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- Newsgroups: rec.puzzles
- Path: sparky!uunet!cs.utexas.edu!hermes.chpc.utexas.edu!news.utdallas.edu!corpgate!crchh327!crchh410!bcash
- From: bcash@crchh410.BNR.CA (Brian Cash)
- Subject: Re: Dividing apples
- Sender: news@news.rich.bnr.ca (news server)
- Message-ID: <C1E5Du.D3B@news.rich.bnr.ca>
- Date: Mon, 25 Jan 1993 03:51:30 GMT
- Reply-To: bcash@crchh410.BNR.CA (Brian Cash)
- References: <19706.2b62bd9d@ecs.umass.edu>
- Nntp-Posting-Host: crchh410
- Organization: Bell-Northern Research Ltd.
- Lines: 74
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- <SPOILER> (maybe...)
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- Is it 351? I figured it this way:
-
- 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.
- I'm rather math-impaired, so I would like to know 1) is this right? 2)
- what is a
- more "scientific" way of finding the answer.
-
- Brian /-|-\
-