Consider an infinite chessboard. You control a horse, standing on a square in the middle of the board. If the horse can perform ten moves, on how many different squares can the horse land?

The category is labeled as 'Out of the box', because it might be too hard to solve this by a logical approach involving maths. First of all, remember that a horse can perform a 2-by-1 move in any direction on this infinite board. From this point onwards, we have to start thinking in terms of 'rows' and 'columns'. Due to the nature of the 2-by-1 move, the horse can only land on evenly spaced squares on the same row. So on one square it can land - on the next it can't - on the next it can - on the next it can't - etcetera. This also count for the vertical direction, so it is also evenly spaced in the columns. Because it has the same effect in the column direction as in the row direction, we get a grid of squares on which it can land that also looks like a chessboard (so, if you take the squares which the horse can land on in a certain row, the row below it has shifted all those squares one to the left/right). The horse can make a maximum horizontal or vertical displacement of $2*10=20$ steps. So now, we have to look at the (41 x 41)-grid in which the horse is able to move ($20+20+ starting \; point=41$ columns/ rows). The next step is to realise that if the horse goes - for example - $10*2=20$ steps upwards, it can go a maximum $10*1=10$ steps sideways. Do you see where this is going? The (41 x 41)-grid will reduce to an octagon, since the horse will not be able to reach the corners of the (41 x 41)-grid. So, the horse is able to land on half of the squares within the octagon, including it's starting point. Therefore, the number of squares the horse can land on is equal to \begin{equation} \frac{41*41 - 4*10*10/2 + 1}{2} = 741 \end{equation} This problem is visualised in the following image:

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