Give me a hand

A professor and their spouse are at a party. At the party there are four more couples (overall five couples).

During the party couples shake hands with the following rules:
a. One does not shake hands with oneself.
b. One does not shake his/her spouse’s hand.

At the end of the party the professor asked all the other guests at the party (including their spouse) how many different people they shook hands with. Each person tells him a different answer (meaning, if one person said “five,” no one else said “five”).

With how many people did the professor shake hands?

Communicated by Michael Geike (who heard it in college in 1999).

 

FIRST THOUGHT

The answer to this riddle follows immediately from a little bit of logic. First, we notice that since nobody shakes hands with his or her partner, each of the party-goers can shake hands with at most 8 people. Since there are 9 people at the party besides the professor, the numbers that we are concerned with in this riddle must be the numbers 0 through 8. We immediately notice that the professor’s spouse is different from the rest of the people, since the professor could have shaken as many hands as he wanted, regardless of how many someone else did. For the rest of the invitees, if one person shook hands with a certain number of people, nobody else did.

 

SECOND THOUGHT

How can we use this information? Perhaps we can try a couple of numbers with the professor’s wife, and see if any of them are not possible. We start with 0, since it is the least. If the professor’s spouse shakes hands with 0 people, then nobody else (excluding the professor) shakes hands with 0 people. Someone else in the crowd, however, shakes hands with all eight people. This contradicts the fact that the professor’s spouse shakes hands with no people. Therefore, the professor’s spouse must shake the hand of at least one person.

 

THIRD THOUGHT

We conclude from the preceding paragraph that someone else in the crowd must be the person who shakes zero hands. Following the same logic as before, if we choose anyone besides zero’s spouse to be the person that shakes all eight hands, then we will reach a contradiction, with Zero shaking someone’s hand. It follows that Zero’s spouse will shake all eight hands.

We repeat this logic several more times to reach the solution. For the sake of clarity, the next step is as follows: The professor’s spouse cannot shake hands with only 1 person because there is someone else in the crowd who has to shake hands with 7 people, excluding Zero. Therefore, someone else must shake hands with only 1 person. If anyone besides One’s spouse shakes hands with 7 people, then One will shake the hand of more than one person, resulting in another contradiction. We conclude that One’s spouse is Seven.

Here, we notice a pattern. We see that 8 + 0 = 0 and 7 + 1 = 8. We could keep going in the way we have before, but it is easy to see that this pattern continues (the logic will continue in the same fashion). Therefore, we can just assign all of the remaining numbers. This leads to Six being paired with Two and Five being paired with Three. The professor’s spouse must then be Four.

To determine how many people the professor shook hands with, we notice that the only people who shook hands with the professor in our process were the ones whose numbers were greater than 4. These people are Five, Six, Seven, and Eight. We conclude that the professor shook hands with 4 people.