Article View: rec.puzzles
Article #2482Re: Dividing apples
From: Brian Cash
Date: Mon, 25 Jan 1993 03:51
Date: Mon, 25 Jan 1993 03:51
74 lines
1327 bytes
1327 bytes
<SPOILER> (maybe...)
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 /-|-\
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Newsgroups: rec.puzzles Path: arc1-ca.pugleaf.net!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