Here is the solution. Only 8 queens or 8 rooks can be placed on the board without attack, while the greatest number of bishops is 14, and of knights 32. But as all these knights must be placed on squares of the same colour, while the queens occupy four of each colour and the bishops 7 of each colour, it follows that only 21 knights can be placed on the same colour in this puzzle. More than 21 knights can be placed alone on the board if we use both colours, but I have not succeeded in placing more than 21 on the "crowded chessboard." I believe the above solution contains the maximum number of pieces, but possibly some ingenious reader may succeed in getting in another knight.
[307.—THE COLOURED COUNTERS.—solution]
The counters may be arranged in this order:—
| R1, | B2, | Y3, | O4, | GS. |
| Y4, | O5, | G1, | R2, | B3. |
| G2, | R3, | B4, | Y5, | O1. |
| B5, | Y1, | O2, | G3, | R4. |
| O3, | G4, | R5, | B1, | Y2. |
[308.—THE GENTLE ART OF STAMP-LICKING.—solution]
The following arrangement shows how sixteen stamps may be stuck on the card, under the conditions, of a total value of fifty pence, or 4s. 2d.:—
If, after placing the four 5d. stamps, the reader is tempted to place four 4d. stamps also, he can afterwards only place two of each of the three other denominations, thus losing two spaces and counting no more than forty-eight pence, or 4s. This is the pitfall that was hinted at. (Compare with No. 43, Canterbury Puzzles.)