Example: $f(112, 9) = 81$. $112 \cdot 8 < 900 < 999 < 112 \cdot 9$. For $9 < k < 81$ is true that $999 < 112k < 9000$, and $112 \cdot 81 = 9072$, hence $f(112, 9) = 81$. Upper bound on $\textcolor{red}{f(n, d) \le 81}$: Let $A$ be a rational number from $1$ to $10$ such that $n = A \cdot 10^k$, where $k$ is a positive integer. If $A < \frac{10}{9}$, then $f(n, d) = d < 81$, if $A > \frac{10}{9}$, then $f(n, 81) > 9 \cdot 10^{\log_{10} n + 1}$, and since $A < 10$, in the sequence $n, 2n, \dots 81n$ there is a number starting with any nonzero digit.