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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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 After a careful reading, the answer is less than 20 | Comment 5 of 34 |
To find the very fewest villagers that meet the conditions of the problem statement, the barbarians' bullets must pass completely over the villagers and kill an opposite barbarian on the other side! Note this interpretation is not prohibited by the ambiguous wording of the problem. Consider the villagers arranged thusly:

```.A.....

......B

..CDE..

..FGH..

..IJK..

L......

.....M.```
This is 13 villagers. Every line of three villagers has two bullets passing over it, one from each direction (for instance CDE and EDC). Since there are 16 lines of three villagers in this arrangement, the 32 barbarians can indeed kill one barbarian each. The 16 lines of three are:

CDE
FGH
IJK
CFI
DGJ
EHK
CGK
EGI
ACJ
BEF
BHJ
LFD
LIH
MJF
MKD
 Posted by Bryan on 2004-01-28 15:25:14

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