If 875A21B is divisible by 9. B is an even number Find the sum of all values of A

Answer

To solve the problem, we need to determine the values of A such that the number 875A21B is divisible by 9, given that B is an even number.

### Step-by-step Solution:

1. **Understand the divisibility rule for 9**: A number is divisible by 9 if the sum of its digits is divisible by 9.

2. **Calculate the sum of the known digits**: The digits in the number 875A21B are 8, 7, 5, A, 2, 1, and B. First, we calculate the sum of the known digits:
\[ 8 + 7 + 5 + 2 + 1 = 23 \]
Therefore, the total sum of the digits can be expressed as:
\[ S = 23 + A + B \]

3. **Set up the condition for divisibility by 9**: We need \( S \) to be divisible by 9:
\[ 23 + A + B \equiv 0 \ (\text{mod} \ 9) \]

4. **Calculate \( 23 \mod 9 \)**:
\[ 23 \div 9 = 2 \quad \text{(remainder 5)} \]
Thus,
\[ 23 \equiv 5 \ (\text{mod} \ 9) \]
Therefore, we need:
\[ 5 + A + B \equiv 0 \ (\text{mod} \ 9) \implies A + B \equiv 4 \ (\text{mod} \ 9) \]

5. **Consider the possible values of B**: Since B is an even number, the possible values for B are 0, 2, 4, 6, and 8.

6. **Evaluate each case for B**:
- **Case 1**: If \( B = 0 \):
\[ A + 0 \equiv 4 \implies A \equiv 4 \quad \text{(A = 4)} \]
- **Case 2**: If \( B = 2 \):
\[ A + 2 \equiv 4 \implies A \equiv 2 \quad \text{(A = 2)} \]
- **Case 3**: If \( B = 4 \):
\[ A + 4 \equiv 4 \implies A \equiv 0 \quad \text{(A = 0)} \]
- **Case 4**: If \( B = 6 \):
\[ A + 6 \equiv 4 \implies A \equiv -2 \quad \text{(not valid, A cannot be negative)} \]
- **Case 5**: If \( B = 8 \):
\[ A + 8 \equiv 4 \implies A \equiv -4 \quad \text{(not valid, A cannot be negative)} \]

7. **Collect the valid values of A**: From the cases evaluated, the valid values of A are:
- A = 4
- A = 2
- A = 0

8. **Calculate the sum of all valid values of A**:
\[ 4 + 2 + 0 = 6 \]

### Final Answer:
The sum of all values of A is **6**.