Money Puzzles

Dudeney opens the book with money, because money was the arithmetic everyone did every day. A shopkeeper who could not make change in his head was soon out of business, and a puzzle about stamps, bananas or a bequest asked for nothing more than the reckoning a customer did at the counter, pushed a little further than usual. Most of these puzzles come down to one or two linear equations. The pleasure is in seeing which quantity to name, and in noticing where the words leave room for more than one answer.

The money is the old English money. Twelve pence (12d.) make a shilling (1s.) and twenty shillings make a pound (£1), so a pound is 240 pence. Below the penny came the halfpenny (½d.) and the farthing (¼d.). A crown was 5s., a half-crown 2s. 6d., a florin 2s. and a “fiver” £5. Amounts are written as Dudeney wrote them, so that £6 13s. is six pounds thirteen shillings and 2s. 8d. is two shillings and eightpence.

No. 1. A Post-Office Perplexity

No. 2. Youthful Precocity

No. 3. At a Cattle Market

No. 4. The Beanfeast Puzzle

No. 5. A Queer Coincidence

No. 6. A Charitable Bequest

No. 7. The Widow’s Legacy

No. 8. Indiscriminate Charity

No. 9. The Two Aeroplanes

No. 10. Buying Presents

No. 11. The Cyclists’ Feast

No. 12. A Queer Thing in Money

No. 13. A New Money Puzzle

No. 14. Square Money

No. 15. Pocket Money

No. 16. The Millionaire’s Perplexity

No. 17. The Puzzling Money-Boxes

No. 18. The Market Women

No. 19. The New Year’s Eve Suppers

No. 20. Beef and Sausages

No. 21. A Deal in Apples

No. 22. A Deal in Eggs

No. 23. The Christmas-Boxes

No. 24. A Shopping Perplexity

No. 25. Chinese Money

No. 26. The Junior Clerk’s Puzzle

No. 27. Giving Change

No. 28. Defective Observation

No. 29. The Broken Coins

No. 30. Two Questions in Probabilities

No. 31. Domestic Economy

No. 32. The Excursion Ticket Puzzle

No. 33. Puzzle in Reversals

No. 34. The Grocer and Draper

No. 35. Judkins’s Cattle

No. 36. Buying Apples

No. 37. Buying Chestnuts

No. 38. The Bicycle Thief

No. 39. The Costermonger’s Puzzle

A Post-Office Perplexity

Count in halfpence and the whole order becomes one equation whose coefficients do the work. Say she gives $a$ twopenny stamps, so $6a$ penny stamps, and $c$ stamps at 2½d. A crown is 60d., or 120 halfpence, so $$4a + 12a + 5c = 120, \qquad\text{that is}\qquad 16a + 5c = 120 .$$ Both 120 and $5c$ are multiples of 5, so $16a$ is too, and $a$ must be a multiple of 5. The gentleman asked for “some” twopenny stamps, so $a \ge 1$, and $a=10$ already costs 160 halfpence. That leaves $a=5$, and then $5c = 40$, so $c=8$. Five twopenny stamps (10d.), thirty penny stamps (30d.) and eight at 2½d. (20d.) make the crown exactly. Dudeney gives the same, and it is the only answer.

Answer 5 twopenny, 30 penny and 8 twopence-halfpenny stamps

Youthful Precocity

The price appears once as a multiplier and once as a divisor, so the equation is a square. Let a banana cost $p$ pence. Sixteen gross, sixteen twelves of twelve, is $16 \times 144 = 2304$ bananas, which cost $2304p$ pence, or $2304p/6 = 384p$ sixpences. A fiver is 1200d., which buys $1200/p$ bananas. Fred says the first number is half the second: $$384p = \frac{600}{p}, \qquad p^2 = \frac{600}{384} = \frac{25}{16}, \qquad p = \frac54 .$$ The price is a penny farthing, and a positive price leaves no other root. As a check, £5 buys 960 bananas at 1¼d., and 2304 bananas cost 2880d., which is 480 sixpences, half of 960. This agrees with Dudeney.

Answer 1¼d.

At a Cattle Market

Each offer is one linear condition, and the three chain round into a single equation in one man’s count. Let Jakes, Hodge and Durrant bring $J$, $H$ and $D$ animals. When Hodge gives six pigs for a horse, Jakes has $J+5$ and Hodge $H-5$; when Durrant gives fourteen sheep for a horse, Hodge has $H+13$ and Durrant $D-13$; when Jakes gives four cows for a horse, Durrant has $D+3$ and Jakes $J-3$. So $$\begin{gather*} J + 5 = 2(H-5), \qquad H + 13 = 3(D-13),\\ D + 3 = 6(J-3). \end{gather*}$$ These give $J = 2H - 15$, $H = 3D - 52$ and $D = 6J - 21$. Substituting each into the one before, $$\begin{align*} J &= 2(3D - 52) - 15 = 6D - 119\\ &= 6(6J - 21) - 119 = 36J - 245, \end{align*}$$ so $35J = 245$ and $J = 7$. Then $D = 21$ and $H = 11$. The equations are linear with a single solution, so there is no other. Jakes brought 7 animals, Hodge 11 and Durrant 21, 39 in all, as Dudeney says.

