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- Path: sparky!uunet!spool.mu.edu!agate!doc.ic.ac.uk!uknet!cam-eng!cmh
- From: cmh@eng.cam.ac.uk (C.M. Hicks)
- Newsgroups: rec.puzzles
- Subject: Re: Dividing apples
- Message-ID: <1993Jan26.114740.11328@eng.cam.ac.uk>
- Date: 26 Jan 93 11:47:40 GMT
- References: <19706.2b62bd9d@ecs.umass.edu> <C1E5Du.D3B@news.rich.bnr.ca>
- Sender: cmh@eng.cam.ac.uk (C.M. Hicks)
- Organization: cam.eng
- Lines: 26
- Nntp-Posting-Host: club.eng.cam.ac.uk
-
- bcash@crchh410.BNR.CA (Brian Cash) writes:
-
- ><SPOILER> (maybe...)
-
- >Is it 351? I figured it this way:
-
- I think so, and I worked it the same way. I'm now trying to find a general
- solution for dividing Q apples among P people - I'll post if I find one.
-
- >Call the number of apples given to person 1 N, person 2 X, and person 3 Y.
- >Now if N = 25, then X and Y must both equal 0. (1 solution)
- >If N = 24, then {X=1,Y=0} and {X=0, Y=1} are valid. (2 solutions)
- >If N = 23, then {X=1, Y=1}, {X=0, Y=2}, {X=2, Y=0} are valid. (3 solutions)
- >If N = 22, then there are 4 solutions.
- >If N = 21, then there are 5 solutions.
- >...
- >If N = 0, then there are 26 solutions (run X from 0 to 25 and Y is
- >forced to a value).
- >So the total number is 1+2+3+4+...+26 or 351.
-
- Christopher Hicks
- --
- ==============================================================================
- Christopher Hicks | If it doesn't fit...
- cmh@uk.ac.cam.eng | ...you need a bigger hammer.
- ==============================================================================
-