Japan Mathematical Olympiad Initial Round

Let $M$ be the maximum number we seek. First we show that $M\le 939$ must hold. Let $m,n$ be positive integers satisfying the conditions of the problem, and let $k$ be the number of digits of $n$ (so, $1\le k\le 1000$). Let us represent $\frac{m}{n}$ as $$\frac{m}{n}=a_{1}10^{b_1}+a_{2}10^{b_2}+\cdots +a_{N}10^{b_N}+ϵ,$$ where $1\le a_i\le 9$, for $i=1,2,\dots ,N$, $0\le b_N<b_{N-1}<\cdots <b_1$, and $0\le ϵ<1$. Then from $10^{999}\le m<10^{1000}$ and $10^{k-1}\le n<10^k$ it follows that we have $b_1\le 1000-k$. We can also show that the following lemma holds:

Lemma: $b_i-b_{i+1}\le k+1$ holds for every $i,1\le i\le N-1$, and $b_N\le k$ holds also.

Proof of Lemma: Suppose for some $i,1\le i\le N-1$, $b_i-b_{i+1}\ge k+2$ is satisfied. Then, if we let $$A=a_{1}10^{b_1-b_{i+1}-k-2}+a_{2}10^{b_2-b_{i+1}-k-2}+\cdots +a_{i}10^{b_i-b_{i+1}-k-2},$$ $$B=a_{i+1}10^{b_{i+1}}+a_{i+2}10^{b_{i+2}}+\cdots +a_{N}10^{b_N}+ϵ,$$ we see that $A$ is a positive integer and $B$ satisfies $0<B<10^{b_i+1+k}$. We see that with $A$ and $B$ as above we can write $\frac{m}{n}=10^{b_i+1+k+2}A+B$. But then, since we have $$10^{b_i+1+k+2}{A}_n<m=10^{b_i+1+k+2}{A}_n+Bn<10^{b_i+1+k+2}{A}_n+10^{b_i+1+k+1},$$ we have to conclude that the $10^{b_i+1+k+1}$-th digit of $m$ is $0$. But this contradicts the assumption that all the digits of $m$ are not zero, and therefore, $b_i-b_{i+1}\le k+1$ must be satisfied for every $i,1\le i\le N-1$.

Next assume that $b_N\ge k+1$ is satisfied. If we let $$C=a_{1}10^{b_1-k-1}+a_{2}10^{b_2-k-1}+\cdots +a_{N}10^{b_N-k-1},$$ then we see that $C$ is a positive integer and we can write $\frac{m}{n}=10^{k+1}C+ϵ$. But then we have $$10^{k+1}{C}_n\le m=10^{k+1}{C}_n+ϵn<10^{k+1}{C}_n+10^k,$$ which implies that the $10^k$-th digit of $m$ is $0$, contradicting the assumption of the problem. Therefore, we must have $b_N\le k$, and this completes the proof of Lemma.

Going back to the proof of the claim that $M\le 939$, we note that the number of $0$ digits in $\lfloor \frac{m}{n}\rfloor$ equals $b_1+1-N$. By summing the corresponding sides of the following $N$ inequalities shown in Lemma above $$b_1-b_2\le k+1,b_2-b_3\le k+1,\dots ,b_{N-1}-b_N\le k+1,b_N\le k,$$ we obtain $b_1\le (k+1)N-1$, which yields $\frac{b_1+1}{k+1}\le N$. Consequently, we get $$\begin{array}{rl}{b}_1+1-N& \le b_1+1-\frac{b_1+1}{k+1}\\ & =\frac{k{b}_1-1}{k+1}+1\le \frac{k(1000-k)-1}{k+1}+1\\ & =1003-\left((k+1)+\frac{1002}{k+1}\right)\le 1003-2\sqrt{1002}<940,\end{array}$$ where we used the fact $(k+1)+\frac{1002}{k+1}\ge 2\sqrt{(k+1)\cdot \frac{1002}{k+1}}=2\sqrt{1002}$. This shows our claim that $M\le 939$.

Finally, we show that $M\ge 939$ by exhibiting a particular choice of $m$ and $n$, satisfying the conditions of the problem, for which $M=939$. For this purpose, consider $$m=\stackrel{31\text{ digits}}{\overbrace{211\cdots 1}}\stackrel{32\text{ digits}}{\overbrace{1266\cdots 6}}\stackrel{32\text{ digits}}{\overbrace{1266\cdots 6}}\stackrel{32\text{ digits}}{\overbrace{1266\cdots 6}}\stackrel{9\text{ digits}}{\overbrace{1266\cdots 6}},n=\stackrel{31\text{ digits}}{\overbrace{211\cdots 1}}$$ Then, we see that $\lfloor \frac{m}{n}\rfloor =1\stackrel{32\text{ digits}}{\overbrace{00\cdots 06}}\stackrel{32\text{ digits}}{\overbrace{00\cdots 06}}\stackrel{32\text{ digits}}{\overbrace{00\cdots 06}}\stackrel{32\text{ digits}}{\overbrace{00\cdots 06}}\stackrel{9\text{ digits}}{\overbrace{00\cdots 0}}$ and, therefore the number of its $0$-digits is $939$. Thus, we conclude that $939$ is the desired answer to the problem.