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 Thirty two barbarians (Posted on 2004-01-28)
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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 I think... | Comment 33 of 34 |
Because all of the barbarians were positioned around the city, they were obviously not in the same place.  Second, because shots were fired simultaneously, there must be a unique set of 3 individuals for every barbarian's bullet.  Thus, 3*32=96 villagers.  Finally, all of the bullets must end up someplace.  So, because we are trying to find the minimum amount of villagers, we can create a scenario in which all of the bullets ended up in one villager.  So 96+1=97 villagers were present in the town.
 Posted by logischer Verstand on 2004-05-02 20:39:07
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