# Problem of the Week Problem C and Solution Take a Seat 1

## Problem

Twelve people are seated around a circular table. They each hold a card with a different integer from $$1$$ to $$12$$ 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 with integers $$1$$, $$3$$, $$a$$, and $$b$$ are seated as shown.

What is the value of $$a + b$$?

Note: The positive difference between two numbers is found by subtracting the smaller number from the larger number.

## Solution

Because two integers that are beside each other must have a positive difference of at most $$2$$, then the possible neighbours of $$1$$ are $$2$$ and $$3$$. Since $$1$$ has exactly two neighbours, then $$1$$ must be between $$2$$ and $$3$$.

Next, consider $$2$$. Its possible neighbours are $$1$$, $$3$$, and $$4$$. The number $$2$$ is already a neighbour of $$1$$ and cannot be a neighbour of $$3$$ (since $$3$$ is on the other side of $$1$$). Therefore, $$2$$ is between $$1$$ and $$4$$. This allows us to update the diagram as shown.

Continuing in this way, the possible neighbours of $$3$$ are $$1$$, $$2$$, $$4$$, and $$5$$. The number $$1$$ is already beside $$3$$, and the numbers $$2$$ and $$4$$ cannot be beside $$3$$. So $$5$$ must be beside $$3$$.

The possible neighbours of $$4$$ are $$2$$, $$3$$, $$5$$, and $$6$$. The number $$2$$ is already beside $$4$$. Numbers $$3$$ and $$5$$ cannot be beside $$4$$. So $$6$$ must be beside $$4$$.

Similarly, we know $$7$$ will be beside $$5$$ and $$8$$ will be beside $$6$$. Thus, $$a=8$$. Continuing this way, we know $$9$$ is beside $$7$$, $$10$$ is beside $$8$$, $$11$$ is beside $$9$$, and $$12$$ is beside $$10$$. Thus, $$b=12$$. The completed circle is shown.

Therefore, $$a + b = 8+12=20$$.