How many ways are there to choose three digits $A,B,C$ with $1 \le A \le 9$ and $0 \le B,C \le 9$ such that $\overline{ABC}_b$ is even for all choices of base $b$ with $b \ge 10$?
To solve the problem, we need to count the number of ways to choose three digits \( A, B, C \) such that \( \overline{ABC}_b \) is even for all choices of base \( b \ge 10 \). This means that the last digit \( C \) must be even, as the parity of the number in any base is determined by the last digit.

1. **Determine the possible values for \( A, B, \) and \( C \):**
   - \( A \) must be between 1 and 9 inclusive: \( 1 \le A \le 9 \).
   - \( B \) and \( C \) must be between 0 and 9 inclusive: \( 0 \le B, C \le 9 \).

2. **Count the number of ways to choose \( A, B, \) and \( C \) such that \( C \) is even:**
   - The even digits are \( 0, 2, 4, 6, 8 \). Therefore, there are 5 choices for \( C \).

3. **Consider the cases for \( A \) and \( B \):**
   - \( A \) can be any digit from 1 to 9, so there are 9 choices for \( A \).
   - \( B \) can be any digit from 0 to 9, so there are 10 choices for \( B \).

4. **Calculate the total number of combinations:**
   - For each choice of \( A \) and \( B \), there are 5 choices for \( C \).
   - Therefore, the total number of ways to choose \( A, B, C \) such that \( \overline{ABC}_b \) is even is:
     \[
     9 \times 10 \times 5 = 450.
     \]

The final answer is \(\boxed{450}\).