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Thirty two barbarians (Posted on 2004-01-28) Difficulty: 3 of 5
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.)

No Solution Yet Submitted by Gamer    
Rating: 3.0000 (6 votes)

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What about this? | Comment 16 of 34 |
I think the fewest is 17 villagers. They are in the shape of an octagon( 8 on each point, and one in between each point) and one in the middle. Let this be reprecented by letters A-Q, with Q as the point in the middle. This will result in 8 lines that contain Q. Example, Line AQI, BQJ, CQK... 8 more lines will be formed by the sides of the octagon. Example, Line ABC, CDE, EFG... This will form 16 lines. With a barbarian on each end of each line the result is 32 barbarians.
The barbarians will killed. Please, let me know if there are any problems with this.

  Posted by York on 2004-01-29 13:39:31
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