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The problem displayed is in base 5 (quinary system). The equation provided is:

\[
442_{\text{five}} - \Box_{\text{five}} = 124_{\text{five}}
\]

We need to find the missing number represented by the box.

### Step-by-step process:

1. **Convert the base-5 numbers to base-10 (decimal system):**
   - \( 442_{\text{five}} \)
     - \( 442_{\text{five}} = 4 \times 5^2 + 4 \times 5^1 + 2 \times 5^0 = 4 \times 25 + 4 \times 5 + 2 = 100 + 20 + 2 = 122_{\text{ten}} \)
   
   - \( 124_{\text{five}} \)
     - \( 124_{\text{five}} = 1 \times 5^2 + 2 \times 5^1 + 4 \times 5^0 = 1 \times 25 + 2 \times 5 + 4 = 25 + 10 + 4 = 39_{\text{ten}} \)

2. **Set up the equation in base 10:**
   \[
   122_{\text{ten}} - \Box_{\text{ten}} = 39_{\text{ten}}
   \]
   Solving for \( \Box_{\text{ten}} \):
   \[
   \Box_{\text{ten}} = 122_{\text{ten}} - 39_{\text{ten}} = 83_{\text{ten}}
   \]

3. **Convert 83 back to base 5:**
   - To convert 83 to base 5, divide by 5 and track the remainders:
     - \( 83 \div 5 = 16 \) remainder \( 3 \)
     - \( 16 \div 5 = 3 \) remainder \( 1 \)
     - \( 3 \div 5 = 0 \) remainder \( 3 \)

   So, \( 83_{\text{ten}} = 313_{\text{five}} \).

Thus, the missing number in the box is \( 313_{\text{five}} \).

Would you like any more details on this process? Here are some related questions:

1. How do you convert other base systems (like base 8) to decimal?
2. What are common methods for subtracting in different base systems?
3. How can you check your answer when working with non-decimal bases?
4. How is base-5 used in real-world applications?
5. What are some tricks for multiplying or dividing in base-5?

**Tip:** When converting between bases, always double-check your place values to avoid mistakes, especially in non-decimal bases like base-5!

Learn how to solve a base-5 subtraction problem: 442_five - □_five = 124_five. This solution includes detailed steps for converting base-5 numbers to base-10 and vice versa. Suitable for Grades 6-8.

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