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Topic: hand shakes (Read 778 times) |
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black_death
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N students are seated at desks in an m x n array, where m, n >= 3. Each student shakes hands with the students who are adjacent horizontally, vertically or diagonally. If there are 81 handshakes, what are the values of N,m,n .... can a generic solution be found for a given number of handshakes?
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towr
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Re: hand shakes
« Reply #1 on: May 5th, 2008, 6:54am » |
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For a given n x m array, you have
4 * 3 +
2 * (m-2) * 5 +
2 * (n-2) * 5 +
(m-2)(n-2) * 8
=
4 - 6m - 6n + 8mn half-handshakes
So, 2 - 3m - 3n + 4mn handshakes (since a handshake takes two people)
I'm not sure if there's a nice way to find the inverse for this (to go from X handshakes to m and n). But in any case {n,m}={4,7} works to get 81 handshakes.
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Wikipedia, Google, Mathworld, Integer sequence DB
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Hippo
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Re: hand shakes
« Reply #2 on: May 5th, 2008, 7:15am » |
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It seems to me not all desks must be occupied ... otherwise only two parameters ... m,n are sufficient.
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| « Last Edit: May 5th, 2008, 7:15am by Hippo » |
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black_death
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Re: hand shakes
« Reply #3 on: May 5th, 2008, 7:22am » |
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on May 5th, 2008, 7:15am, Hippo wrote:| It seems to me not all desks must be occupied ... otherwise only two parameters ... m,n are sufficient. |
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ya actually the question in the original form was a MCQ so that you can work backwords .... but when my friend asked me he didn't give me choices so was wondering is there a way to find the values with only one equation ...
btw towr's answer is perfect 
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Eigenray
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Re: hand shakes
« Reply #4 on: May 6th, 2008, 1:44pm » |
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We can rewrite the formula as (4m-3)(4n-3) = 4X+1. There is only a solution when 4X+1 has a divisor d=1 mod 4 with 5<d<(4X+1)/5, so for X = 20, 29, 38, 42, 47, 55, 56, 65, 68, 72, 74, 81, 83, 89, 92, 94, ...
The set of such X has density 1 in the natural numbers.
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| « Last Edit: May 6th, 2008, 1:58pm by Eigenray » |
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