Foxes and Geese

The miscellany goes on, Nos. 87 to 100: a toy train, a poultry farm, a primrose, a round table and Nelson’s Column among them. The chapter takes its name from No. 94, an old game played on a board shaped like a cross.

No. 87. The Chifu-Chemulpo Puzzle

No. 88. The Eccentric Market-woman

No. 89. The Primrose Puzzle

No. 90. The Round Table

No. 91. The Five Tea Tins

No. 92. The Four Porkers

No. 93. The Number Blocks

No. 94. Foxes and Geese

No. 95. Robinson Crusoe’s Table

No. 96. The Fifteen Orchards

No. 97. The Perplexed Plumber

No. 98. The Nelson Column

No. 99. The Two Errand Boys

No. 100. On the Ramsgate Sands

The Chifu-Chemulpo Puzzle

Twenty moves cannot be done; the fewest possible is twenty-six. The picture settles how the track works. The siding leaves the main line at one set of points and rejoins it at another, and beyond each set of points is a short dead end with room for one vehicle, where car 1 and car 8 stand. Because 7 can go onto the siding while 8 stays at the end, a vehicle can stand on the points themselves and back onto the siding from there. Along the main line or along the siding, vehicles can shuffle up and down for nothing, but they can never pass one another, so what matters is the order of the vehicles on each track, and every change of order costs a trip over the points.

The table gives one route of twenty-six moves. Each line shows the order along the main line from left to right, and along the siding from its left end, after the move of that number. A vehicle leaving the main line at the left goes onto the left end of the siding, and one coming off at the right leaves from the right end.

Dudeney gives a different route, also of twenty-six moves. No route takes fewer.

Answer 26 moves

The Eccentric Market-woman

Try a small sale: from 29 eggs she sells 15 and leaves 14. Adding one to each stock turns 30 into 15, an exact halving. This suggests keeping track of the eggs-plus-one. If she has $x$ eggs and sells half of them and half an egg more, she is left with $x-\tfrac{x+1}{2}=\tfrac{x-1}{2}$, so the number she has plus one halves. In the same way the sale of a third and a third of an egg leaves her two-thirds of her eggs-plus-one, and so on. Starting with $N$ eggs, after the four sales she has $$(N+1)\times\frac12\times\frac23\times\frac34\times\frac45-1=\frac{N+1}{5}-1 .$$ The four sales are $(N+1)/2$, $(N+1)/6$, $(N+1)/12$ and $(N+1)/20$ eggs. For all four to be whole numbers, $N+1$ must be divisible by their denominators. Their least common multiple is 60, so write $N+1=60k$. What is left, $12k-1$, must be shared among thirteen friends. The smallest $k$ that makes $12k-1$ a multiple of 13 is $k=12$, giving $12\times 12-1=143=13\times 11$.

So she took $N=60\times 12-1=719$ eggs. She sold 360, 120, 60 and 36, kept 143, and each friend received 11.

Answer 719 eggs

The Primrose Puzzle

Number the nine flowers round the garland.

A jump lands two flowers away from where it starts, so the difficulty comes at the end: the last letter must go on a flower that still has an empty flower two places from it on one side or the other, and as the garland fills, that becomes impossible for most flowers. A word can be written in the order its letters are read only if some letters can be written out of turn, in places where an equal letter belongs.

Dudeney’s words are BLUEBELL and PEARTREE. For BLUEBELL his moves, each giving the flower touched and the flower written on, are B from 3 to 1, L from 6 to 8, U from 5 to 3, E from 4 to 6, B from 7 to 5, E from 2 to 4, L from 9 to 7, L from 9 to 2. The garland then reads BLUEBELL from flower 1 to flower 8. The word works because its second and eighth letters are both L and its fourth and sixth both E, so those letters can change places without spoiling the word. PEARTREE has the same two coincidences, E and R.

Of all the ways the eight letters can be placed, the one that asks least of the word is Dudeney’s: letters 2 and 8 equal, and letters 4 and 6 equal. Every other way needs more letters to coincide, three the same or more pairs. Dudeney adds that MARITIMA, the sea pink, would also do if it were English.

Answer BLUEBELL, PEARTREE

The Round Table

Each man has six companions,

so there are $\binom62=15$ possible pairs of neighbours for him, and every sitting uses one of them. Fifteen sittings is therefore the most possible, and a scheme for fifteen days must give every man every pair of neighbours exactly once. Here is Dudeney’s, each seating read round the table, with the last man sitting next to Adams:

Dudeney first set the problem for six people over ten days, in 1905. For $n$ people the most is $(n-1)(n-2)/2$ sittings, and he records that schemes for an even number of people were found by others, while odd numbers proved much harder; he says he eventually found a method for every case and wrote out schemes up to 25.

