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My wife and I recently attended a party at which there were four other married couples. Various handshakes took place. No one shook hands with oneself, nor with one's spouse, and no one shook hands with the same person more than once. After all the handshakes were over, I asked each person, including my wife, how many hands he (or she) had shaken. To my surprise each gave a different answer. How many hands did my wife shake?

Solution

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution

Among the five married couples no one shook more than eight hands. Therefore, if nine people each shake a different number of hands, the numbers must be 0, 1, 2, ..., and 8. The person who shook 8 hands has to be married to the person who shook 0 hands (otherwise that person could have shaken only seven hands.) Similarly, the person who shook seven hands is bound to be married to the person who shook 1 hand. So that the married couples shook hands in pairs 8/0, 7/1, 6/2, 5/3. The only person left who shook hands with 4 is my wife.

Reference

  1. M. Gardner, The Colossal Book of Short Puzzles and Problems, W. W. Norton, 2006

    Copyright © 1996-2008 Alexander Bogomolny

29628731Page copy protected against web site content infringement by Copyscape


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