Various Arithmetical and Algebraical Problems

This chapter is a miscellany, and its puzzles have little in common beyond the algebra that settles them: a quadratic here, a pair of simultaneous equations there, a count that collapses into a perfect square. Several of them hide a second answer, or rest on a phrase that can be read two ways, and those are noted as they come.

Measurements are in feet and inches (twelve inches to the foot) and pounds weight, and the money is pounds sterling. Land is measured in rods, a rod (also called a pole) being $5\tfrac12$ yards; 160 square rods make an acre, and 320 rods a mile.

No. 97. The Spot on the Table

No. 98. Academic Courtesies

No. 99. The Thirty-Three Pearls

No. 100. The Labourer’s Puzzle

No. 101. The Trusses of Hay

No. 102. Mr Gubbins in a Fog

No. 103. Painting the Lamp-Posts

No. 104. Catching the Thief

No. 105. The Parish Council Election

No. 106. The Muddletown Election

No. 107. The Suffragists’ Meeting

No. 108. The Leap-Year Ladies

No. 109. The Great Scramble

No. 110. The Abbot’s Puzzle

No. 111. Reaping the Corn

No. 112. A Puzzling Legacy

No. 113. The Torn Number

No. 114. Curious Numbers

No. 115. A Printer’s Error

No. 116. The Converted Miser

No. 117. A Fence Problem

No. 118. Circling the Squares

No. 119. Rackbrane’s Little Loss

No. 120. The Farmer and His Sheep

No. 121. Heads or Tails

No. 122. The See-Saw Puzzle

No. 123. A Legal Difficulty

No. 124. A Question of Definition

No. 125. The Miners’ Holiday

No. 126. Simple Multiplication

No. 127. Simple Division

No. 128. A Problem in Squares

No. 129. The Battle of Hastings

No. 130. The Sculptor’s Problem

No. 131. The Spanish Miser

No. 132. The Nine Treasure Boxes

No. 133. The Five Brigands

No. 134. The Banker’s Puzzle

No. 135. The Stonemason’s Problem

No. 136. The Sultan’s Army

No. 137. A Study in Thrift

No. 138. The Artillerymen’s Dilemma

No. 139. The Dutchmen’s Wives

No. 140. Find Ada’s Surname

No. 141. Saturday Marketing

The Spot on the Table

Put the corner at the origin with the walls along the axes. A table of radius $r$ touching both walls has its centre at $(r,r)$, and the spot is at $(8,9)$. The crux is simply that the spot lies on the circle: $$\begin{gather*} (r-8)^2+(r-9)^2=r^2,\\ \text{so}\quad r^2-34r+145=(r-5)(r-29)=0. \end{gather*}$$ The radius is 29 inches or 5 inches, and the diameter 58 inches or 10 inches. Both are genuine circles touching both walls with the spot on the edge. With the 58-inch table the spot is on the arc facing the corner, where the boy is pointing in Dudeney’s picture; with the 10-inch table it is on the far side, away from the corner. Dudeney finds both and rules out the second as absurd for a large table, which is fair: the puzzle says the table is large. His arithmetic is the quadratic formula in disguise, $r=17\pm\sqrt{2\cdot8\cdot9}=17\pm12$.

Answer 58 inches (the algebra also allows a 10-inch table, spot on the far side)

Academic Courtesies

The crux is to count from the point of view of one child. With $n$ children in all, each bows to the other $n-1$ children and to the teacher, which is $n$ bows. So the morning’s bows number $n\times n=n^2=900$, and $n=30$. Twice as many girls as boys makes 10 boys and 20 girls.

The careless calculation counts a girl bowing to a boy and the boy bowing back as one bow, or forgets the teacher. Dudeney’s breakdown, $380+90+400+30=900$, is the same count done piece by piece, and his answer is the only one.

Answer 10 boys (and 20 girls)

The Thirty-Three Pearls

Measure every pearl against the big one: the string is then thirty-three copies of the big pearl less a discount that does not depend on its value at all. The big pearl is the seventeenth from either end, with sixteen pearls on each side. Call its value $B$. On one side the pearls fall by £100 at each step away from the centre and on the other by £150, so the total is $$\begin{align*} B+\sum_{k=1}^{16}(B-100k)+\sum_{k=1}^{16}(B-150k)&=33B-250\cdot\frac{16\cdot17}{2}\\ &=33B-34{,}000 . \end{align*}$$ Setting this to 65,000 gives $33B=99{,}000$ and $B=3{,}000$. The end pearls are worth £1,400 ($3{,}000-1{,}600$) and £600 ($3{,}000-2{,}400$). Dudeney agrees.

