[359.—COUNTER SOLITAIRE.—solution]

Play as follows: 3—11, 9—10, 1—2, 7—15, 8—16, 8—7, 5—13, 1—4, 8—5, 6—14, 3—8, 6—3, 6—12, 1—6, 1—9, and all the counters will have been removed, with the exception of No. 1, as required by the conditions.

[360.—CHESSBOARD SOLITAIRE.—solution]

Play as follows: 7—15, 8—16, 8—7, 2—10, 1—9, 1—2, 5—13, 3—4, 6—3, 11—1, 14—8, 6—12, 5—6, 5—11, 31—23, 32—24, 32—31, 26—18, 25—17, 25—26, 22—32, 14—22, 29—21, 14—29, 27—28, 30—27, 25—14, 30—20, 25—30, 25—5. The two counters left on the board are 25 and 19—both belonging to the same group, as stipulated—and 19 has never been moved from its original place.

I do not think any solution is possible in which only one counter is left on the board.

[361.—THE MONSTROSITY.—solution]

WhiteBlack,
1.P to KB 4P to QB 3
2.K to B 2Q to R 4
3.K to K 3K to Q sq
4.P to B 5K to B 2
5.Q to K sqK to Kt 3
6.Q to Kt 3Kt to QR 3
7.Q to Kt 8P to KR 4
8.Kt to KB 3R to R 3
9.Kt to K 5R to Kt 3
10.Q takes BR to Kt 6, ch
11.P takes RK to Kt 4
12.R to R 4P to B 3
13.R to Q 4P takes Kt
14.P to QKt 4P takes R, ch
15.K to B 4P to R 5
16.Q to K 8P to R 6
17.Kt to B 3, chP takes Kt
18.B to R 3P to R 7
19.R to Kt sqP to R 8 (Q)
20.R to Kt 2P takes R
21.K to Kt 5Q to KKt 8
22.Q to R 5K to R 5
23.P to Kt 5R to B sq
24.P to Kt 6R to B 2
25.P takes RP to Kt 8 (B)
26.P to B 8 (R)Q to B 2
27.B to Q 6Kt to Kt 5
28.K to Kt 6K to R 6
29.R to R 8K to Kt 7
30.P to R 4Q (Kt 8) to Kt 3
31.P to R 5K to B 8
32.P takes QK to Q 8
33.P takes QK to K 8
34.K to B 7Kt to KR 3, ch
35.K to K 8B to R 7
36.P to B 6B to Kt sq
37.P to B 7K takes B
38.P to B 8 (B)Kt to Q 4
39.B to Kt 8Kt to B 3, ch
40.K to Q 8Kt to K sq
41.P takes Kt (R)Kt to B 2, ch
42.K to B 7Kt to Q sq
43.Q to B 7, chK to Kt 8

And the position is reached.