I hoped Mike A could have found the complicated answer before the solutions… As I said, his 200/75 was very close to the solution 175/75. I’ll not brag for having found that solution: once committed to trials, to find the right solution was a question of luck or perseverance if there are no mistakes made. I reached it but I could haven’t. If the 200/75 was thought (just) second by Mike A for some logical reason, then he has my admiration.

But the solution 150/125 is much more interesting and a true revelation to me. To understand why, I’ll need to return to the 150/25 solution. That one is obvious. All the distances are a multiple of 25, so we only need a reasonable long distance club (multiple of 25) and one 25. We simply shot the big distances with the first one and do small corrections with the second. We could even have done everything with the small one. Based on that, 150/125 is an immediate possibility too because 150-125=25, meaning every two shots can reproduce the action of the previous 25 yards “universal” club. This appears to be a bad trade but there is an advantage in it: if some big distances are around a multiple of the 150 club, then to get the immediate smaller distances is enough to substitute 150 shots by 125 shots to get that multiple, minus 25, minus 50, et cetera, distances… where with the 150/25, I had to shoot a multiple of 150 and then shoot back multiples of 25, here to get the same effect, I only have to substitute 150 shots by 125… saving shots. If by some reason, there is advantages in having a “long” club not “long” enough to get the many big distances in one shot, but in several instead, then we have enough long shots to be substituted and flexibility to use this to our advantage. Once I thought that, it became obvious that solutions type 150/125, 175/150, et cetera, are strong candidates to the solution too. And so it was.