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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 \(a + b\)?

Twelve chairs are evenly spaced around a
circle. Starting at the chair labelled 3 and moving counterclockwise
around the circle, there is a chair labelled 1, then three chairs with
no labels, then a chair labelled a, then a chair with no label, then a
chair labelled b, then four chairs with no labels, before arriving back
at the chair labelled 3.

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.

Moving counterclockwise from the chair labelled 1, there is
a chair labelled 2, then a chair labelled 4.

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.

Starting at the chair labelled 3 and moving
counterclockwise, the twelve chairs are labelled 3, 1, 2, 4, 6, 8, 10,
12, 11, 9, 7, then 5.

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