Answer £3,000

The Labourer’s Puzzle

Work in inches. The man is 70 inches tall; if the hole is now $d$ deep, his head is $70-d$ above ground, and when the hole is finished at depth $D$ it will be $D-70$ below. So $$D-70=2(70-d).$$ Everything turns on what “going twice as deep” fixes, and the words allow two readings.

Dudeney reads it as going twice as far again as he has gone already, so that $D=3d$. Then $3d-70=140-2d$, $d=42$, and the hole is now 3 ft 6 in deep, the man’s head 2 ft 4 in above ground. Finished, it is 10 ft 6 in deep, and his head 4 ft 8 in below the surface.

The other natural reading is “going down to twice the present depth”, $D=2d$. Then $2d-70=140-2d$, $d=52\tfrac12$, the hole is now 4 ft $4\tfrac12$ in deep, and finished it is 8 ft 9 in, with his head 2 ft 11 in below ground against 1 ft $5\tfrac12$ in above it now. Dudeney defends his reading by contrasting “twice as deep” with “as deep again”, but the phrase does not force it, and many readers would take the second.

Answer 10 ft 6 in on Dudeney’s reading; 8 ft 9 in if “twice as deep” means twice the present depth

The Trusses of Hay

Every truss appears in four of the ten pairs, so the ten weights add up to four times the total. They add up to 1,156, and the five trusses together weigh 289 lb.

Now the crux: the extreme sums can be identified without being told the pairs. Call the weights $a\le b\le c\le d\le e$. The lightest pair must be $a+b=110$. The next lightest must be $a+c=112$, because every other pair is at least as heavy: a pair containing $a$ with anything but $b$ weighs at least $a+c$, and a pair without $a$ weighs at least $b+c$. In the same way the heaviest pair is $d+e=121$ and the next heaviest $c+e=120$. Then $$c=289-(a+b)-(d+e)=289-110-121=58,$$ and so $a=54$, $b=56$, $e=62$, $d=59$. These five do give exactly the ten weights reported, and since every step was forced, no other five do.

Dudeney gives the same weights.

Answer 54, 56, 58, 59 and 62 lb

Mr Gubbins in a Fog

Measure each candle as a fraction of its full length. After $t$ hours the four-hour candle has $1-\tfrac t4$ left and the five-hour candle $1-\tfrac t5$. The crux is to decide which stub is the longer: the five-hour candle burns more slowly, so its stub is the long one, and $$1-\tfrac t5=4\bigl(1-\tfrac t4\bigr)=4-t,\qquad \tfrac45t=3,\qquad t=\tfrac{15}4 .$$ The candles burnt for three hours and three-quarters, leaving $\tfrac1{16}$ of one and $\tfrac4{16}$ of the other. The other way round, $1-\tfrac t4=4\bigl(1-\tfrac t5\bigr)$, gives $t=\tfrac{60}{11}$, which is longer than the four-hour candle could burn, so it is no answer. Dudeney’s three hours and three-quarters is right and is the only answer.

Answer 3 hours 45 minutes

Painting the Lamp-Posts

Say each side has $n$ posts. Pat paints the south side except Tim’s three, then six on the north: $(n-3)+6=n+3$. Tim paints three on the south and the north side except Pat’s six: $3+(n-6)=n-3$. The difference is always six, whatever $n$ is, which is the crux and the charm of the puzzle: if each man had kept to his own side they would have tied, and every post painted on the wrong side adds one to one man’s count and takes one from the other’s, a swing of two. Pat’s six posts swing the count twelve his way, and Tim’s three swing six back. Dudeney’s answer is the same.

Answer Pat, by six posts

Catching the Thief

Measure distance in the thief’s steps. The constable’s step is $\tfrac52$ of them. The crux is to compare the two men over the time the constable takes five steps: he covers $5\times\tfrac52=12\tfrac12$ thief-steps while the thief takes 8. He gains $4\tfrac12$ thief-steps every five of his own steps. A start of 27 thief-steps is closed after $27\div4\tfrac12=6$ such spells, that is after 30 of the constable’s steps. The thief has then taken 48 steps, and $27+48=75=30\times\tfrac52$, as Dudeney says.

The counsel’s “twenty-seven steps ahead of you” does not say whose steps, and Dudeney silently takes the thief’s. If the start was 27 of the constable’s own steps, it is $67\tfrac12$ thief-steps, which takes $67\tfrac12\div4\tfrac12=15$ spells, or 75 constable steps.

