must allow $d_{3} \geq 1$ with a difference of at least 2).
- If $d_{2} = 5$ , then $d_{3}$ can be 3, 2, or 1. (3 possibilities: 7-5-3, 7-5-2, 7-5-1)
- If $d_{2} = 4$ , then $d_{3}$ can be 2 or 1. (2 possibilities: 7-4-2, 7-4-1)
- If $d_{2} = 3$ , then $d_{3}$ can only be 1. (1 possibility: 7-3-1)
Total for $d_{1} = 7$ is $3 + 2 + 1 = 6$ combinations.

• If $d_{1} = 6$ :
The possible values for $d_{2}$ are 4 and 3.
- If $d_{2} = 4$ , then $d_{3}$ can be 2 or 1. (2 possibilities: 6-4-2, 6-4-1)
- If $d_{2} = 3$ , then $d_{3}$ can only be 1. (1 possibility: 6-3-1)
Total for $d_{1} = 6$ is $2 + 1 = 3$ combinations.

• If $d_{1} = 5$ :
The only possible value for $d_{2}$ is 3.
- If $d_{2} = 3$ , then $d_{3}$ can only be 1. (1 possibility: 5-3-1)
Total for $d_{1} = 5$ is 1 combination.

For any value of $d_{1} < 5$ , it is impossible to choose $d_{2}$ and $d_{3}$ such that the difference between each adjacent digit is at least 2 (for example, if $d_{1} = 4$ , the maximum possible $d_{2}$ is 2, which leaves no valid option for $d_{3}$ since $d_{3} \leq 0$ , but the PIN only contains digits from 1 to 7).

Adding all the valid combinations together:

$Total combinations = 6 + 3 + 1 = 10$
Since there are 10 possible unique PINs that satisfy the given conditions, one would need a maximum of 10 attempts to find out the PIN with absolute certainty.