We would like to secure the curtain to the rod. We can ensure equal distances simply if, after marking the two outer clips, the middle one is in the remainder, and we even require that this be true for every further division, after securing the middle clip. How many clips can we attach to the rod if we want to secure the curtain this way?
Solution. First, we place the two outer clips, which define one segment. Then, by placing the middle clip, we bisect this segment, so the adjacent clips now define two segments. In the third step, we bisect both segments, thus the adjacent clips now define four segments. In subsequent steps, we continue to bisect the existing segments, so the number of segments doubles with each step; after the $k$-th step, the clips divide the edge of the curtain into $2^{k-1}$ segments. The number of clips is one more than the number of these segments, that is, $2^{k-1}+1$.