Puzzles

The product of the numbers in the red boxes is 120.

Between 100 and 109 (inclusive), there are 5 integers whose digits add up to an even number, and 5 whose digits add up to an odd number. Between 110 and 119 (inclusive), there are 5 integers whose digits add up to an even number, and 5 whose digits add up to an odd number... In general, between \(10n\) and \(10n+9\) (inclusive), there are 5 integers whose digits add up to an even number, and 5 whose digits add up to an odd number.

The integers from 100 to 999 (inclusive) can be split into 45 sets of integers from \(10n\) to \(10n+9\) (and the digits of 1000 don't add to an even number), so there are 450 integers between 100 and 1000 whose digits add up to an even number.

Let \(a\) be the height of the rectangle. As the total of the blue lines is 50, the width of the rectangle is \(50-a\). As the total of the red lines is 50, each red line segment is 25.

Using Pythagoras's theorem in one of the right-angled triangles, we see that:

\(a\) is not zero, and so \(a=20\). This means that the area of the rectangle is 20×30=600.

The total is 729.

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

The number of ways to form sets like this containing the numbers 1 to \(n\) is the \((n+2)\)th Fibonacci number. Therefore there are 987 ways for the numbers from 1 to 14.

Can you show why the solution is the Fibonacci numbers in a similar way to in the solution to the puzzle on 6 December?

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.