Answer Jakes 7, Hodge 11, Durrant 21; 39 animals

The Beanfeast Puzzle

The three comparisons are really comparisons between whole parties, and seeing that turns the puzzle into a sharing in fixed ratios. Five cobblers spend as much as four tailors, so 25 cobblers spend as much as 20 tailors: the cobblers’ party and the tailors’ party spend the same. Twelve tailors spend as much as nine hatters, so one hatter spends $\tfrac43$ of a tailor, and 18 hatters spend as much as 24 tailors. Six hatters spend as much as eight glovers, so a glover spends $\tfrac34$ of a hatter, which is exactly one tailor, and 12 glovers spend as much as 12 tailors.

Measured in tailors, the four parties spend $20$, $20$, $24$ and $12$, in the ratio $5:5:6:3$. That is 19 shares of £6 13s., which is 133s., so a share is 7s. The cobblers spent 35s., the tailors 35s., the hatters 42s. and the glovers 21s., in agreement with Dudeney.

One small curiosity. A tailor and a glover each spend 1s. 9d. and a hatter 2s. 4d., but a cobbler spends $\tfrac75$ of a shilling, 1s. 4$\tfrac45$d., which is not a whole number of farthings. Dudeney asked only for what each party spent, and those come out in whole shillings.

Answer Cobblers 35s., tailors 35s., hatters 42s., glovers 21s.

A Queer Coincidence

Follow one player at a time: his money is doubled at every game he loses, and the total never changes, so his one win is the only step that needs thought. Work in farthings. Each man ends with 2s. 8d., which is 128 farthings, so the seven hold $T = 7 \times 128 = 896$ between them throughout, since every payment moves money from one pocket to another.

Take the player who wins game $k$, and say he starts with $s$. He loses the first $k-1$ games, so his money is doubled each time and reaches $2^{k-1}s$. In game $k$ he pays every other player what that player holds, which is everything not in his own pocket, $T - 2^{k-1}s$. He is left with $$2^{k-1}s - \bigl(T - 2^{k-1}s\bigr) = 2^k s - T .$$ He then loses the remaining $7-k$ games and ends with $2^{7-k}(2^k s - T) = 128 s - 2^{7-k}T$. Setting this equal to 128 and putting $T = 7 \times 128$, $$128 s = 128 + 2^{7-k} \times 7 \times 128, \qquad s = 7 \times 2^{7-k} + 1 .$$ So Adams ($k=1$) began with 449 farthings, Baker 225, Carter 113, Dobson 57, Edwards 29, Francis 15 and Gudgeon 8. In money: 9s. 4¼d., 4s. 8¼d., 2s. 4¼d., 1s. 2¼d., 7¼d., 3¾d. and 2d. The starting amounts add to 896, as they must, and each is determined, so the answer is unique.

This is Dudeney’s general rule in disguise. With $n$ players each ending with $m \cdot 2^n$, the same argument gives the player who wins game $k$ a start of $m(2^{n-k}n + 1)$: the last winner $m(n+1)$, the one before $m(2n+1)$, and so on up to $m(2^{n-1}n + 1)$ for the first. Here $n = 7$ and $2^7 = 128$ farthings is exactly 2s. 8d., so $m = 1$. The odd sum of 2s. 8d. was chosen to make the farthings come out whole.

Answer In farthings: Adams 449, Baker 225, Carter 113, Dobson 57, Edwards 29, Francis 15, Gudgeon 8

A Charitable Bequest

The arithmetic gives seven ways; Dudeney’s answer of six rests on a quibble about the word “men”. In pence, $w$ women at 18d. and $m$ men at 30d. must share 660d.: $$18w + 30m = 660, \qquad\text{that is}\qquad 3w + 5m = 110 .$$ Since $110 - 5m$ must be a multiple of 3, and 110 and 5 are each 2 more than a multiple of 3, $m$ must be 1 more than a multiple of 3: $m = 1, 4, 7, \ldots, 22$. The value $m = 22$ gives $w = 0$, no women at all, which the bequest does not allow. The rest give seven ways: $$\begin{gather*} (w, m) = (35, 1),\ (30, 4),\ (25, 7),\ (20, 10),\\ (15, 13),\ (10, 16),\ (5, 19). \end{gather*}$$ Dudeney throws out $(35, 1)$ on the ground that “a single man is not men”, and answers six years. That reading is defensible but not forced: the will speaks of “a number of women” in the same breath, and a number can be one. Taken plainly, with at least one man and at least one woman, the charity runs for seven years; on Dudeney’s reading it runs for six; and if a year with no women were allowed, it would run for eight.

Answer 7 years (6 if a lone man is not “men”, as Dudeney rules)

The Widow’s Legacy

Measure every share in widow’s shares and the legacy is a single division. A daughter gets two widow’s shares and a son three daughters’ shares, which is six. The widow, four daughters and five sons therefore hold $1 + 4 \times 2 + 5 \times 6 = 39$ shares, and the widow’s share is $$\frac{£8000}{39} = £205\tfrac{5}{39}.$$ Now $\tfrac{5}{39}$ of a pound is $\tfrac{100}{39} = 2\tfrac{22}{39}$ shillings, and $\tfrac{22}{39}$ of a shilling is $\tfrac{264}{39} = 6\tfrac{10}{13}$ pence. The widow receives £205 2s. 6d. and $\tfrac{10}{13}$ of a penny, exactly as Dudeney says. The fraction is the joke: 39 does not divide 8000, and no division of the money in real coin can obey the will to the letter.

Answer £205 2s. 6$\tfrac{10}{13}$d.

Indiscriminate Charity

Run the walk backwards: each gift leaves half the money less a fixed sum, so each step undoes by adding the sum back and doubling. After the third beggar he has $x/2 - 3$, if he had $x$ before, and this is 1d., so he had 8d. Before the second beggar he had $y$ with $y/2 - 2 = 8$, so $y = 20$d. Before the first he had $z$ with $z/2 - 1 = 20$, so $z = 42$d.

