Problem 1315

To solve the problem, we need to find the maximum six-digit number $\overline{ABCDEF}$ with distinct digits between 1 and 8 that is divisible by 99. A number is divisible by 99 if and only if it is divisible by both 9 and 11.

1. Divisibility by 9:
A number is divisible by 9 if the sum of its digits is divisible by 9. The digits are distinct and between 1 and 8, so the sum of all possible digits is:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$$
Since 36 is divisible by 9, any permutation of these digits will satisfy the divisibility by 9 condition.

2. Divisibility by 11:
A number is divisible by 11 if the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is a multiple of 11 (including 0). Let’s denote the digits as $A, B, C, D, E, F$:
$$(A + C + E) - (B + D + F) \equiv 0 \pmod{11}$$
or
$$(A + C + E) - (B + D + F) \equiv 11 \pmod{11}$$

3. Maximizing the number:
To maximize $\overline{ABCDEF}$, we should start with the largest digits and check if they satisfy the divisibility conditions. We will try to place the digits in descending order and check the conditions.

- Start with the largest digits: 8, 7, 6, 5, 4, 3, 2, 1.
- Check the sums of the odd and even positions:
$$\text{Odd positions: } A, C, E \quad \text{Even positions: } B, D, F$$
$$\text{Odd positions sum: } 8 + 6 + 4 = 18$$
$$\text{Even positions sum: } 7 + 5 + 3 = 15$$
$$18 - 15 = 3 \quad (\text{not a multiple of 11})$$

- Adjust the digits to satisfy the divisibility by 11 condition. We need to find a permutation where the difference is 0 or 11.

4. Finding the correct permutation:
- Try different permutations while keeping the digits as large as possible.
- After several trials, we find that the permutation 87653412 satisfies both conditions:
$$\text{Odd positions sum: } 8 + 6 + 3 + 1 = 18$$
$$\text{Even positions sum: } 7 + 5 + 4 + 2 = 18$$
$$18 - 18 = 0 \quad (\text{multiple of 11})$$

Thus, the maximum possible value of $\overline{ABCDEF}$ that satisfies the given conditions is $\boxed{87653412}$.