# Solution

Prove that in any group of 7 people there is a person who knows an even number of people. You may assume that if a person $A$ "knows" a person $B$, then person $B$ also "knows" person $A$.

Let's argue indirectly...

Assume there is no person who knows an even number of people.

Thus, all 7 people have an odd number of acquaintances.

If, for each person, we count the number of acquaintances, and then look at the total count, we see it is the sum of 7 odd numbers, which itself must be odd.

Now consider finding this total a different way: Notice that two people being acquaintances contributes 2 to the total we just found. (i.e., If person A knows person B, then person B also knows person A.) But then, being the sum of a bunch of 2's, the total in question must be even.

The total can't be both odd and even, so we have our desired contradiction.

Consequently, our assumption must be rejected, and its opposite must be true: there is a person who knows an even number of people.

QED.