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THE KNIGHT'S CENTURY.

THEN Dr. Bates drew on a large sheet of paper a reduced chessboard of twenty-five squares, and numbered it in the manner shown in our diagram. He then placed a chess knight on the central square, which was not numbered, and asked us to make a series of knight's moves (as many as we chose, only never using the same square twice), so that the numbers on the squares used should add up to exactly one hundred.

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Kenneth got very near it by taking the course indicated by the dotted line, for the numbers 42, 13, 13, 33, add up to 101, only one too many. A number of rough copies of the diagram were made, and the party worked for some considerable time, making haphazard trials and counts, all without success. But after a little figuring of my own, apart from the diagram, I was able to show them without any difficulty what is obviously the only possible solution. Perhaps the reader can find it.

Here is a copy of the chessboard without the dotted lines of the path example:

enter image description here

This puzzle was published by Henry Dudeney in The Strand Magazine in December 1922.

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2 Answers 2

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The path is as follows:

enter image description here

I found this path after

reducing the entire board modulo 11:

 +2 -2 -2 =0 =0

 +2 =0 -2 +2 =0

 -2 +2 NA +2 +2

 -2 +2 =0 -2 =0

 =0 =0 +2 +2 +2

We can't reach 99+1, unless we stack enough +2's or -2's to wrap around to +12 or -10, and since all the -2's come from 31's and 42's, we can't hope to get enough before our sum exceeds 100. Therefore, we snag as many +2's from 13's as we can, and the sum works out.

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    $\begingroup$ +1 Your reduction idea yields a very simple and elegant chessboard. $\endgroup$
    – Will.Octagon.Gibson
    Commented Apr 20 at 20:52
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The digits of the numbers are both odd or both even.
That means that the last digits cannot add up to 10 or 30 while the first digits add up to 9 or 7 respectively.

If the first digits add up to 8:
If we do use even numbers, 4?+1?+1?+1?+1? is the only one that may lead to 80+20, but even with using 13s, we would need 48, which does not exist.
If we do not use even numbers, we need at least 7 numbers (since odd numbers only end in 1 or 3) and thus all numbers must be <30.

The numbers below 30 $(6*13+2*11)$ can indeed be used to form 100. (and all are needed)

The blue lines are mandatory; they can be connected by either the green or purple ones.
enter image description here

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  • $\begingroup$ This is an alternative argument on how to find the path. Maybe add a drawing or some other description of the final solution to make this a complete answer. $\endgroup$
    – quarague
    Commented Apr 20 at 9:46
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    $\begingroup$ Dudeney is mistaken. There are two solutions, albeit involving the same squares. I was delighted to discover this but less so when I saw that your answer already took account of both paths. $\endgroup$
    – David Conrad
    Commented Apr 21 at 0:24

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