5. 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 25?
Solution. The number must be divisible by 25, so “$\lambda$A” equals 25, 50, or 75 (00 cannot be, as the letters are different). If “LA” equals 50, then for the other letters (“G”, “V”, “T”, “E”, “M”) there are $A_{8}^{5}$ options; otherwise, for the other letters there are $7 \cdot A_{7}^{4}$ options. In total, $8!/ 6 + 2 \cdot 7 \cdot 7!/ 6 = 18480$ ways.

Criteria. It is explicitly stated that the option 00 is not suitable for the letters “LA”, and all 3 cases are demonstrated - 3 points. If the 2 variants are further analyzed without errors, then another 4 points. If the answer is given as an expression not fully calculated - do not deduct points.