Each step is a linear equation with one root, so 42d. is the only starting sum. Forwards: he gives 22d. and keeps 20d., gives 12d. and keeps 8d., gives 7d. and keeps the penny. He left with 3s. 6d., in agreement with Dudeney.

Answer 3s. 6d.

The Two Aeroplanes

The two percentages are taken of different costs, so they do not cancel. The machine sold at a loss of 20 per cent fetched four-fifths of its cost, so it cost $600 \times \tfrac54 = £750$. The one sold at a gain of 20 per cent fetched six-fifths of its cost, so it cost $600 \times \tfrac56 = £500$. He paid £1,250 and took £1,200, a loss of £50 on the whole. The loss is charged on the dearer machine, the gain on the cheaper one, which is why equal percentages leave him out of pocket. Each cost is fixed by its own equation, and Dudeney’s answer agrees.

Answer A loss of £50

Buying Presents

Home is half of the start, so twice the home sum equals the start, and in shillings that gives a single equation with a small solution. Say he set out with $P$ pounds and $S$ shillings, with $S$ less than 20. He took home $\tfrac{S}{2}$ pounds and $P$ shillings, so $S$ is even and $P$ is less than 20. In shillings, $$2\Bigl(20 \cdot \tfrac{S}{2} + P\Bigr) = 20P + S, \qquad\text{that is}\qquad 18P = 19S .$$ As 18 and 19 have no common factor, 19 divides $P$ and 18 divides $S$. With both below 20 and not both zero, $P = 19$ and $S = 18$. He set out with £19 18s., took home £9 19s., and spent £9 19s., as Dudeney says.

That relies on Jorkins’s sums being in pounds and shillings only, which is how Dudeney means it. If he also had some pence, and took home half as many, the pounds and shillings are still £19 18s., but the amount spent grows by those half-pence: starting with £19 18s. 4d., say, he spends £9 19s. 2d. and takes home £9 19s. 2d., which fits every word he said.

Answer £9 19s.

The Cyclists’ Feast

Two ways of sharing the same 80 shillings differ by 2s. a head, and that is a quadratic in the number of riders. With $n$ riders the fair share is $80/n$ shillings, and the $n-2$ who stayed paid $80/(n-2)$. So $$\begin{gather*} \frac{80}{n-2} - \frac{80}{n} = 2, \qquad 80n - 80(n-2) = 2n(n-2),\\ n^2 - 2n - 80 = 0, \end{gather*}$$ which factorises as $(n-10)(n+8) = 0$. The only positive root is $n = 10$. Ten riders should have paid 8s. each; the eight honest ones paid 10s. This agrees with Dudeney.

Answer 10 cyclists

A Queer Thing in Money

Casting out nines turns the condition into a congruence that fixes the length of the pounds, and only then do the actual digits need checking. A number and its digit sum leave the same remainder on division by 9. So if a sum of $P$ pounds, $S$ shillings and $D$ pence has the same digit sum as $240P + 12S + D$, then $$\begin{gather*} P + S + D \equiv 240P + 12S + D \pmod 9,\\ \text{that is}\quad 5P + 2S \equiv 0 \pmod 9, \end{gather*}$$ since $239 = 26 \times 9 + 5$ and $11 = 9 + 2$. Now let every part be made of the figure $f$, with $k$ figures in the pounds. Then $P \equiv fk$, and taking $S = f$ (the choice of 11s. when $f = 1$ goes the same way), for $f$ not a multiple of 3 the congruence becomes $5k + 2 \equiv 0 \pmod 9$, which means $k$ is 5, 14, 23 and so on. For $f = 3$ or $6$ it needs only $k \equiv 2 \pmod 3$, and $f = 9$ passes it for every $k$.

The congruence is necessary, not sufficient, and the actual digit sums finish the job. With $f = 6$ and $k = 2$ we have Dudeney’s example. With $f = 4$ and $k = 5$, £44,444 4s. 4d. is 10,666,612 pence, and both sides add up to 28. That is the other answer, and Dudeney’s.

Why no longer pounds work: adding one more figure to the pounds adds one more copy of a single repeated digit in the middle of the pence (15,918, then 159,918, then 1,599,918 for the sixes), so each extra figure changes the gap between the two digit sums by the same step. For the sixes each extra figure adds 6 to one side and 9 to the other, and the gap, zero at two figures, never closes again. The same happens for every figure, and for $f = 9$ the gap stays at 9 throughout.

Answer £44,444 4s. 4d. (10,666,612 pence)

A New Money Puzzle

The smallest sum is the one with the fewest figures in the pounds, so the question is how many figures the other three parts can soak up. Pence can take only one figure, since 10d. needs a nought and 11d. repeats a figure. The farthings always take two. Shillings can take two only as a number from 12 to 19, which uses the figure 1. So at most five figures sit outside the pounds, and the pounds have at least four.

Four is reached only with two-figure shillings, which spend the 1, so the farthings cannot be ¼ or ½ and must be ¾, spending 3 and 4. The pounds then take the smallest four figures left, and they are smallest when the shillings and pence use up the two largest figures, 8 and 9. That leaves 2, 5, 6 and 7 for the pounds, written £2,567. Between 18s. 9d. and 19s. 8d. the first is smaller. The least sum is £2,567 18s. 9¾d., and it is the only one of that value.

