The difference of the digits of a two-digit number is 8. The sum of its digits can be ____

Options

A. 8
B. 10
C. Either (a) or (b)
D. Neither (a) or (b)

Explanation

To solve the problem, we need to find the possible sums of the digits of a two-digit number given that the difference of its digits is 8. Let's denote the digits of the two-digit number as \( x \) (the tens place) and \( y \) (the units place).

### Step-by-step Solution:

1. **Understanding the Problem**:
We know that the difference between the digits \( x \) and \( y \) is 8. This can be expressed as:
\[ |x - y| = 8 \]
This means either \( x - y = 8 \) or \( y - x = 8 \).

2. **Case 1: \( x - y = 8 \)**:
- Here, \( x \) is greater than \( y \).
- Since \( x \) and \( y \) are digits (0 to 9), the maximum value for \( y \) can be calculated:
\[ y = x - 8 \]
- If \( x \) is at its maximum value of 9, then:
\[ y = 9 - 8 = 1 \]
- The valid pairs of \( (x, y) \) are:
- \( (9, 1) \)
- The sum of the digits for this pair is:
\[ x + y = 9 + 1 = 10 \]

3. **Case 2: \( y - x = 8 \)**:
- Here, \( y \) is greater than \( x \).
- We can express this as:
\[ y = x + 8 \]
- Since \( y \) must also be a digit (0 to 9), the maximum value for \( x \) can be calculated:
\[ x \leq 1 \quad (\text{because } 1 + 8 = 9) \]
- The valid pairs of \( (x, y) \) are:
- \( (1, 9) \)
- The sum of the digits for this pair is:
\[ x + y = 1 + 9 = 10 \]

4. **Conclusion**:
The possible sums of the digits \( x + y \) for the two-digit number with the given condition are:
- From Case 1: 10
- From Case 2: 10

Thus, the sum of the digits can be **10**.