I pointed out that it was impossible to get all the letters into the box under the conditions, but the puzzle was to place as many as possible. This requires a little judgment and careful investigation, or we are liable to jump to the hasty conclusion that the proper way to solve the puzzle must be first to place all six of one letter, then all six of another letter, and so on. As there is only one scheme (with its reversals) for placing six similar letters so that no two shall be in a line in any direction, the reader will find that after he has placed four different kinds of letters, six times each, every place is occupied except those twelve that form the two long diagonals. He is, therefore, unable to place more than two each of his last two letters, and there are eight blanks left. I give such an arrangement in Diagram 1. [Illustration: 1] [Illustration: 2] The secret, however, consists in not trying thus to place all six of each letter. It will be found that if we content ourselves with placing only five of each letter, this number (thirty in all) may be got into the box, and there will be only six blanks. But the correct solution is to place six of each of two letters and five of each of the remaining four. An examination of Diagram 2 will show that there are six each of C and D, and five each of A, B, E, and F. There are, therefore, only four blanks left, and no letter is in line with a similar letter in any direction. 306.--THE CROWDED CHESSBOARD. [Illustration] 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. 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. 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.:-- [Illustration] 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_.) 309.--THE FORTY-NINE COUNTERS.