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). He got the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 5?
Solution. The number must be divisible by 5, so the letter "A" is equal to 0 or 5. If it is equal to 0, then for the other letters ("G", "V", "T", "E", "M", "L") there are $A_{9}^{6}=9!/ 3$! options; if "A" is equal to 5, then for the other letters there are $8 \cdot A_{8}^{5}=8$!/3! options, since "G" cannot be zero. In total, $9!/ 6+8 \cdot 8!/ 6=114240$ ways.

Criteria. It is explicitly stated that for the letter "A" there are 2 options: 5 or $0-2$ points. If the two cases are further analyzed, that's another 5 points. If the answer is given as an expression not fully calculated to the end - do not deduct points.