Dudeney gives the same.

Answer £2,567 18s. 9¾d.

Square Money

Sum equals product rearranges to $(x-1)(y-1) = 1$, so the two amounts are $1 + u$ and $1 + 1/u$ pence, and the coinage decides which $u$ are allowed. Both amounts must be whole numbers of farthings. If $u$ were negative, one of the two amounts would be less than a penny and the other negative, so $u > 0$. Write $x = a/4$ with $a$ a whole number of farthings; then $u = (a-4)/4$ and $y = 1 + 4/(a-4)$, which is a whole number of farthings exactly when $16/(a-4)$ is a whole number. So $a - 4$ is 1, 2, 4, 8 or 16, and the pairs (smaller first) are $$\begin{gather*} (2, 2) \text{ giving } 4, \qquad (1\tfrac12, 3) \text{ giving } 4\tfrac12,\\ (1\tfrac14, 5) \text{ giving } 6\tfrac14 . \end{gather*}$$ These are all the pairs there are. After Dudeney’s fourpence, the next result is 4½d., from 1½d. and 3d.: they add to 4½d., and 1½ times 3 is 4½. This agrees with Dudeney.

Answer 1½d. and 3d., giving 4½d.

Pocket Money

Parity carries the answer:

every coin but the threepenny bit is a whole number of sixpences, so one threepenny bit can never help to make ten shillings, and it rides free on top of the largest safe pile of the other coins. Count in threepences, so the coins are worth 1, 2, 4, 8, 10 and 20 and ten shillings is 40.

First the big coins alone. Five florins make 40, four half-crowns make 40, a crown with two half-crowns makes 40, so do two crowns, and no other mixture of florins, half-crowns and crowns does. So the most they can safely hold is four florins with either three half-crowns or one half-crown and a crown, worth 62, or 15s. 6d. Now add one threepenny bit. Any selection that uses it has an odd total, and 40 is even, so the pile is still safe, and it is worth 63, or 15s. 9d.

Showing that nothing beats 15s. 9d. takes a search: add any other small coin, or trade big coins for small ones, and some selection makes exactly ten shillings. The two pocketfuls are a crown, a half-crown, four florins and a threepenny bit, or three half-crowns, four florins and a threepenny bit, exactly as Dudeney says.

Answer 15s. 9d.

The Millionaire’s Perplexity

At most six gifts of each size is exactly the rule that makes the counts the digits of the million in base seven, and base-seven digits are unique. The number of one-dollar gifts must leave the same remainder as a million on division by 7, since every other gift is a multiple of 7, and it is at most 6, so it is that remainder. Take those dollars away, divide the rest by 7, and the same argument fixes the number of seven-dollar gifts; and so on up. This is repeated division by 7, and the remainders are $$1{,}000{,}000 = 1 \cdot 7^7 + 1 \cdot 7^6 + 3 \cdot 7^5 + 3 \cdot 7^4 + 3 \cdot 7^3 + 3 \cdot 7^2 + 1 \cdot 7 + 1 ,$$ that is, 11333311 in base seven. He gives one gift each of $7^7 = 823{,}543$ and $7^6 = 117{,}649$ dollars, three gifts each of $7^5$, $7^4$, $7^3$ and $7^2$ dollars, one gift of seven dollars and one of a single dollar: eighteen gifts in all, and no other way exists. Dudeney gives the same, and says so.

Answer One gift each of $7^7$, $7^6$, $7$ and $1$ dollars; three each of $7^5$, $7^4$, $7^3$, $7^2$

The Puzzling Money-Boxes

Call the common amount $x$ shillings and every box is a simple expression in it; the six coins then only confirm the answer, apart from one coin they force. John has $x - 2$, William $x + 2$, Charles $x/2$ and Thomas $2x$, so $$(x - 2) + (x + 2) + \tfrac{x}{2} + 2x = \tfrac{9x}{2} = 45, \qquad x = 10 .$$ John has 8s., William 12s., Charles 5s. and Thomas £1. In the fewest coins these are two double florins, a half-sovereign and a florin, a crown, and a sovereign: six coins, as the brothers said, and the amounts are forced by the equation alone. Dudeney gives the same.

The coin count earns its place in one box. Without the double florin, 8s. needs three coins at least (a crown, a florin and a shilling, say), and the total would be seven. So the six coins require the double florin, which Dudeney counts as a current coin here, just as in Pocket Money, where he has to rule it out by name.

Answer John 2 double florins; William a half-sovereign and a florin; Charles a crown; Thomas a sovereign

The Market Women

In farthings the takings are 105, and each woman’s price times her weight must give 105; so if the weights are whole pounds, the prices are the divisors of 105, and they number eight. Since $105 = 3 \times 5 \times 7$, its divisors are the products of any selection of those three primes, $2^3 = 8$ of them: 1, 3, 5, 7, 15, 21, 35 and 105 farthings a pound, with 105, 35, 21, 15, 7, 5, 3 and 1 lb. sold. That is Dudeney’s answer, eight women.

The whole-pound weights are Dudeney’s assumption, not the puzzle’s. The statement asks only that each price be payable in coin. A woman selling $2\tfrac13$ lb. at 11¼d. a pound (45 farthings) takes exactly 105 farthings too, and any price at all in whole farthings goes with some weight. Read literally, there is no greatest number. The answer of eight needs each woman to sell a whole number of pounds, which is a fair guess at what a market woman did, but a guess.