Answer 30 steps (75 if the start is measured in the constable’s steps)

The Parish Council Election

A vote for $k$ candidates is a choice of $k$ names from 23, which can be made in $\binom{23}{k}$ ways, and the crux is only to remember that each size of vote from one to nine is allowed. The counts are $$\begin{gather*} 23,\ 253,\ 1{,}771,\ 8{,}855,\ 33{,}649,\\ 100{,}947,\ 245{,}157,\ 490{,}314,\ 817{,}190, \end{gather*}$$ each found from the one before by $\binom{23}{k}=\binom{23}{k-1}\cdot\frac{24-k}{k}$, and they add up to 1,698,159, Dudeney’s total. He does not count a voter who turns up and votes for nobody. If a blank paper counts as a way of voting, add one: 1,698,160.

Answer 1,698,159 ways (1,698,160 counting a blank paper)

The Muddletown Election

Raise every loser to the winner’s level. Adding the three majorities, $18+146+575=739$, to the poll gives $5{,}473+739=6{,}212$, which is four times the Liberal’s vote. So the Liberal had 1,553, and subtracting the majorities gives the Conservative 1,535, the Independent 1,407 and the Socialist 978. That is Dudeney’s rule and his answer, and the equation $4L-739=5{,}473$ allows no other.

Answer Liberal 1,553, Conservative 1,535, Independent 1,407, Socialist 978

The Suffragists’ Meeting

Let $n$ be at the meeting and $x$ leave. The chair-woman stayed, so her leaving would have made $x+1=\tfrac23n$. The two friends were among those who left, so their staying would have made $x-2=\tfrac12n$. The crux is that the two remarks differ by exactly three people and by $\tfrac23-\tfrac12=\tfrac16$ of the meeting, so $\tfrac16n=3$ and $n=18$. Then $x=11$: eleven left, twelve would have been two-thirds, and nine would have been half. This agrees with Dudeney.

Answer 18 (of whom 11 left)

The Leap-Year Ladies

As Dudeney says, the difficulty is in stating the problem. Let there be $w$ widows. The crux is that every other quantity can be written as a multiple of $w$, leaving the 1,221 declined spinsters as the only fixed number to fix the scale.

The widows are all accepted: $\tfrac{35}{44}w$ by bachelors and $\tfrac9{44}w$ by widowers. The spinsters accepted by bachelors number $7\times\tfrac{35}{44}w=\tfrac{245}{44}w$. Let $A$ be the proposals accepted by widowers. The men to be married number $\tfrac{35}{44}w+\tfrac{245}{44}w+A$, and one-eleventh of them are the $A$ widowers, so $$11A=\tfrac{280}{44}w+A,\qquad A=\tfrac{28}{44}w .$$ Of these, $\tfrac9{44}w$ are widows, so widowers accepted $\tfrac{19}{44}w$ spinsters. Accepted proposals to widowers are four-fifths of all proposals to them, so widowers declined $\tfrac14A=\tfrac7{44}w$, all spinsters. The widows are one-eighth of the women, so there are $7w$ spinsters, and counting them by what happened to them, $$7w=\tfrac{245}{44}w+1{,}221+\tfrac{19}{44}w+\tfrac7{44}w=\tfrac{271}{44}w+1{,}221,$$ so $\tfrac{37}{44}w=1{,}221$ and $w=33\times44=1{,}452$. The women who proposed number $8w=11{,}616$. In full: of 10,164 spinsters, 8,085 married bachelors, 627 married widowers, 1,221 were declined by bachelors and 231 by widowers; of the 1,452 widows, 1,155 married bachelors and 297 widowers. The equations are linear, so the answer is unique, and every count comes out whole. Dudeney agrees.

Answer 11,616 women

The Great Scramble

First the trap Dudeney points out. If each of the three boys took one-fifth of the whole upset heap, they would take three-fifths of it while Bob and Andrew recovered five-eighths, more than the heap. So “one-fifth of the same” must mean one-fifth of the five-eighths, that is one-eighth of the heap each, and then $\tfrac58+\tfrac38$ accounts for it exactly.

