Maths Olympiad Prep

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, 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 22. The integers 3, 4, xx, and yy are placed as shown.


What is the value of x+yx + 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:

[[IMAGE0]]

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:

[[IMAGE1]]

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

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

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.