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We are given a 6 m × 48 m apartment building and three types of units (1-bedroom, 2-bedroom, and 3-bedroom) with different dimensions:

- 1-bedroom units: \( 6 \, \text{m} \times 6 \, \text{m} \)
- 2-bedroom units: \( 6 \, \text{m} \times 12 \, \text{m} \)
- 3-bedroom units: \( 6 \, \text{m} \times 16 \, \text{m} \)

We need to determine how the apartment building, with an area of \( 6 \, \text{m} \times 48 \, \text{m} = 288 \, \text{m}^2 \), can be subdivided into a combination of these units.

### Step-by-Step Solution:

Let:
- \( x \) be the number of 1-bedroom units.
- \( y \) be the number of 2-bedroom units.
- \( z \) be the number of 3-bedroom units.

The total area occupied by each type of unit is:
- Each 1-bedroom unit occupies \( 6 \, \text{m} \times 6 \, \text{m} = 36 \, \text{m}^2 \).
- Each 2-bedroom unit occupies \( 6 \, \text{m} \times 12 \, \text{m} = 72 \, \text{m}^2 \).
- Each 3-bedroom unit occupies \( 6 \, \text{m} \times 16 \, \text{m} = 96 \, \text{m}^2 \).

The total area of the apartment building is \( 288 \, \text{m}^2 \), so the equation to represent the total area is:

\[
36x + 72y + 96z = 288
\]

### Simplifying the Equation:
We can divide the entire equation by 12 to simplify:

\[
3x + 6y + 8z = 24
\]

This is a Diophantine equation (integer solution problem) where \( x \), \( y \), and \( z \) must be non-negative integers. We will now look for integer solutions to this equation.

### Finding the Solutions:
We can systematically check for values of \( z \) (number of 3-bedroom units) and find corresponding values of \( x \) and \( y \).

1. **If \( z = 0 \):**
   \[
   3x + 6y = 24
   \]
   Dividing by 3:
   \[
   x + 2y = 8
   \]
   Possible solutions for \( x \) and \( y \) are:
   - \( x = 8, y = 0 \)
   - \( x = 6, y = 1 \)
   - \( x = 4, y = 2 \)
   - \( x = 2, y = 3 \)
   - \( x = 0, y = 4 \)

2. **If \( z = 1 \):**
   \[
   3x + 6y = 16
   \]
   Dividing by 3:
   \[
   x + 2y = \frac{16}{3} \quad \text{(no integer solutions)}
   \]

3. **If \( z = 2 \):**
   \[
   3x + 6y = 8
   \]
   Dividing by 3:
   \[
   x + 2y = \frac{8}{3} \quad \text{(no integer solutions)}
   \]

4. **If \( z = 3 \):**
   \[
   3x + 6y = 0
   \]
   This gives \( x = 0 \) and \( y = 0 \).

### Solutions:
The possible combinations of 1-bedroom, 2-bedroom, and 3-bedroom units that can fit in the apartment are:
1. \( x = 8, y = 0, z = 0 \) (8 one-bedroom units)
2. \( x = 6, y = 1, z = 0 \) (6 one-bedroom units and 1 two-bedroom unit)
3. \( x = 4, y = 2, z = 0 \) (4 one-bedroom units and 2 two-bedroom units)
4. \( x = 2, y = 3, z = 0 \) (2 one-bedroom units and 3 two-bedroom units)
5. \( x = 0, y = 4, z = 0 \) (4 two-bedroom units)
6. \( x = 0, y = 0, z = 3 \) (3 three-bedroom units)

These are the ways the apartment building can be subdivided into different combinations of 1-bedroom, 2-bedroom, and 3-bedroom units.

Would you like further details or have any questions?

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Here are five follow-up questions to explore:
1. What if the dimensions of the building were different, such as 10 m × 50 m?
2. How would the problem change if the area of each unit type was altered?
3. Can you calculate the total cost if each unit type has a different price per square meter?
4. How would you solve this if there were an additional 4-bedroom unit?
5. What happens if you need to reserve space for common areas?

**Tip:** When solving Diophantine equations, checking the range of integer solutions methodically can save time.

A 6 m × 48 m apartment building may be divided into 1-, 2-, or 3-bedroom units, or combinations thereof. 1-bedroom units are each 6 m × 6 m, 2-bedroom units are each 6 m × 12 m, and 3-bedroom units are each 6 m × 16 m. How may the apartment building be subdivided into one or more types of units?

This problem involves dividing a 6 m × 48 m apartment building into 1-, 2-, or 3-bedroom units based on their respective areas. Learn how to solve the Diophantine equation and find the possible combinations of unit types. Suitable for Grades 10-12.

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