Dripread - Book

307.?THE COLOURED COUNTERS.? solution

The counters may be arranged in this order:?

R1, B2, Y3, 04, GS.

Y4, 05, G1, R2, B3.

G2, R3, B4, Y5, 01.

B5, Y1, 02, G3, R4.

03, 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. 2c/.:?

If, after placing the four 5d. stamps, the reader is tempted to place four 4c/. 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.? solution

The counters may be arranged in this order:?

A1, B2, C3, D4, E5, F6, G7.

F4, G5, A6, B7, C1, D2, E3.

D7, E1, F2, G3, A4, B5, C6.

B3, C4, D5, E6, F7, G1, A2.

G6, A7, B1, C2, D3, E4, F5.

E2, F3, G4, A5, B6, C7, D1.

C5, D6, E7, F1, G2, A3, B4.

310.?THE THREE SHEER? solution

The number of different ways in which the three sheep may be placed so that every pen shall always be either p 9 218 occupied or in line with at least one sheep is forty-seven.

The following table, if used with the key in Diagram 1, will enable the reader to place them in all these ways:?

This, of course, means that if you place sheep in the pens marked A and B, then there are seven different pens in which you may place the third sheep, giving seven different solutions. It was understood that reversals and reflections do not count as different.

If one pen at least is to be not in line with a sheep, there would be thirty solutions to that problem. If we counted all the reversals and reflections of these 47 and 30 cases respectively as different, their total would be 560, which is the number of different ways in which the sheep may be placed in three pens without any conditions. I will remark that there are three ways in which two sheep may be placed so that every pen is occupied or in line, as in Diagrams 2, 3, and 4, but in every case each sheep is in line with its companion. There are only two ways in which three sheep may be so placed that every pen shall be occupied or in line, but no sheep in line with another. These I show in Diagrams 5 and 6. Finally, there is only one way in which three sheep may be placed so that at least one pen shall not be in line with a sheep and yet no sheep in line with another. Place the sheep in C, E, L. This is practically all there is to be said on this pleasant pastoral subject.

311 ?? THE FIVE DOGS PUZZLE.?solution

The diagrams show four fundamentally different solutions. In the case of A we can reverse the order, so that the single dog is in the bottom row and the other four shifted up two squares. Also we may use the next column to the right and both of the two central horizontal rows. Thus A gives 8 solutions. Then B may be reversed and placed in either diagonal, giving 4 solutions. Similarly C will give 4 solutions. The line in D being symmetrical, its reversal will not be different, but it may be disposed in 4 different directions. We thus have in all 20 different solutions.

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312.?THE FIVE CRESCENTS OF BYZANTIUM.? solution

If that ancient architect had arranged his five crescent tiles in the manner shown in the following diagram, every tile would have been watched over by, or in a line with, at least one crescent, and space would have been reserved for a perfectly square carpet equal in area to exactly half of the pavement. It is a very curious fact that, although there are two or three solutions allowing a carpet to be laid down within the conditions so as to cover an area of nearly twenty-nine of the tiles, this is the only possible solution giving exactly half the area of the pavement, which is the largest space obtainable.

313.?QUEENS AND BISHOP PUZZLE.? solution

Amusements in Mathematics (PDF)