If the number $87m6203m$ is divisible by 6, then find the sum of all possible values of *m*.

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**Solution:**

1. For a number to be divisible by 6, it must be divisible by both 2 and 3.

2. **Divisibility by 2:** The last digit of the number must be even.
   - The last digit is *m*.
   - So, *m* must be one of {0, 2, 4, 6, 8}.

3. **Divisibility by 3:** The sum of the digits must be divisible by 3.
   - The digits are: 8, 7, m, 6, 2, 0, 3, m
   - Sum of digits = 8 + 7 + m + 6 + 2 + 0 + 3 + m = (8 + 7 + 6 + 2 + 0 + 3) + 2m = 26 + 2m

4. We want $26 + 2m$ to be divisible by 3.

5. Check for each possible *m* from step 2:
   - For $m=0$: $26 + 0 = 26$, $26 mod 3 = 2$ (not divisible)
   - For $m=2$: $26 + 4 = 30$, $30 mod 3 = 0$ (divisible)
   - For $m=4$: $26 + 8 = 34$, $34 mod 3 = 1$ (not divisible)
   - For $m=6$: $26 + 12 = 38$, $38 mod 3 = 2$ (not divisible)
   - For $m=8$: $26 + 16 = 42$, $42 mod 3 = 0$ (divisible)

6. Possible values of *m* are 2 and 8.

7. Sum of all possible values of *m* = $2 + 8 = 10$.

**Answer:** $10$

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