# Problem of the Week Problem E and Solution Take a Seat 3

## Problem

Twelve people are sitting, equally spaced, around a circular table. They each hold a card with a 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 $$5$$ and $$6$$ are seated as shown. The person opposite the person holding the $$6$$ is holding the integer $$x$$. What are the possible values of $$x$$?

## Solution

We will start with the card with the integer $$6$$. We are given that $$5$$ is on one side of the $$6$$. Let $$a$$ be the integer on the other side of the $$6$$.

Since each card contains a different integer and the positive difference between the integers on two cards beside each other is no more than $$2$$, then $$a$$ must be $$4$$, $$7$$, or
$$8$$. We will consider these three cases.

Case 1: $$a=7$$

Since the number on each card is different and we know that someone is holding a card with a $$5$$ and someone is holding a card with a $$6$$, then the integer to the right of the $$7$$ must be $$8$$ or $$9$$. Furthermore, every integer to the right of $$7$$ must be greater than $$7$$. Similarly, the integer to left of $$5$$ is either $$3$$ or $$4$$. Furthermore, any integer to the left of $$5$$ must be less than $$5$$.

Since $$x$$ is both to the right of $$7$$ and to the left of $$5$$, it must be both greater than $$7$$ and less than $$5$$. This is not possible.

Therefore, when $$a=7$$, there is no solution for $$x$$.

Case 2: $$a=4$$

Since the number on each card is different and we know that someone is holding a card with a $$5$$ and someone is holding a card with a $$6$$, then the integer to the right of $$4$$ must be $$3$$ or $$2$$. We will look at these two subcases.

• Case 2a: The card with integer $$3$$ is to the right of the card with integer $$4$$.

Notice then that every integer to the right of the $$3$$ must be less than $$3$$. Also the integer to the left of $$5$$ must be $$7$$ and every integer to left of the $$7$$ must be greater than $$7$$. Since $$x$$ is both to the right of $$3$$ and to the left of $$7$$, it must be both greater than $$7$$ and less than $$3$$. This is not possible.

Therefore, when $$a=4$$ and the integer to the right of it is $$3$$, there is no solution for $$x$$.

• Case 2b: The card with integer $$2$$ is to the right of the card with integer $$4$$.

Now, the integer to the left of $$5$$ can be either $$7$$ or $$3$$.

If the integer is $$7$$, then using a similar argument to that in Case 2a, there is no solution for $$x$$.

If the integer to the left of $$5$$ is $$3$$, the only possible integer to the left of $$3$$ is $$1$$. This means the only possible integer to the right of $$2$$ is $$0$$. Which leads to the only possible integer to the left of $$1$$ is $$-1$$. Furthermore, the only possible integer to the right of $$0$$ is $$-2$$. Continuing in this manner, we get the table set up shown below.

From here, the only possible solution is $$x=-5$$.

Therefore, when $$a=4$$, the solution is $$x=-5$$.

Case 3: $$a=8$$

Since the number on each card is different and we know that someone is holding a card with a $$6$$, then the integer to right of the $$8$$ must be $$7$$, $$9$$, or $$10$$. Furthermore, since someone is already holding a $$5$$ and someone is already holding a $$6$$, every other integer to the the right of $$8$$ must be $$7$$ or greater.

The integer to the left of $$5$$ is either $$3$$, $$4$$, or $$7$$. If it is $$3$$ or $$4$$, then since someone is already holding the $$5$$ and someone is already holding the $$6$$, every integer to the left of $$5$$ must be less than $$5$$. Since $$x$$ is both to the right of $$8$$ and to the left of $$5$$, if there is a $$3$$ or a $$4$$ to the left of $$5$$, then $$x$$ must be both $$7$$ or greater and less than $$5$$. This is not possible.

Therefore, if a solution exists when $$a=8$$, then the integer to the left of $$5$$ must be
$$7$$. The integer to the left of $$7$$ must be $$5$$, $$6$$, $$8$$, or $$9$$. Since the $$5$$, $$6$$, and $$8$$ are already placed, then the only possible integer to the left of $$7$$ is $$9$$. Similarly, the only possible integer to the right of $$8$$ is $$10$$. Thus, the integer to the left of $$9$$ must be $$11$$. Continuing in this manner, we get the table set up shown below.

From here, the only possible solution is $$x=15$$.

Therefore, when $$a=8$$, the solution is $$x=15$$.

Therefore, the possible values for $$x$$ are $$x=-5$$ or $$x=15$$.