Japan 2007

Let $f(n)=n-\lfloor \frac{n}{a}\rfloor -\lfloor \frac{n}{b}\rfloor -\lfloor \frac{n}{c}\rfloor$ for integer $n$. For $n$ positive, $f(n)$ is equal to the contestants with no medals on an $n$-people contest. Since $x-1<[x]\le x$, it follows that $Sn\le f(n)<Sn+3$ where $S=1-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$. To satisfy the conditions $S$ must be positive.

It can be conjectured that $S=\frac{1}{2}$, because $f(n)$ increases in constant speed $S$ and takes every integer larger than or equal to $3$ twice. We will prove a stronger fact and prove this conjecture from that.

Let $L$ be the L.C.M. of $a,b,c$. Let $S=1-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}=\frac{M}{L}$. $M$ is integer, and is positive because $S>0$. For any integer $n$, $$\begin{array}{rl}f(n+L)& =n+L-\lfloor \frac{n+L}{a}\rfloor -\lfloor \frac{n+L}{b}\rfloor -\lfloor \frac{n+L}{c}\rfloor \\ & =n+L-\lfloor \frac{n}{a}\rfloor -\frac{L}{a}-\lfloor \frac{n}{b}\rfloor -\frac{L}{b}-\lfloor \frac{n}{c}\rfloor -\frac{L}{c}\\ & =f(n)+L\left(1-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)=f(n)+M\end{array}$$ and so $f(n+tL)=f(n)+tM$.

Let $\bar{f}(n)$ be the remainder of $f(n)$ divided by $M$. From the last equality, $\bar{f}$ has period $L$. Now we prove the following lemma.

Lemma. Let $L$ and $M$ be positive integers and $g(n)$ be a function from integers to integers such that $g(n+tL)=g(n)+tM$ for all $n,t$. Let $\bar{g}(n)$ be the remainder of $g(n)$ divided by $M$. Then for $0\le k<m$ we have the following: if there are exactly $q$ integers $n$ with $0\le n<L$ and $\bar{g}(n)=k$, for all integer $k^{\prime }$ which is congruent to $k$ modulo $M$ there are exactly $q$ integers $n$ with $g(n)=k^{\prime }$.

Proof of Lemma. Let there be exactly $q$ integers $0\le n_1,\dots ,n_q<L$ with $\bar{g}(n_i)=k$. Take $k^{\prime }=k+uM$. If we write $g(n_i)=k+t_{i}M$ by integer $t_i$ and let $n_i^{\prime }=n_i+(u-t_i)L$, $g(n_i^{\prime })=k^{\prime }$ and all $n_i^{\prime }$ are different since their remainder modulo $L$ is different. Now we are going to prove that possible cases for $n$ with $g(n)=k^{\prime }$ are only $n_1^{\prime },\dots ,n_q^{\prime }$. Suppose $g(n)=k^{\prime }$ then we have $\bar{g}(n)=k$. Let $\bar{n}$ be the remainder of $n$ divided by $L$. Since $\bar{g}$ has a period $L$ it follows that $\bar{g}(\bar{n})=k$ and $\bar{n}$ is equal to some $n_i$. Writing $n=n_i^{\prime }+sL$ we have $g(n_i^{\prime })=g(n)=g(n_i^{\prime }+sL)=g(n_i^{\prime })+sM$ and therefore $s=0,n=n_i^{\prime }$.

Corollary. The condition in the problem is equivalent to the condition that for any $0\le k<M$ there are exactly 2 integers with $\bar{f}(n)=k$ among $0,\dots ,L-1$. And if it is satisfied, $S=\frac{1}{2}$.

Proof of Corollary. If $f(n)\ge 3$, $n$ must be positive, so the first statement follows from the lemma. Then it must be satisfied that $L=2M$ and $S=\frac{M}{L}=\frac{1}{2}$.

Now we are going to solve the problem with this corollary. First, determine all $(a,b,c)$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$. Since $a\ge b\ge c>0$, $\frac{1}{2}>\frac{1}{c}\ge \frac{1}{6}$ and $3\le c\le 6$. If $c=3$, $\frac{1}{a}+\frac{1}{b}=\frac{1}{6}$. By $a\ge b$ we get $\frac{1}{b}\ge \frac{1}{12}$. Trying all possible $b$, we get $(a,b)=(42,7),(24,8),(18,9),(15,10),(12,12)$. Checking other possible $c$ in the same way, we get all $(a,b,c)$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$ as $(a,b,c)=(42,7,3),(24,8,3),(18,9,3),(15,10,3),(12,12,3),(20,5,4),(12,6,4),(8,8,4),(10,5,5),(6,6,6)$.

Checking these possibilities by the corollary, we get the answer to the problem: $(a,b,c)=(6,6,6),(8,8,4),(10,5,5),(12,6,4)$.