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.)

Assuming that in fact nobody can be killed twice, one way of arranging the people allows for a population of 40:

Form an 8 pointed star by overlaying two concentric squares with one angled at 45 degrees to the other. Place a person at each of the 8 corners of squares and at each of the 8 intersections of lines from the two squares. The 16 people are now in 8 rows of 4 people. The inner 8 people can be the outer 8 people on a similar smaller concentric scheme. When there are 4 sets of outer people, or 8x5=40 people in all (there being one set of 8 inner people who are not further matched with even-more-inner people), there are 32 sight lines (say aimed counterclockwise each) in which three people are passed over and a final fourth is shot dead.

The town must be in a valley so that this is possible--otherwise the mix of tall and short people wouldn't allow some to be shot by some bullets and passed over by others.

Posted by Charlie on 2004-01-28 15:02:06