Answer a scheme of fifteen sittings

The Five Tea Tins

First the tins alone. With No. 1 fixed at the end, the other four can stand in $4!=24$ orders, of which 12 put Nos. 3 and 5 side by side, leaving 12. Each of the five tins can show any of its six faces at the front, so if all thirty pictures were different there would be $12\times 6^5=93\,312$ rows of pictures.

Some of these rows are the same. A row can be repeated only if two tins that share pictures change places while showing shared pictures. No. 1 cannot move from its end, so the only exchange is between Nos. 3 and 4, which share two pictures. For the rows to match, each must show one of those two pictures (four ways), the other three tins can show anything ($6^3=216$ ways), and the order of the tins must stay allowed after the exchange. The exchange keeps an order allowed only if No. 5 stands next to neither No. 3 nor No. 4, which puts it between No. 1 and No. 2 or at the far end beside No. 2. Only two pairs of orders do that: 1 5 2 3 4 with 1 5 2 4 3, and 1 3 4 2 5 with 1 4 3 2 5. So $2\times 4\times 216=1728$ rows are counted twice, and the number of different rows is $93\,312-1728=91\,584$.

Answer 91 584

The Four Porkers

Read the pigs as chess queens. Every sty in line with a pig means every square is attacked, and no pig in line with another means no two queens attack each other, so the question asks for the ways four queens that leave each other alone can cover a six-by-six board. A search of every placing finds seventeen arrangements; the count here comes from that search.

Turning and reflecting them gives 120 in all: fourteen of the seventeen give eight arrangements each, one (201605) gives only four because it has some symmetry of its own, and two (260015 and 306104) give only two each, being more symmetrical still. So $14\times 8+4+2\times 2=120$. Writing each arrangement as the row of the pig in each column (0 for a column with no pig), the seventeen are

The one in the picture is 160025. Dudeney notes that Jaenisch, in 1862, quoted twenty-one, which he thought must count some reflections twice. With three pigs, if they are allowed to be in line with one another, every sty can still be covered, in one way only: 105030.

Answer 17 (120 counting turns and reflections)

The Number Blocks

Arrange the blocks into two multiplications using disjoint sets of figures, then compare their products. Larger leading figures are worth trying first; a complete search through every split shows that the largest product is $58\,560$: $$915\times 64=732\times 80=58\,560 .$$ The second sum can also be written $7320\times 8$. Large products want the large figures in the leading places of both factors, and the two sums must balance exactly, which is what makes the search awkward by hand. The children’s smallest, $3485\times 2=6970\times 1$, is confirmed as the least.

Answer 58 560

Foxes and Geese

The lines on the board are the moves of a chess knight on a board three squares wide and four deep, as Dudeney’s Diagram A in Figure 10.1 shows. Untangle the strings and the board becomes Diagram B: a ring of ten discs with two cross-pieces, one through disc 11 and one through disc 2. The geese start on 1, 3 and the cross-piece at 2; the foxes on 10, 12 and the cross-piece at 11.

Seen this way the play is simple. A goose on 1 or 3 must step to 8 to let the fox on 11 onto the ring; then all six creatures walk round the ring the same way, keeping their distance, until the last moves take the foxes to 1, 2 and 3 and the geese to 10, 11 and 12. It takes eleven moves for the foxes and eleven for the geese, twenty-two in all, and a search shows that no shorter solution exists. Here is Dudeney’s, played round by round, the fox moving first in each:

Answer 22 moves

Robinson Crusoe’s Table

The holes lie in three rows of five. Dudeney cuts the timber into two pieces, E and F, along a stepped line (Figure 10.2), and throws away the shaded strips that carry the holes. The pieces then fit together, E into F, to make the square ABCD. The steps are what let two pieces do the work of several: each row of holes is taken out as a strip, and the stepped edges close the gaps the strips leave.

Answer two pieces, E and F

The Fifteen Orchards

Call the number each man would end with $N$. Then every orchard except five held $N$ apples, and each orchard’s total is a multiple of its number of trees, because every tree bore the same. So $N$ is divisible by 1, 2, 3, 4, 5, 6, 8, 10, 12 and 15, that is, by 120. The other five held $N-1$ (seven trees), $N-3$ (nine), $N+1$ (eleven), $N-3$ (thirteen) and $N+6$ (fourteen), so $N$ must leave remainder 1 on division by 7, 3 by 9, 10 by 11, 3 by 13 and 8 by 14. The number 120 does all of these: $120=17\times 7+1=13\times 9+3=11\times 11-1=9\times 13+3=9\times 14-6$.

