Answer to the Problem of the Week
for the week of January 1, 2006

The Hand Shakin' Problem
From Car Talk


 

Solution to the Problem:

First, assign each person a letter from A through J.


Now, let's say that A shakes eight hands. Let's say he shakes hands C through J.


It's given that A doesn't shake hands with his spouse (call her B). But in order to get eight handshakes, he has to shake hands with all of the others. That means B must be both the one he doesn't shake hands with AND his spouse — one and the same. Since everyone else has shaken hands, B is the only person who can answer "0" when asked how many hands she has shaken. So 8 is married to 0.


So A is the person who shook hands with eight people.

Okay. So, now, A and B are accounted for, and B, we know, is 0.

We know someone is 1, one hand shake. Let's say that person is D.

D already has his or her handshake — remember? That came from person A.

Now, we know that someone has to shake hands seven times. Let's call that person C. The only way C can shake hands seven times, you'll notice, is if C doesn't shakes hands with D.

Since C doesn't shake hands with his or her spouse, D must be that spouse. So 7 is married to 1.


Repeat this sequence, and you'll see that the person who shakes hands six times is married to the person who shakes hands two times...


And the person who shakes hands five times is married to the person who shakes hands three times.


The person who shakes hands four times... is married to another person who shakes hands four times.

Guess what? We didn't hear the number four called off twice when Elvis asked the other nine people how many times they shook hands. So, Elvis must have shaken hands four times, and therefore, his wife Priscilla also shook hands four times.