Answer 8, if each sells a whole number of pounds (otherwise no limit)

The New Year’s Eve Suppers

Take away eighteenpence a head for everyone and what is left is a small equation with one solution. Let there be $l$ single ladies, $m$ single men and $p$ couples, so $l + m + 2p = 25$, and the takings in pence are $18l + 30m + 126p = 1200$. If all twenty-five had paid 18d. the café would have taken 450d. Each single man adds 12d. to that, and each couple adds $126 - 36 = 90$d., so $$12m + 90p = 750, \qquad 2m + 15p = 125 .$$ Then $15p$ is odd, so $p$ is odd, and $p = 1, 3, 5, 7$ give $m = 55, 40, 25, 10$. Only the last leaves room in twenty-five people: $l = 25 - 10 - 14 = 1$. The company was seven couples, ten single men and one single lady, the only answer, as Dudeney says.

Answer 7 couples, 10 single men, 1 single lady

Beef and Sausages

Everything scales with the weight she bought, so the two pounds she lost measure that weight directly. Say she bought $q$ lb. of each, spending $24q + 18q = 42q$ pence. Half of that, $21q$ pence, buys $21q/24 = \tfrac78 q$ lb. of beef, and the other half buys $21q/18 = \tfrac76 q$ lb. of sausages, together $$\tfrac78 q + \tfrac76 q = \tfrac{49}{24} q ,$$ which is $\tfrac{1}{24}q$ more than the $2q$ she actually took home. So $q/24 = 2$ and $q = 48$. She bought 48 lb. of each and spent $42 \times 48 = 2016$d., which is £8 8s. Split evenly, £4 4s. buys 42 lb. of beef and 56 lb. of sausages, 98 lb. against 96. The equation is linear, so the answer is unique, and it agrees with Dudeney.

Answer £8 8s.

A Deal in Apples

The price per dozen is $144/n$ pence when a shilling buys $n$ apples, and a fall of one penny turns into a product of two numbers two apart. With $n$ apples offered and $n + 2$ taken, $$\frac{144}{n} - \frac{144}{n+2} = 1, \qquad 288 = n(n+2) .$$ Since $16 \times 18 = 288$, $n = 16$, and the other root, $-18$, means nothing here. The first price was 9d. a dozen, the second 8d., and the shilling bought eighteen apples. This agrees with Dudeney and is the only answer.

Answer 18 apples

A Deal in Eggs

Subtracting the count from the cost removes the halfpenny eggs and leaves one short equation; the equal pair then picks one of its eleven solutions. In halfpence, with $a$ eggs at 5d., $b$ at 1d. and $c$ at ½d., $$\begin{gather*} a + b + c = 100, \qquad 10a + 2b + c = 200,\\ \text{so}\quad 9a + b = 100 . \end{gather*}$$ Then $b = 100 - 9a$ and $c = 8a$, and some of each kind means $a$ runs from 1 to 11: eleven ways to buy the hundred eggs. The two equal counts settle it. $a = c$ would need $a = 8a$; $b = c$ would need $17a = 100$; so $a = b$, which gives $10a = 100$. He bought ten new-laid eggs, ten fresh ones and eighty at a halfpenny, the only answer, as Dudeney says.

Answer 10 at 5d., 10 at 1d., 80 at ½d.

The Christmas-Boxes

Everything turns on the total, 361 pence, which is $19^2$.

Every person received the same number of pence, so the number of people divides 361, and the only divisors are 1, 19 and 361. A single recipient is no sharing out at all, and 361 people cannot share 100 coins. So nineteen people each received nineteen pence.

Here the odd penny earns its place. The silver coins of 1917 below a florin were the threepence, the sixpence and the shilling, all multiples of threepence, and 19 is not a multiple of 3. No gift of nineteen pence can be made from them. The coin that rescues the puzzle is the fourpenny piece, the old groat, which had once circulated as ordinary money and is why Dudeney says “some years ago”. With threepences and fourpences to hand there are five ways to make up 19d.: $$\begin{gather*} 12+4+3, \qquad 6+6+4+3, \qquad 6+4+3+3+3,\\ 4+4+4+4+3, \qquad 4+3+3+3+3+3, \end{gather*}$$ using 3, 4, 5, 5 and 6 coins. Nineteen gifts must use 100 coins in all, an average of a little over five, so the man mixed the patterns. Dudeney gives fourteen people four fourpences and a threepence (70 coins) and five people five threepences and a fourpence (30 coins): 100 coins and $19 \times 19 = 361$ pence. That is one sharing among many. Counting how many people get each pattern, there are 147 ways to spend exactly 100 coins. The puzzle asks only how he could have done it, and the number of people is fixed.

Answer 19 people, 19d. each (for example 14 given 4+4+4+4+3, 5 given 3+3+3+3+3+4)

A Shopping Perplexity

Without change, a bill must be paid exactly, so the question for each amount is the least number of coins that make it up exactly.

The coins current in 1917 were the farthing, halfpenny, penny, threepence, sixpence, shilling, florin, half-crown and crown, and the double florin was still legal tender. Working in farthings keeps everything whole.

