CEMC Pascal, 2016

Combinatorics Difficulty 2.6 Junior Find the answer Canada

Each integer from 1 to 12 is to be placed around the outside of a circle so that the positive difference between any two integers next to each other is at most $2$ . The integers 3, 4, $x$ , and $y$ are placed as shown.

What is the value of $x + y$ ?

Pick one

Solution

Because two integers that are placed next to each other must have a 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 follows:

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

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

Continuing to complete the circle in this way, we obtain:

Note that when the even numbers and odd numbers meet (with 12 and 11) the conditions are still satisfied.

Therefore, $x=8$ and $y=12$ and so $x+y=8+12=20$ .

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