There is a small town situated by a barbarian colony. The population in this town is very small, but they live well. Upon seeing the villagers in this town so happy, a group of thirty two barbarians sneak up and position themselves around the city. All the barbarians fired at exactly the same time, and every bullet went over 3 villager's heads before it killed another person, including anyone who may have been shot already. If no villager was at the same place at the time (and all villagers were in the town) when the simultaneous shooting occurred, what is the fewest amount of villagers in the town? (Note: "Around" means actually around. A line going around the city would work, but one going out of the city would not.)

(In reply to What about this? by York)

I think your answer is still too high: you can have 16 lines through the center and 8 through the edges. A hexagon with villagers at the points and centers of the sides would give 12 lines through the center and 6 along the edges, which is more than enough and only requires 13 villagers. But this still seems too high as it allows for up to 36 barbarians.

You've pointed out something that I think bears repeating:

Gamer carefully reworded this puzzle after earlier confusion. He specifically states that each bullet went over 3 villager's heads, and killed a person. There is no requirement that that person be a villager. I somehow like the idea that all the barbarians die, and the villagers are all spared.

I also fully agree with Charlie's point that a person can only be killed once, so no person is shot twice.

Posted by Brian Wainscott on 2004-01-29 18:23:48