The fractions of a penny force coins on their own. A bill ending in ¾d. needs a halfpenny and a farthing, and one ending in ½d. or ¼d. needs one coin for that. The whole pence below threepence need one coin each, and so on up. Dudeney’s two bills are $$\begin{align*} \text{1s.~5¾d.} &= 1\text{s.} + 3\text{d.} + 1\text{d.} + 1\text{d.} + ½\text{d.} + ¼\text{d.},\\ \text{1s.~11½d.} &= 1\text{s.} + 6\text{d.} + 3\text{d.} + 1\text{d.} + 1\text{d.} + ½\text{d.}, \end{align*}$$ and together they make 3s. 5¼d. $=$ half-crown $+$ 6d. $+$ 3d. $+$ 1d. $+$ 1d. $+$ ¼d. Each needs six coins and no fewer. In the first, the ¾d. takes two coins and the remaining 1s. 5d. cannot be made in three. With a shilling, the other two coins would have to make 5d., and no two coins do; without one, three coins of sixpence or less make 17d. only with two sixpences and a 5d. piece, which does not exist. In the second, the ½d. takes one coin and 1s. 11d. cannot be made in four: with a shilling, three coins would have to make 11d., and from 6d., 3d. and 1d. three coins make 10d. or 12d. but never 11d.; without one, four coins of sixpence or less make at most 2s., and the next largest they make is 1s. 9d., so 23d. is missed. The total, 3s. 5¼d., needs a farthing and then 41d. in four coins: a half-crown leaves 11d. for three, just ruled out; a florin leaves 17d. for three, and a shilling plus two of 6d., 3d. and 1d. never makes 5d.; and four coins of a shilling or less make 4s., 3s. 6d., 3s. 3d. or less, never 41d.

That these are the smallest is a matter of counting, since the least number of coins jumps about irregularly as the amount grows. A program works out the fewest coins for every amount below five shillings and tries every pair of different bills. The least total that works is 165 farthings, 3s. 5¼d., and only Dudeney’s pair reaches it. Counting the double florin as current or not makes no difference.

Answer 3s. 5¼d. (1s. 5¾d. and 1s. 11½d.)

Chinese Money

The ching-chang is defined in terms of itself, and that is the crux: if a ching-chang $c$ is twopence plus four-fifteenths of itself, then the other eleven-fifteenths of it are worth twopence, so $c = \tfrac{30}{11}$d. Eleven ching-changs are therefore thirty pence, exactly a half-crown.

The three coins are worth $\tfrac{15}{11}$, $\tfrac{16}{11}$ and $\tfrac{17}{11}$ ching-changs, so $r$ round, $s$ square and $t$ triangular coins make eleven ching-changs when $$15r + 16s + 17t = 121 .$$ Write the left side as $15(r+s+t) + s + 2t$. The number of coins $n = r+s+t$ must then satisfy $15n \le 121 \le 17n$, which allows only $n = 8$; and $n = 8$ leaves $s + 2t = 1$, so $s = 1$, $t = 0$, $r = 7$. The change is seven round-holed coins and one square-holed coin, and there is no other way.

Answer 7 round-holed coins and 1 square-holed coin

The Junior Clerk’s Puzzle

The reason Snoggs asked for his rise half-yearly is the whole puzzle: it pays him more. Moggs’s salary rises by £10 a year, so each of his half-yearly payments goes up by £5 once a year: $$25, 25, 30, 30, 35, 35, 40, 40, 45, 45 \quad\text{pounds, total £350.}$$ Snoggs’s payment goes up by £2 10s. every half-year: $$25, 27\tfrac12, 30, 32\tfrac12, \dots, 47\tfrac12 \quad\text{pounds, total £362~10s.}$$ He is level with Moggs at the start of each year and ahead of him in the middle of it, so over five years he gains five payments of £2 10s., which is £12 10s.

If Moggs saves a proportion $p$ and Snoggs $2p$, then $350p + 2 \times 362\tfrac12\, p = 1075p$ pounds, and this is £268 15s. $= 268\tfrac34$. So $p = \tfrac14$. Moggs saved a quarter of £350, which is £87 10s., and Snoggs half of £362 10s., which is £181 5s. They add up to £268 15s.

A reader who takes Snoggs’s rise to mean that his yearly rate goes up £2 10s. each half-year gets a smaller total for him, £306 5s., and a proportion of $\tfrac{43}{154}$, which gives savings in fractions of a farthing. That reading cannot be what was meant.

Answer Moggs £87 10s., Snoggs £181 5s.

Giving Change

The method is to pool every coin and ask what each person must take away. The shopkeeper came in with 75 cents and must leave with 109, his own money plus the 34 of the sale. The buyer came with 105 and must leave with 71. The stranger must leave with the 28 he brought. The ten coins on the counter are 100, 50, 25, 10, 10, 5, 3, 2, 2 and 1.

Now the large coins settle themselves. Only the shopkeeper wants as much as 100, so he takes the dollar and needs 9 more. The half-dollar is then too big for anyone but the buyer, who needs 21 more; the quarter goes to the stranger, who needs 3 more. The buyer’s 21 must include both dimes, since 5, 3, 2, 2 and 1 make only 13, and then he needs 1: the cent. The stranger’s 3 is the three-cent piece. The shopkeeper’s 9 is what is left, 5, 2 and 2.

So the shopkeeper takes 100, 5, 2 and 2; the buyer takes 50, 10, 10 and 1; the stranger takes 25 and 3. Each step was forced, so the settlement is unique, and, as Dudeney points out, nobody leaves with a single one of the coins he brought.

Answer Shopkeeper 100, 5, 2, 2; buyer 50, 10, 10, 1; stranger 25, 3

Defective Observation

On the pennies of Dudeney’s day the date sat under Britannia, on the tail.

