Here is an extension of the last game of solitaire. All you need is a chessboard and the thirty-two pieces, or the same number of draughts or counters. In the illustration numbered counters are used.
The puzzle is to remove all the counters except two, and these two must have originally been on the same side of the board; that is, the two left must either belong to the group 1 to 16 or to the other group, $17$ to $32.$ You remove a counter by jumping over it with another counter to the next square beyond, if that square is vacant, but you cannot make a leap in a diagonal direction. The following moves will make the play quite clear: $3—11,$ $4—12,$ $3—4,$ $13—3.$ Here $3$ jumps over $11,$ and you remove $11;$ $4$ jumps over $12,$ and you remove $12;$ and so on. It will be found a fascinating little game of patience, and the solution requires the exercise of some ingenuity.
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