# Answer: 31

Solution. The sum of two two-digit numbers is no more than 199, so $\overline{\text { YAG }}$ is a three-digit number starting with 1, Y $=1$. Let's look at the last digit in each number, A. It is added twice and subtracted once, so the value of the expression is $\mathrm{A}$, and $\mathrm{A} \neq 0$. $\mathrm{A} \neq 1$, since Y = 1. In addition, B $\geq 5, B+B=\overline{Y \Gamma}$. If B $=5$, then A can be any digit except 1, 0, and 5 (7 options). Further, if B $=6,7,8,9$, then A can be any digit except 0, 1, B, and $\Gamma$, and $\Gamma$ is determined as the units digit of the number 2B (it is not equal to 1, 0, or B). Thus, in these cases, there are 6 options left. In total, there are $7+4 \cdot 6=7+24=31$ options.
7.1 In the example of addition and subtraction, the student replaced the digits with letters according to the rule: identical letters are replaced by identical digits, different letters are replaced by different digits. From how many different examples could the record $0<\overline{\overline{Б A}}+\overline{\text { БА }}-\overline{\text { ЯГА }}<10$ be obtained?