Coming and Going - Futility Closet

A puzzle by Pierre Berloquin:

In my house are a number of rooms. (A hall separated from the rest of the house by one or more doors counts as a room.) Each room has an even number of doors, including doors that lead outside. Is the total number of outside doors even or odd?

	Select Click for Answer
Call a door internal if it opens between rooms or external if it opens outside. And designate each room either O (if it has an odd number of internal doors) or E (if even). Each E room has an even number of external doors, since all rooms have an even number of doors and even – even = even. So the E rooms, taken as a group, have a total number of external doors that’s even. The total number of O rooms is either odd or even. If it’s odd, then the total number of doors in the O rooms is odd, since the sum of an odd number of odd numbers is odd. But this is impossible: Every internal door must open on two rooms, and this would leave one door that has no second room to open on. So the total number of O rooms is even. Since each O room has an odd number of internal doors and an even number of total doors, each must have an odd number of external doors. Because the total number of O rooms is even, this means that the O rooms taken as a group have an even number of external doors (since the sum of an even number of odd numbers is even). Since both groups, the E rooms and the O rooms, have an even number of external doors, so does the whole house. L.A. Graham provided a simpler solution in 1959.

Select Click for Answer

Call a door internal if it opens between rooms or external if it opens outside. And designate each room either O (if it has an odd number of internal doors) or E (if even).

Each E room has an even number of external doors, since all rooms have an even number of doors and even – even = even. So the E rooms, taken as a group, have a total number of external doors that’s even.

The total number of O rooms is either odd or even. If it’s odd, then the total number of doors in the O rooms is odd, since the sum of an odd number of odd numbers is odd. But this is impossible: Every internal door must open on two rooms, and this would leave one door that has no second room to open on. So the total number of O rooms is even.

Since each O room has an odd number of internal doors and an even number of total doors, each must have an odd number of external doors. Because the total number of O rooms is even, this means that the O rooms taken as a group have an even number of external doors (since the sum of an even number of odd numbers is even).

Since both groups, the E rooms and the O rooms, have an even number of external doors, so does the whole house.

L.A. Graham provided a simpler solution in 1959.