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Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2

This is an interesting investigation with many possible ways of solving and generalising the problem, and several people wrote in giving an answer of 66 moves.

Isabelle from Lathallan School investigated what happened with different sized grids, and spotted a pattern, which she used to get an answer of 66.

The smallest numbers of moves for 99 squares is 66 moves. I worked out the smallest number of moves for a 3 square grid which was 4. I then worked out the number of moves for a 4 square (which was 2) and a 5 square which was 4. The 6 square was also 4 and so was the 7 square. And then because the moves have to go from a white square and end up on a white square then the number of moves cannot ever be an odd number. So then I tried 8, 9 and 10 square grids which all needed6 moves. So then when I got to 15 squares I needed 10 moves. So then I worked out that a 30 square needs 20 moves. The number of squares is 1.5 times the number of required moves. I tested this at 45 squares (30 moves) and therefore at 99 squares it was 66 moves.

Emily from Durham Johnston Comprehensive School and Tom from Bristol Grammar School both used algebra to explain Isabelle's pattern. Here is Tom's solution:

Firstly, we establish the (obvious) fact that a knight can move down 3 and across 3 in only two moves, and that this is the most efficient way (or certainly no less efficient than any other way) to generally move in the direction from the top left to the bottom right.
Proof: one move is either (2,1) or (1,2). The above move is (3,3), which is the furthest distance you can travel with 2 knight moves, and in precisely the correct direction. We also establish that an even number of moves will be required, because the knight finishes on a square of the same colour to that upon which it started. Therefore, our problem is reduced to the most efficient way to get out of, and into, the corners. Take n> 3 (3 is shown, and less than 3 is impossible). We have three cases.

Case 1: n=3k+1 for some integer k. In this case, we can get across using only the kind of move established above, and we also established that it was most efficient. So we are already done. 2k moves will be required.

Case 2: n=3k-1 for some integer k. In this case, we can reduce it to a 5x5 board using (3,3) moves. We can solve this problem with only 4 moves. Since we know an even number of moves is needed, and that the knight cannot possibly cover the ground in two moves, this must be the most efficient way, so 2k moves are needed here, 2(k-2) to get us to the 5x5 case, and 4 more to complete.

Case 3: n=3k for some integer k. This reduces to a 6x6 square (since n> 3). Again, this can be solved using four moves (see diagram), and cannot be solved with fewer moves by the same logic as in case 2, so 2k is the minimum number of moves needed here: 2(k-2) to get to the 6x6 case, and 4 to finish from there.