15. Lorenza is on a track shaped like a regular polygon with 2007 sides, whose vertices are numbered from 1 to 2007 in a counterclockwise direction. Lorenza, starting from vertex 6, jumps over 4 vertices each time and lands on the fifth one ahead (for example, from 20 she jumps to 25), but she jumps back 2 vertices when she lands on a vertex identified by a power of 2 (for example, after a possible jump from 27 to 32, she must jump back to 30). After how many jumps will Lorenza have passed vertex 1 for the first time?
15. The answer is 405. Lorenza starts from vertex 6; after just 2 jumps, she reaches vertex $16=2^{4}$, so she goes back to 14. At this point, Lorenza will alternately jump to an odd vertex (which cannot be a power of 2) and an even one; in particular, between one even vertex and the next, she will take two jumps, corresponding to an increase of 10. Therefore, the next power of 2 must end with the digit 4, like 14; it will thus be 64, and Lorenza will take 10 jumps to reach it. Starting now from 62, the smallest power of 2 greater than 62 and ending with the digit 2 is 512, and to reach it, Lorenza will have to make $\frac{512-62}{5}=90$ jumps. From 510, Lorenza will no longer encounter any obstacles. Indeed, no power of 2 greater than this ends with a 0 (like 510); Lorenza will then take 300 jumps to reach vertex 2010, which is the 3, and thus to surpass the vertex 1 for the first time. Therefore, in total, Lorenza will have jumped $2+10+90+300=402$ times forward and 3 times backward, for a total of 405 jumps.