, 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 $n$,
$$\begin{aligned} f(n+L) &= n+L - \left\lfloor \frac{n+L}{a} \right\rfloor - \left\lfloor \frac{n+L}{b} \right\rfloor - \left\lfloor \frac{n+L}{c} \right\rfloor \\ &= n+L - \left\lfloor \frac{n}{a} \right\rfloor - \frac{L}{a} - \left\lfloor \frac{n}{b} \right\rfloor - \frac{L}{b} - \left\lfloor \frac{n}{c} \right\rfloor - \frac{L}{c} \\ &= f(n) + L \left( 1 - \frac{1}{a} - \frac{1}{b} - \frac{1}{c} \right) = f(n) + M \end{aligned}$$
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 $k'$ which is congruent to $k$ modulo $M$ there are exactly $q$ integers $n$ with $g(n) = k'$.

Proof of Lemma. Let there be exactly $q$ integers $0 \le n_1, \dots, n_q < L$ with $\bar{g}(n_i) = k$. Take $k' = k + uM$. If we write $g(n_i) = k + t_i M$ by integer $t_i$ and let $n'_i = n_i + (u - t_i)L$, $g(n'_i) = k'$ and all $n'_i$ are different since their remainder modulo $L$ is different. Now we are going to prove that possible cases for $n$ with $g(n) = k'$ are only $n'_1, \dots, n'_q$. Suppose $g(n) = k'$ 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 $n = n'_i + sL$ we have $g(n'_i) = g(n) = g(n'_i + sL) = g(n'_i) + sM$ and therefore $s = 0, n = n'_i$.

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: $(a,b,c) = (6,6,6), (8,8,4), (10,5,5), (12,6,4)$.