Difference between revisions of "1975 Canadian MO Problems/Problem 6"

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== Solution ==  
 
== Solution ==  
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(i)
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Notice that for each person there will only be one configuration(achieved through rotation) for which the name card matches the guest therefore for the <math>15</math> possible configurations, there will be <math>15</math> correct placed cards. But since one of the configurations does not hold any correct placed card, we shall distribute <math>15</math> correct placed cards to <math>14</math> configurations, hence using the [https://artofproblemsolving.com/wiki/index.php/Pigeonhole_Principle?srsltid=AfmBOorTkqnZSbEc86ZG31ET425xJCLwSsjujr7QRLtcfy4C3L5jJKAb Pigeonhole Principle], there is at least two correct placed cards in one single configuration. 
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(ii)
 
None yet!
 
None yet!
 
 
{{Old CanadaMO box|num-b=5|num-a=7|year=1975}}
 
{{Old CanadaMO box|num-b=5|num-a=7|year=1975}}

Revision as of 12:21, 30 January 2025

Problem 6

(i)$15$ chairs are equally place around a circular table on which are name cards for $15$ quests. The guests fail to notice these cards until after they have sat down, and it turns out that no one is sitting in the correct seat. Prove that the table can be rotated so that at least two of the guests are simultaneously correctly seated.
(ii) Give an example of an arrangement in which just one of the 15 quests is correctly seated and for which no rotation correctly places more than one person.

Solution

(i) Notice that for each person there will only be one configuration(achieved through rotation) for which the name card matches the guest therefore for the $15$ possible configurations, there will be $15$ correct placed cards. But since one of the configurations does not hold any correct placed card, we shall distribute $15$ correct placed cards to $14$ configurations, hence using the Pigeonhole Principle, there is at least two correct placed cards in one single configuration. (ii) None yet!

1975 Canadian MO (Problems)
Preceded by
Problem 5
1 2 3 4 5 6 7 8 Followed by
Problem 7