Six pennies can surround a penny, every one touching it. The centres of the outer coins lie on a circle whose radius is two penny-radii, and two neighbours may not overlap, so their centres are at least two radii apart. On that circle a chord of two radii subtends exactly 60 degrees at the centre, which is the equilateral triangle formed by three touching pennies. So the outer centres are at least 60 degrees apart, at most $360/60 = 6$ fit, and six do, each touching its neighbours too. The argument uses only the ratio of sizes, so it holds for any coin.

The half-crown is the real test.

Two coins of diameter $d$ placed side by side inside a coin of diameter $D$ need $D \ge 2d$: their centres are at least $d$ apart, and each must lie within $(D-d)/2$ of the centre of the large coin, so $d \le D - d$. A silver threepence was 16.20 mm across and a half-crown 32.31 mm. Two threepences need 32.40 mm, and the half-crown falls short by nine hundredths of a millimetre. Only one fits. It is a far narrower miss than the dinner-table answers of three, four or more suggest.

Answer The date is on the tail; six pennies; one threepence

The Broken Coins

Because the same fraction is lost from each coin, the same fraction is lost from their total. The three coins were worth $240 + 12 + 1 = 253$ pence and are now worth 240, so the loss is $\tfrac{13}{253}$ of the whole, and therefore of each coin.

Answer $\tfrac{13}{253}$ of each coin

Two Questions in Probabilities

Five pennies can fall in $2^5 = 32$ equally likely ways. All five alike happens in 2 of them. Exactly four alike means choosing which coin is the odd one out, 5 ways, and then whether the four are heads or tails, so 10 ways. That is 12 favourable falls out of 32, a chance of $\tfrac38$, or odds of 3 to 5 against.

For the bag, a fair price is the average value of a draw. Three draws in four give a sovereign, 240 pence, and one in four gives a shilling, 12 pence: $$\tfrac14(3 \times 240 + 12) = 183 \text{ pence} = 15\text{s.~}3\text{d.}$$ The usual wrong answer, 15s., counts the sovereigns and forgets that the worst draw still pays a shilling.

Answer $\tfrac38$ (3 to 5 against); 15s. 3d.

Domestic Economy

Everything depends on reading Mrs Perkins exactly. The shares add up to $\tfrac13 + \tfrac12 + \tfrac19 = \tfrac{17}{18}$ of a year’s income $I$. The hasty reading has this spent every year, so that two years leave $2 \times \tfrac{1}{18} I = \tfrac19 I = 190$, and the income is £1,710. But she says that since they set up house they have spent a third of his yearly income on rent, a half on the house and a ninth on other things: in the two years together, not in each. Then $$2I - \tfrac{17}{18} I = \tfrac{19}{18} I = 190, \qquad I = 180 .$$ In two years he earned £360, and they spent £60, £90 and £20, leaving £190. Dudeney takes this exact reading, and it is the answer his warning points to.

Answer £180 a year (£1,710 on the careless reading)

The Excursion Ticket Puzzle

Dudeney printed the answer, 458,908,622, and declined to give his method as out of proportion to its interest.

The method is short once it is put the right way round. Count, for every amount up to the fare, the number of ways it can be paid with the smallest coin alone; then let the coins in one at a time. When a coin worth $c$ farthings joins, an amount $v$ can be paid either without it, in the ways already counted, or with at least one of it, which is the same as paying $v - c$ with all the coins admitted so far. So $$W_{\text{new}}(v) = W_{\text{old}}(v) + W_{\text{new}}(v - c),$$ and a single pass up the amounts brings in each coin. Nothing needs to be listed.

The current coins of 1917, in farthings, were the farthing (1), halfpenny (2), penny (4), threepence (12), sixpence (24), shilling (48), florin (96), half-crown (120), double florin (192) and crown (240), with the half-sovereign and sovereign too large to matter for a fare of 948 farthings. Running the rule gives 458,908,622 exactly, Dudeney’s figure.

He counted the double florin, which the Royal Mint struck only from 1887 to 1890 but which remained legal tender. Without it the count is 357,549,696.

His table of the ways to change each coin, using only smaller coins, is correct in every entry:

So are the formulas he gave in passing. The simplest can be proved in two lines: with pennies, halfpennies and farthings, an amount of $n$ pence can be paid in $(n+1)^2$ ways. Choose the number of pennies, $n - j$ for some $j$ from 0 to $n$; the remaining $j$ pence, which is $2j$ halfpennies’ worth, can be made with $0, 1, \dots, 2j$ halfpennies and farthings for the rest, which is $2j + 1$ ways. Adding, $$\sum_{j=0}^{n} (2j+1) = (n+1)^2 .$$ Dudeney stated it for multiples of threepence, but the argument shows it holds for every whole number of pence. His two longer formulas, for threepences and then sixpences admitted, check exactly for every multiple of threepence up to 297d., and the rule for the constant (216 for multiples of sixpence, 324 otherwise) is right as well.

Answer 458,908,622 ways

Puzzle in Reversals

The crux is that the difference depends only on the gap $k$ between the pounds and the pence. An amount of $P$ pounds, $S$ shillings and $D$ pence is $240P + 12S + D$ pence, and its reversal is $240D + 12S + P$. The shillings cancel, and the difference is $$239(P - D) = 239k = 240(k-1) + (240 - k) \text{ pence.}$$ When $1 \le k \le 12$, the second part is $228 + (12 - k)$, so the difference is $£(k-1)$ 19s. $(12-k)$d. Reversing it gives $£(12-k)$ 19s. $(k-1)$d., and the two add up to $$240 \times 11 + 2 \times 228 + 11 = 3107 \text{ pence} = £12~18\text{s.}~11\text{d.}$$ whatever $k$ is. That is the rule, and it needs only $1 \le k \le 12$, not pounds below twelve.

