SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Obviously $m>9$. If $m=\overline{ab}$, where $a\ge 1$ then we must have $b=1$ to ensure last digit of $1\cdot m$ is equal $1$. In this case regardless of value of $a$ the second last digit of $\underset{k}{\underbrace{11\dots 1}}\cdot m$ is equal to the last digit of $a+1$ and can't be equal $1$. So $m\ge 100$. Obviously $m=100$ doesn't satisfy to the condition, but $m=101$ satisfies, since $11\cdot 101=1111$. Hence, the answer to this problem is $m=101$. $\square$