7.1. A certain number was written on the board, multiplied by 7, the last digit of the resulting product was erased, the resulting number was multiplied by 9, the last digit of the resulting number was erased again, and the result was 8. What numbers could have been written initially?
Answer: 13,14 (All answers)

Solution. Let $a$ be the initially written number. $7 a=10 x+y$, then after erasing the last digit, $x .9 x=10 p+q$ remains, where $q$ is the last digit, and after erasing, $p . p=8$ remains. Substitute into the number before erasing: $9 x=80+q$. Since $80+q$ is divisible by 9, then $q=1.9 x=81, x=9,7 a=90+y$, where $y$ is the last digit. Since $90+y$ is divisible by $7, y=1$ or 8. Hence, the answer is: 13 or 14.