The 35th Japanese Mathematical Olympiad

Let $n$ be an integer with $1\le n\le 1000$, and for each integer $i$ with $2\le i\le 6$, let $r_i$ denote the remainder when $n$ is divided by $i$. We require that $r_2,r_3,r_4,r_5,r_6$ be pairwise distinct. Since $0\le r_i\le i-1$, in particular $r_2$ can only be $0$ or $1$.

Case 1: Assume $r_2=0$. Then $n$ is even, so $r_4\in \{0,2\}$, and $r_2\ne r_4$ forces $r_4=2$. Similarly $r_6\in \{0,2,4\}$ and $r_6\ne r_2,r_4$ force $r_6=4$, which in turn gives $r_3=1$. Finally $r_5\notin \{0,1,2,4\}$ forces $r_5=3$. Hence the only possible tuple is $$(r_2,r_3,r_4,r_5,r_6)=(0,1,2,3,4),$$ which, by the Chinese remainder theorem, holds precisely when $n\equiv 58(\mathrm{mod}60)$.

Case 2: Assume $r_2=1$. Then $n$ is odd, so $r_4\in \{1,3\}$, and $r_2\ne r_4$ forces $r_4=3$. Similarly $r_6\in \{1,3,5\}$ and $r_6\ne r_2,r_4$ force $r_6=5$, which in turn gives $r_3=2$. Finally $r_5\notin \{1,2,3,5\}$ forces $r_5\in \{0,4\}$. Hence the only possible tuples are $$(r_2,r_3,r_4,r_5,r_6)=(1,2,3,0,5)\text{and}(1,2,3,4,5),$$ which, by the Chinese remainder theorem, hold precisely when $n\equiv 35(\mathrm{mod}60)$ or $n\equiv 59(\mathrm{mod}60)$.

Therefore it suffices to count the integers between $1$ and $1000$ whose remainder modulo $60$ is $35$, $58$, or $59$. There are $17$, $16$, and $16$ such integers, respectively. Hence the total number is $49$.