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Nine kids (Posted on 2004-09-09) Difficulty: 3 of 5
My neighbor has nine kids (all under 21), and their ages are such that you can never pick three kids so that the middle one is the average of the ages of the extreme ones. What are their ages?

(Note that you could also say that you can never find three kids whose ages are in arithmetic progression.)

  Submitted by e.g.    
Rating: 4.1250 (8 votes)
Solution: (Hide)
The basic solution, with all ages between 1 and 20, is 1, 2, 6, 7, 9, 14, 15, 18, 20. Replacing all ages with their difference to 21 also works (20, 19, 15, 14, 12, 7, 6, 3, 1) but is essentially the same solution.

If you allow kids to be 0 years old (babies?) then there are many more solutions, as Penny showed.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some ThoughtsProblem SolutionK Sengupta2007-04-20 15:21:27
No Subjectnathan2004-09-23 04:27:22
Solutionre: No Subjectnathan2004-09-15 04:17:51
No Subjectnathan2004-09-15 04:04:05
re(2): Nine Kidsnathan2004-09-14 11:19:13
SolutionNo Subjectnathan2004-09-14 11:17:55
re: Nine KidsJuggler2004-09-14 11:17:11
Nine Kidsnathan2004-09-14 11:10:01
SolutionMany different solutionsPenny2004-09-09 10:22:04
Some Thoughtsre: Brute force solutions (spoilers)Oskar2004-09-09 10:14:51
Some ThoughtsDuplicate ages?Oskar2004-09-09 10:13:35
re: maybee.g.2004-09-09 09:10:58
Solutionre(3): Solution...With a twist of SyzygyJuggler2004-09-09 09:03:22
SolutionBrute force solutions (spoilers)Charlie2004-09-09 09:01:05
maybeMartin2004-09-09 08:46:40
re(2): Solution...With a twist of Syzygye.g.2004-09-09 08:43:46
re: Solution...With a twist of SyzygyJuggler2004-09-09 08:25:22
Some Thoughtsre: (Not a) SolutionFederico Kereki2004-09-09 08:14:10
Some Thoughtsre: Valid Solution ? - I don't think so !Federico Kereki2004-09-09 08:13:19
re: Valid Solution ? - I think so !Juggler2004-09-09 08:10:17
SolutionJuggler2004-09-09 08:07:47
SolutionValid Solution ? - I think so !Syzygy2004-09-09 07:53:06
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