MMM #14 Winner

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The results are in, and the winner for Monday Math Madness #14 is none other than Chao Xu! Honorable mention goes out to Will Sun and Fred Shipley for their cool answers as well. Hope everyone enjoyed the problem. Get ready for MMM #15, which will be hosted over at wildaboutmath.com on Monday!

The Answer by Jeff Hatley

This might be more easily seen with a picture, but here is a written explanation.

First of all, since there are 5 couples total, there are 10 people total. Clearly, the least number of hands that one could shake was 0. The most was 8, since no one is shaking their own hand or their date’s hand.

When I ask everyone how many hands they shook, I get a different answer from each of the 9 people, which means that I hear every number from 0-8.

Now, consider the person who shook 8 hands (call him A). They must have shaken hands with everybody except their date (call her B). Since hand-shaking is symmetric, this means that everybody except B has shaken at least one hand, so B is the only person who could have possibly shaken 0 hands. Thus, we have A=8 and B=0.

Now consider, from the remaining 8 people, the person who shook 7 hands (call him C). He has one handshake from A, and he can’t shake hands with B, himself, or his date (call her D). That means he must have shaken hands with every other person, and thus everybody except for D (and, of course, B) has shaken at least 2 hands. So the only person who could have possibly shaken 1 hand is D. Thus, C=7 and D=1.

Continuing in this way, we see that date/date pairs always shake a number of hands whose sum is 8. So we will get the pairs
(8,0)
(7,1)
(6,2)
(5,3)
(4,4)
I know that I represent one of those people, but which? Well, since everybody gave a different answer when I asked them how many hands they shook, I must have shaken 4 hands.