Puzzles

Let \(a_n\) be the number of ways to tile a \(n\times2\) rectangle. It is easy to check that \(a_1=1\) (ie there is 1 way to tile a 1×2 rectangle) and \(a_2=2\) (ie there are 2 ways to tile a 2×2 rectangle).

For an \(n\times2\) rectangle, from the left the tiling either starts with a vertical tile, or a pair of horizontal tiles. If it starts with a vertical tile, then there are \(a_{n-1}\) ways to tile the remaining \((n-1)\times2\) rectangle. If it starts with a pair of horizontal tile2, then there are \(a_{n-2}\) ways to tile the remaining \((n-2)\times2\) rectangle. Therefore, \(a_n=a_{n-1}+a_{n-2}\).

(And so the number of ways to tile a \(n\times2\) rectangle is the \((n+1)\)th Fibonacci number.)

Therefore, the number of ways to tile a 12×2 rectangle is 233.

Each number will appear 7 times: one time paired with the numbers six numbers 5 to 10, plus an extra appearance on the tile containing the same number twice. The total of all the numbers is therefore 7×(5+6+...+10)=7×45=315.

Let \(a_n\) be the number of ways to tile an \(n\times2\) rectangle.

There is one way to tile a 1×2 rectangle; and there are three ways to tile a 2×2 rectangle. Therefore \(a_1=1\) and \(a_2=3\).

In a \(n\times2\) rectangle, the rightmost column will either contain a vertical 2×1 domino, two horizontal 2×1 dominoes, or a 2×2 square. Therefore \(a_n=a_{n-1}+2a_{n-2}\).

Using this, we find that there are 341 ways to tile a 9×2 rectangle.