86. At some point on a straight line, there is a particle. In the first second, it splits in half, and the halves move in opposite directions to a distance I from the previous position. In the next second, the resulting particles again split in half, and the halves move in opposite directions to a distance 1 from their previous positions. Upon colliding, any two particles are destroyed, so, for example, after two seconds, only two particles remain. How many particles will remain after 129 seconds?
86. The number and position of particles at moments $t=0$, $1,2,3$ and 4 sec are shown in Fig. 44. We will prove that after

![](https://cdn.mathpix.com/cropped/2024_05_21_7d0959e1f193b6338b84g-69.jpg?height=407&width=854&top_left_y=944&top_left_x=59)

Fig. 44.

$2^{n}-1$ sec there will be $2^{n}$ particles, arranged in a row at a distance of 2 from each other, with the outermost particles being at a distance of $2^{n}-1$ from the initial position, and after $2^{n}$ sec, only two particles will remain at a distance of $2^{n}$ from the initial position. The validity of this statement for $n=1$ and $n=2$ is clear from Fig. 44. Let this statement be already proven for $n=k$. We will prove it for $n=k+1$.

According to the induction hypothesis, at the moment $t=2^{k}$, only two particles will remain at a distance of $2^{k}$ from the initial position. Then, during the time $2^{k}-1$ sec, the "offspring" of the first particle will not interact with the "offspring" of the second, and therefore, as follows from the induction hypothesis, at the moment $t=2^{k}+2^{k}-1=2^{k+1}-1$ sec, $2 \cdot 2^{k}=2^{k+1}$ particles will form, arranged in a row at a distance of 2 from each other. It is clear that at the moment $t+1=2^{k+1}$, only two particles will remain at a distance of $2^{k+1}$ from the initial position.

For $k=7$, we get from this that after 128 sec, there will be two particles at a distance of 128 from the initial position, and the number of particles after 129 sec is 4.