#
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, \(a + b = 8+12=20\).