Japan Mathematical Olympiad

2519 Note that $[x]>x-1$ for any real number $x$. In particular, for an integer $1\le k\le 10$ we have $\lfloor \frac{n}{k}\rfloor >\frac{n}{k}-1$, thus $k\lfloor \frac{n}{k}\rfloor >n-k$. Since both sides are integers, we have $k\lfloor \frac{n}{k}\rfloor \ge n-k+1$, thus $\lfloor \frac{n}{k}\rfloor \ge \frac{n-k+1}{k}$. Since $0<k\le n$ implies $\frac{n-k+1}{k}>0$, multiplying both sides of $\lfloor \frac{n}{k}\rfloor \ge \frac{n-k+1}{k}$ for $k=1,2,\dots ,10$ yields $$\lfloor \frac{n}{1}\rfloor \lfloor \frac{n}{2}\rfloor \dots \lfloor \frac{n}{10}\rfloor \ge \frac{n}{1}\cdot \frac{n-1}{2}\dots \frac{n-9}{10}=\left(\genfrac{}{}{0ex}{}{n}{10}\right).$$ Therefore the equation in the problem holds if and only if $\lfloor \frac{n}{k}\rfloor =\frac{n-k+1}{k}$ for all integers $1\le k\le 10$. If $\lfloor \frac{n}{k}\rfloor =\frac{n-k+1}{k}$ holds, $k$ divides $n-k+1$ because the right side is an integer. Conversely, if $k$ divides $n-k+1$ then $\lfloor \frac{n}{k}\rfloor =\lfloor \frac{n-k+1}{k}\rfloor +\frac{k-1}{k}=\frac{n-k+1}{k}$ holds. Therefore the equation $\lfloor \frac{n}{k}\rfloor =\frac{n-k+1}{k}$ holds if and only if $k$ divides $n-k+1$, that is, $n+1$ is a multiple of $k$. Therefore, for an integer $n\ge 10$, the equation in the problem holds if and only if $n+1$ is a multiple of $k$ for any integer $1\le k\le 10$, that is, $n+1$ is a common multiple of $1,2,\dots ,10$. Since the least common multiple of $1,2,\dots ,10$ is 2520, the smallest possible $n$ is $2520-1=2519$.