3. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). The result was the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 8?
Solution. For a number to be divisible by 8, "АЛА" must be divisible by 8, with "А" -

![](https://cdn.mathpix.com/cropped/2024_05_06_817b153e53d7268a6f99g-1.jpg?height=60&width=1730&top_left_y=2260&top_left_x=186)
the expression in parentheses is clearly divisible by 8, so it is sufficient to require that ("А" + 2 * "Л") : 8. By enumeration, we find 11 options: (0,4), (0,8), (2,3), (2,7), (4,2), (4,6), (6,1), (6,5), (6,9), (8,0), (8,4). In three of them, where there is a zero, for the remaining five letters ("Г", "В", "Т", "Е", "М") there are \(A_{8}^{5} = 8!/3!\) options; in the other eight \(7 \cdot A_{7}^{4} = 7 \cdot 7!/3!\). In total: \(3 \cdot 8!/3! + 8 \cdot 7 \cdot 7!/3! = 67200\).

Criteria. The divisibility rule for 8 is formulated and it is explicitly stated that "А" is an even digit - 1 point. It is proven that "А" + 2 "Л" or 5 "А" + 10 "Л" is divisible by 8 - 3 points. For each