Now $3107 = 13 \times 239$, which settles the other side. The result is the difference, $239k$, plus its reversal, which is positive. For $k \ge 13$ the difference alone is already $3107$ or more, so the total can never be £12 18s. 11d. The rule therefore holds exactly when the pounds exceed the pence by one to twelve.

The lowest amount that fails is the first with a gap of 13: £13 with nought shillings and nought pence. The highest that works has the largest pounds allowed by a gap of 12 and the largest pence, 11d.: £23 19s. 11d. Both agree with Dudeney. As he notes, the condition that the pounds exceed the pence already rules out sums below £1 and sums like £2 16s. 2d.

Answer Lowest failing £13; highest working £23 19s. 11d.

The Grocer and Draper

The interruptions come first. If the grocer lost $t$ minutes, the draper lost $17t$ and $18t = 9$, so the grocer lost half a minute and the draper eight and a half. The grocer’s 48 parcels at two a minute take 24 minutes, so he finishes in 24 minutes 30 seconds.

The draper’s trap is the fencepost: 48 yard pieces need only 47 cuts, because the last yard is what remains after the 47th. At three cuts a minute that is 15 minutes 40 seconds, and with his delays 24 minutes 10 seconds. The draper wins by twenty seconds. A solver who makes 48 cuts gets 24 minutes 30 seconds for him as well, and a dead heat.

Answer The draper wins by 20 seconds

Judkins’s Cattle

The number of animals divides equally into five droves and among eight dealers, so it is a multiple of 40. At two dollars at the least per animal, $301 buys at most 150, so the number is 40, 80 or 120. With $x$ oxen, $y$ pigs and $z$ sheep, $$17x + 4y + 2z = 301, \qquad x + y + z = 120 .$$ Subtracting twice the second from the first gives $15x + 2y = 61$. So $x$ is odd and at most 4: $x = 1$ gives $y = 23$ and $z = 96$, and $x = 3$ gives $y = 8$ and $z = 109$. Both work, and 120 is the greatest possible number.

Dudeney chooses between them with his favourite grammatical point: the droves held “oxen”, and a single ox is not oxen. That leaves 3 oxen, 8 pigs and 109 sheep. The reading is fair, and the puzzle has two answers without it.

Answer 120 animals: 3 oxen, 8 pigs, 109 sheep

Buying Apples

Work in sixths of a penny, so that the apples cost 6, 3 and 2 and the sevenpence is 42. If each of $n$ children receives $a$ large, $b$ middling and $c$ small apples, then $$n(6a + 3b + 2c) = 42 ,$$ so $n$ divides 42. As many boys as girls makes $n$ even: 2, 6, 14 or 42. For 42 each child’s share would cost one sixth of a penny, which no apple does. For 14 it costs 3, which is exactly one middling apple. For 6 it costs 7, and $6a + 3b + 2c = 7$ forces $a = 0$ and then $3b + 2c = 7$, so $b = 1$, $c = 2$. For 2 there are ten different shares.

Dudeney then reads the words closely, as he does all through the book. One boy and one girl are not “boys and girls”, and one apple each is not “apples”. That removes 2 and 14, and leaves six children, three boys and three girls, each given one halfpenny apple and two small ones, worth 1$\tfrac16$d. a child and sevenpence in all. With the plurals taken seriously, this is the only answer.

Answer Six children, each given one middling and two small apples

Buying Chestnuts

Read literally, the two men contradict each other: if five chestnuts are one short, a pennyworth is six, and six cannot be five too many. Both can be right only if their words mean something shorter. The buyer ought to have “a sixth”, meaning a sixth of a chestnut more; the shopman’s extra chestnut would give him “five too many”, meaning five-sixths of a chestnut too many. Then a pennyworth is $5\tfrac16$ chestnuts, and indeed $6 - 5\tfrac16 = \tfrac56$. Half a crown is thirty pence, so it buys $30 \times 5\tfrac16 = 155$ chestnuts.

This is a puzzle about language more than arithmetic, and Dudeney’s reading is the only one under which both men are right.

Answer 155 chestnuts

The Bicycle Thief

The salesman cannot have lost more than the thief gained, and the thief rode off with a bicycle that cost £11 and £10 of change, for a worthless piece of paper. The loss is £21.

The other transactions cancel. The neighbour gave £25 and got it back, and the friend lent £25 and is owed it. Following the salesman’s purse confirms this: £25 in from the neighbour, £10 out to the cyclist, £25 in from the friend and £25 back to the neighbour leave him £15 in hand and £25 in debt, which is £10 down, and the bicycle has gone as well. The £4 profit he expected is a disappointment, not money out of his pocket.

Answer £21

The Costermonger’s Puzzle

A drop in price of fourpence a hundred has to turn into exactly five more oranges for the same ten shillings, and that is a quadratic whose roots are two factors four apart. Let Bill pay $x$ pence a hundred, after beating the price down from $x + 4$. Ten shillings is 120 pence, so it buys $12{,}000/x$ oranges now and $12{,}000/(x+4)$ before, and the difference is five: $$\frac{12{,}000}{x} - \frac{12{,}000}{x+4} = 5, \qquad x(x+4) = 9{,}600 .$$ Since $96 \times 100 = 9{,}600$, $x = 96$, and the other root is negative. Bill paid 96d., which is 8s. a hundred: ten shillings buys 125 oranges at his price and 120 at the old one of 8s. 4d.

Answer 8s. a hundred