Question

Amit has forgotten his 4-digit locker key. He remembers that all the digits are positive integers and are different from each other. Moreover, the fourth digit is the smallest and the maximum value of the first digit is 3. Also, he recalls that if he divides the second digit by the third digit, he gets the first digit. How many different combinations does Amit have to try for unlocking the locker?

• A. 2
• B. 1
• C. 4
• D. 5
• E. 3

Answer 1

Answer: (E)

To solve the problem of finding the number of different combinations of a 4-digit locker key, we need to consider and apply the given conditions systematically.

1. All the digits are positive integers and are different from each other.
2. The fourth digit (let's denote it as \(D_4\)) is the smallest digit.
3. The maximum value of the first digit (denote it as \(D_1\)) is 3.
4. The division of the second digit by the third digit gives the first digit, i.e., \(\frac{D_2}{D_3} = D_1\).

Let's analyze each condition and derive possible values for each digit:

 

1. Since \(D_1\) can be at most 3, \(D_1\) can take values {1, 2, 3}.

 

2. For the division condition (\(\frac{D_2}{D_3} = D_1\)) to hold true, \(D_2\) must be a multiple of \(D_3\). We'll explore combinations:

• If \(D_1 = 1\), then \(\frac{D_2}{D_3} = 1\) implies \(D_2 = D_3\). However, as digits must be different, this is not possible.
• If \(D_1 = 2\), then \(\frac{D_2}{D_3} = 2\) implies \(D_2 = 2 \times D_3\). Possible pairs for \((D_2, D_3)\) are (2,1) and (4,2).
• If \(D_1 = 3\), then \(\frac{D_2}{D_3} = 3\) implies \(D_2 = 3 \times D_3\). Possible pairs for \((D_2, D_3)\) are (3,1), (6,2), (9,3).

3. Additionally, \(D_4\) being the smallest number implies it cannot take any value among \({D_1, D_2, D_3}\) if any of these is 1 since \(D_4\) must be less than these.

 

4. Now let us check all feasible combinations:

• For \((D_1, D_2, D_3, D_4) = (2, 4, 2, 1)\): Does not work as \(D_3\) and \(D_2\) must be different.
• For \((D_1, D_2, D_3, D_4) = (2, 4, 2, \text{and any number not being 2 and minimum})\): Only \(D_4 = 1\).
• For \((D_1, D_2, D_3, D_4) = (3, 3, 1, \text{any number not being 1 and minimum})\), invalid as \(D_2\) and \(D_3\) need to be different.
• For valid combinations:
  • (3, 6, 2, 1) works
  • (3, 9, 3, 1) works

 

Taking into account similar reasoning and ensuring unique and appropriate combinations, we find the following possible solutions: (2, 4, 2, 1), (3, 6, 2, 1), (3, 9, 3, 1)

Thus, the total number of different combinations Amit needs to try is:

\(3\)

Answer 2

Amit needs a 4-digit locker code where all digits are different positive integers. Let's break down the conditions:

1. The digits are positive integers with a maximum value of 9.
2. All digits are distinct.
3. The first digit has a maximum value of 3.
4. The fourth digit is the smallest.
5. The second digit divided by the third digit equals the first digit.

Given the constraints, let's denote the digits as \(a, b, c,\) and \(d\) where: \(a\) is the first digit, \(b\) is the second digit, \(c\) is the third digit, \(d\) is the fourth digit. According to the problem, \(a = \frac{b}{c}\) and the smallest number \(d\) should take the smallest available value that satisfies all conditions:

1. The possible values of \(a\) are 1, 2, or 3.
2. The smallest possible value for \(d\) is 1.

We need to try different permutations for \(a = 1, 2, 3\):

• For \(a = 3\):
  • \(b = 6, 9\), \(c = 2, 3\)
  • Possible 4th digit is 1 (the smallest available).
• For \(a = 2\):
  • \(b = 2, 4, 6, 8\), \(c = 1, 2, 3, 4\)
• For \(a = 1\):
  • \(b = 1, 2, 3, 4, 5, 6, 7, 8, 9\), \(c = b\)

Counting valid sequences from these choices:

• Combination with \(a = 3\), \(b = 9, c = 3, d = 1\) is possible.
• Combination with \(a = 3\), \(b = 6, c = 2, d = 1\) is possible.
• Combination with \(a = 2\), \(b = 4, c = 2, d = 1\) is possible.

Therefore, the total possible unique combinations are:

3 solutions.