# Problem of the Week Problem D and Solution Take a Seat 2

## Problem

Twelve people are seated, equally spaced, around a circular table. They each hold a card with different integer on it. For any two people sitting beside each other, the positive difference between the integers on their cards is no more than $$2$$. The people holding the integers $$3$$, $$4$$, and $$8$$ are seated as shown. The person opposite the person holding $$8$$ is holding the integer $$x$$. What are the possible values of $$x$$?

## Solution

Let $$a$$ represent the integer on the card between the card numbered $$4$$ and the card numbered $$8$$, and let $$b$$ and $$c$$ represent the integers on the cards between the card numbered $$8$$ and the card numbered $$3$$, as shown in the diagram.

The integer $$6$$ is the only integer that is within $$2$$ of both $$4$$ and $$8$$. Therefore, $$a = 6$$. Now, $$b$$ can be either $$7$$, $$9$$, or $$10$$. (We cannot have $$b=6$$ since each person has a card with a different integer on it.) If $$b=9$$ or $$b=10$$, then for $$c$$ there is no integer that is within $$2$$ of $$b$$ and $$3$$. Therefore, $$b=7$$. Furthermore, the integer $$5$$ is the only integer that is within $$2$$ of both $$3$$ and $$7$$. Therefore, $$c = 5$$.

Next, we again consider the card numbered $$4$$. The possible card numbers for its neighbours are $$2$$, $$3$$, $$5$$, and $$6$$. It is already beside the card numbered $$6$$, and the integers $$3$$ and $$5$$ are on cards that are not beside the card numbered $$4$$. Therefore, the card on the other side of the card numbered $$4$$ must be numbered $$2$$.

Next, we again consider the card numbered $$3$$. The possible card numbers for its neighbours are $$1$$, $$2$$, $$4$$, and $$5$$. It is already beside the card numbered $$5$$, and the integers $$2$$ and $$4$$ are on cards that are not beside the card numbered $$3$$. Therefore, the card on the other side of the card numbered $$3$$ must be numbered $$1$$.

We continue in this way to determine that the other card beside the card numbered $$2$$ must be numbered $$0$$. Then, the other card beside the card numbered $$1$$ must be numbered $$-1$$. Then, the other card beside the card numbered $$0$$ must be numbered $$-2$$.

Finally, the possible card numbers for the neighbours of the card numbered $$-2$$ are
$$0$$, $$-1$$, $$-3$$, and $$-4$$. Also, the possible card numbers for the neighbours of the card numbered $$-1$$ are $$1$$, $$0$$, $$-2$$, and $$-3$$. Thus, since $$x$$ is a neighbour of both the card numbered $$-2$$ and the card numbered $$-1$$, we must have $$x=0$$ or $$x=-3$$. Since $$0$$ is already on another card, then $$x=-3$$.