Puzzles

The total is 729.

If you started with the numbers 1 to \(2n-1\), what would the total be? Why?

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.

The product of the numbers in the red boxes is 378.

We can split this fraction into:

As long as \(n\) is larger than 123, \(n-123\) is positive and less than \(n\). This means that \(n!/(n-123)\) is an integer, as \(n-123\) will be one of the numbers multiplied together in \(n!\).

(\((124!-123)/(124-123)\) is an integer, so the answer is at least 124 and \(n!/(2-123)\) is an integer when \(n\) is the answer.)

We now need to answer the simpler question: What is the largest value of \(n\) for which \(123/(n-123)\) is an integer? This will be when \(n-123\) is equal to 123, and so \(n\) is 246.

195.

If we add up the first \(n\) even numbers then add on half of the next even number, we get \((n+1)^2\). This means that Holly added up the first \(\sqrt{465124}-1\) or 681 even numbers.

Can you show why adding up the first \(n\) even numbers and half of the next even number gives \((n+1)^2\)?

These numbers all have 3 digits are there are 5 choices for each digit (5, 6, 7, 8, or 9). The number of numbers is therefore 5×5×5=125.

The next terms in the sequence are 107, 707, 713, 323.