The next number that works is $120+360\,360=360\,480$, since 360 360 is the least number divisible by all of 1 to 15. As Dudeney says, no young apple tree bears apples by the tens of thousands, so 120 is the only sensible answer.

Answer 120 apples

The Perplexed Plumber

A long, narrow base uses more wall than a square base of the same area. For example, bases $5\times20$ and $10\times10$ have the same area, but perimeters 50 and 40 feet. In general, $a+b\ge2\sqrt{ab}$ follows from $(a-b)^2\ge0$. Replacing a rectangular base by a square of the same area leaves the volume, height, floor and roof unchanged, and can only reduce the four walls. So the best base is square.

A cube of side 10 feet has the required volume. To compare all square bases with it, write the base side as $10t$, where $t$ is positive. Keeping the volume at 1000 cubic feet makes the height $10/t^2$. Floor and roof use $200t^2$ square feet and the walls $400/t$, giving a total $S=200t^2+400/t$. Subtract the cube’s 600 square feet: $$S-600=\frac{200(t-1)^2(t+2)}{t}\ge0.$$ Every factor is non-negative, and the excess vanishes only when $t=1$. Thus the best closed cistern is a cube of side 10 feet, using 600 square feet of zinc.

For an open cistern, place two identical cisterns rim to rim. Their combined metal is exactly the metal of one closed cistern of twice the volume. The argument above works at any scale, so that closed cistern is best when it is a cube. Each open cistern is therefore half a cube, with a square base whose side is twice its depth.

For 1000 cubic feet the base is $\sqrt[3]{2000}\approx 12.599$ feet square and the depth about 6.300 feet. Dudeney suggested $12.6\times 12.6\times 6.3$, which holds 1000.19 cubic feet, a little too much, to the buyer’s advantage.

Answer cube of 10 ft; open: half a cube

The Nelson Column

Roll a sheet of paper with a diagonal drawn on it into a tube, and the diagonal becomes one turn of a spiral (Figure 10.4). Unroll the column in the same way and each of the five turns of the garland is the long side of a right-angled triangle whose other sides are a fifth of the height, 40 feet, and the circumference, $16\tfrac23$ feet. These are in the ratio $12:5$, so the triangle is the familiar 5, 12, 13 triangle enlarged, and each turn measures $\tfrac{13}{12}\times 40=43\tfrac13$ feet. The whole garland is five times as long, $216\tfrac23$ feet, or 216 feet 8 inches. By a pleasant accident of the figures, that is exactly the height and the circumference added together.

Answer 216 ft 8 in

The Two Errand Boys

At the first meeting the two boys have together run the distance between the shops once, and the baker’s boy has done 720 yards of it. At the second meeting they have together run it three times: out, and back as far as the meeting place. Each has also waited ten minutes, but they waited the same time, so at the second meeting each has been running for the same length of time, and the baker’s boy has again done his fixed share of the running. Three times the distance means three times his 720 yards, 2160 yards. He has run the whole distance to the butcher’s and then 400 yards back, so the distance is $2160-400=1760$ yards, exactly a mile.

Dudeney’s rule, three times the first distance less the second, is the same reasoning. It needs both boys to have left on the way back before they pass again; with these figures they have.

Answer 1760 yards, a mile

On the Ramsgate Sands

Each child has twelve others who might hold its hands, and every ring uses two of them, so no child can take part in more than six rings. Six is possible. Here are Dudeney’s six, each read round the ring with the ends joined:

Dudeney credits Lucas with a simple mechanical method for $2n+1$ children and $n$ rings. Lucas himself credits it to M. Walecki, a teacher at the Lycée Condorcet in Paris. Put one child in the middle and the rest on a circle; each ring runs from the middle child in a zigzag across the circle, and turning the zigzag one place round the circle gives the next ring.

Answer 6 rings

Sources. The puzzles are retold from Nos. 87 to 100, and the illustrations reproduced from, Henry Ernest Dudeney, The Canterbury Puzzles and Other Curious Problems (1907; revised edition 1919), in the text of Project Gutenberg eBook 27635. On the rings: Édouard Lucas, Récréations mathématiques, volume II (second edition, 1896), the sixth recreation. The numerical answers are found again by the programs in the verify folder accompanying this book.