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:

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-078.jpg?height=439&width=455&top_left_y=534&top_left_x=889)

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:

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-078.jpg?height=433&width=442&top_left_y=1277&top_left_x=901)

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 $x+y=8+12=20$.

ANSWER: (D)
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 $x+y$ ?
(A) 17
(B) 18
(C) 19
(D) 20
(E) 21

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-046.jpg?height=434&width=442&top_left_y=862&top_left_x=1270)