, 2016

The values in question are all integers $r$ greater than $1$.

We start with a general observation about two particles $P_1$ and $P_2$ moving on $AB$ by the given rules, with different constant speeds $v_1$ and $v_2$, $v_1 > v_2$. Suppose that they are at the same point $Q$ of $AB$ at a certain moment $t$. There are two possibilities for the distances $v_1 t$ and $v_2 t$ the particles have traveled until that moment. If $P_1$ and $P_2$ are moving in the same direction when they simultaneously reach $Q$, then the integer parts of $v_1 t$ and $v_2 t$ have the same parity, and their fractional parts are equal. Hence $v_1 t - v_2 t$ is an even positive integer. And if $P_1$ and $P_2$ are moving in opposite directions when they meet at $Q$, then the integer parts of $v_1 t$ and $v_2 t$ have different parity, and the sum of their fractional parts is $1$. Therefore $v_1 t + v_2 t$ is an even positive integer.

Now let the rational $r > 1$ have the stated property, for any number $n+1$, $n \ge 1$, of particles with speeds $1, r, r^2, \dots, r^n$. Let $t$ be a moment when all of them are at the same point. Then all of them are at the same point of $AB$ at instant $t$. Apply the observation to the first and the last particle, with speeds $v_1 = r^n$ and $v_2 = 1$. We infer that $(r^n - 1)t$ or $(r^n + 1)t$ is an integer. Because $r$ is rational, $t$ is rational too. Write $r$ and $t$ as irreducible fractions: $r = \frac{a}{b}$, $t = \frac{c}{d}$. Then $(r^n \pm 1)t = \frac{(a^n \pm b^n)c}{b^n d}$. Since $a^n \pm b^n$ and $b^n$ are coprime, it follows that $b^n$ divides $c$. Moreover, the latter holds for each $n > 1$ by hypothesis. This is possible only if $b = 1$, that is, if $r > 1$ is an integer.

Conversely, every integer $r > 1$ is a solution. Let $r \ge 3$ be odd and $n \ge 1$ arbitrary. Then all $n+1$ particles with speeds $1, r, r^2, \dots, r^n$ will be at the midpoint of $AB$ at $t = \frac{1}{2}$. Indeed, $r^k \cdot \frac{1}{2}$ has fractional part $\frac{1}{2}$ for each $k = 0, 1, 2, \dots, n$ since $r^k$ is odd.

Let $r = 2m$, $m \ge 1$, be even and $n \ge 1$ arbitrary. Then at the moment $t = \frac{2m}{2m+1}$ all $n+1$ particles with speeds $1, r, r^2, \dots, r^n$ will be at the point $Q$ at distance $\frac{2m}{2m+1}$ from $A$. It is enough to prove that for each $k = 0, 1, 2, \dots$ the next equality holds:

$$(2m)^k \frac{2m}{2m+1} = \begin{cases} 2q + \frac{2m}{2m+1} & \text{con } q = 0, 1, 2, \dots \text{ si } k \ge 0 \text{ es par;} \\ 2q+1 + \frac{1}{2m+1} & \text{con } q = 0, 1, 2, \dots \text{ si } k \ge 1 \text{ es impar.} \end{cases}$$

Indeed, these relations mean that, for $k$ even, the particle with speed $r^k$ will be moving from $A$ towards $B$ at the moment $t = \frac{2m}{2m+1}$, and it will be at distance $\frac{2m}{2m+1}$ from $A$, that is, at point $Q$.

For $k$ odd, the particle with speed $r^k$ will be moving from $B$ towards $A$ at the moment $t = \frac{2m}{2m+1}$, and it will be at distance $\frac{1}{2m+1}$ from $B$, hence at point $Q$ again.

So it remains to prove the displayed equalities, which we do by induction in $k$. The case $k=0$ is obvious. Proceed to the inductive step $k \to k+1$. The induction hypothesis yields

$$k \text{ odd: } (2m)^{k+1} \frac{2m}{2m+1} = 2m(2q+1) + \frac{2m}{2m+1} = 2q' + \frac{2m}{2m+1}, \quad q' = 0, 1, 2, \dots;$$
$$k \text{ even: } (2m)^{k+1} \frac{2m}{2m+1} = 4mq + \frac{4m^2}{2m+1} = 4mq + 2m - 1 + \frac{1}{2m+1} = 2q' + 1 + \frac{1}{2m+1}, \quad q' = 0, 1, 2, \dots$$

This completes the induction and the solution.