Now follow a parcel of $N$ plums, writing each boy’s gain as a fraction of $N$. Andrew takes $\tfrac23N$; Bob takes $\tfrac38$ of that, $\tfrac14N$, and Charlie $\tfrac3{10}$, $\tfrac15N$. Andrew is left with $\tfrac{13}{60}N$, of which Edgar gets a seventh, $\tfrac{13}{420}N$, and David the rest. Bob drops $\tfrac18N$, and David and Edgar gain $\tfrac1{16}N$ each. From Charlie’s $\tfrac15N$, Andrew gathers $\tfrac1{20}N$, Bob $\tfrac1{15}N$, David $\tfrac2{35}N$, and Charlie and Edgar $\tfrac{11}{840}N$ each. David then scatters three-quarters of Andrew’s $\tfrac1{20}N$ and Bob’s $\tfrac1{15}N$, together $\tfrac7{80}N$: Andrew and Bob recover $\tfrac7{256}N$ each, and Charlie, David and Edgar take $\tfrac7{640}N$ each. The untouched third of the parcel gives each boy $\tfrac1{15}N$.

The crux is which of these fractions bind. Every one must be a whole number of plums. Since 11 is prime to 840, $\tfrac{11}{840}N$ forces 840 to divide $N$; since 7 is prime to 256, $\tfrac7{256}N$ forces 256 to divide $N$. So $N$ is a multiple of $\operatorname{lcm}(840,256)=26{,}880$. Every other denominator above (3, 8, 16, 20, 60, 70, 80, 420, 640 and the rest) divides 26,880, and $N=26{,}880$ works. Adding each boy’s gains and losses then gives Andrew 2,863, Bob 6,335, Charlie 2,438, David 10,294 and Edgar 4,950, which total 26,880.

All of this agrees with Dudeney.

Answer 26,880 plums: Andrew 2,863, Bob 6,335, Charlie 2,438, David 10,294, Edgar 4,950

The Abbot’s Puzzle

Doubling the corn equation removes the half-bushels: $6m+4w+c=200$. Subtracting $m+w+c=100$ leaves the crux, $$5m+3w=100 .$$ Since $5m$ and 100 are multiples of 5, so is $w$; writing $w=5j$ gives $m=20-3j$. So the whole family of answers is $(m,w,c)=(20-3j,\ 5j,\ 80-2j)$ for $j=0,1,\dots,6$: seven in all, one of them ($j=0$) with no women. That confirms Dudeney’s count of six.

With five times as many women as men, $5j=5(20-3j)$, so $j=5$: 5 men, 25 women and 70 children. The men receive 15 bushels, the women 50 and the children 35. Dudeney’s answer is right and is the only one.

One historical slip: Dudeney calls Alcuin Abbot of Canterbury. Alcuin was a Yorkshireman, as he says, but the abbey he ruled, from 796, was St Martin’s at Tours.

Answer 5 men, 25 women and 70 children

Reaping the Corn

Let the field be $s$ rods square. Cutting a strip one rod wide all round leaves a square of side $s-2$, and the farmer’s claim is that it holds half the field: $$(s-2)^2=\tfrac12s^2,\qquad\text{so}\qquad s-2=\frac{s}{\sqrt2}.$$ The crux is to take the square root at once rather than expand: the inner side is the outer side divided by $\sqrt2$, as it must be for half the area. Then $s(1-1/\sqrt2)=2$ and $$s=\frac{2\sqrt2}{\sqrt2-1}=2\sqrt2(\sqrt2+1)=4+2\sqrt2\approx6.8284\ \text{rods}.$$ (The negative square root would give $s=4-2\sqrt2$, less than two rods, too small to have a one-rod strip cut from both sides.) The field is $s^2=24+16\sqrt2\approx46.627$ square rods, about $0.2914$ of an acre, and the square left standing has side $2+2\sqrt2\approx4.8284$ rods and area $12+8\sqrt2\approx23.314$ square rods.

Dudeney’s figures are right to the precision that matters, with one small slip. His inner square of 23.313 square rods is the true $23.3137\ldots$ cut short, but his field of 46.626 square rods is wrong in the last place: the true area is $46.6274\ldots$, twice the inner square.

Answer $24+16\sqrt2\approx46.627$ square rods, about 0.29 of an acre

A Puzzling Legacy

The will is itself awkward: a third, a quarter and a fifth add up to $\tfrac{47}{60}$, not the whole. Dudeney reads the fractions as proportions, so that the three sons were to share the whole hundred acres in the ratio $\tfrac13:\tfrac14:\tfrac15=20:15:12$. With Charles gone, the crux is that the ratio between the survivors does not change: $\tfrac13:\tfrac14=4:3$. Alfred takes $\tfrac47$ of the land, $57\tfrac17$ acres, and Benjamin $\tfrac37$, $42\tfrac67$ acres.

The answer depends on that reading. If the will means a third of the hundred acres to Alfred and a quarter to Benjamin outright, they receive $33\tfrac13$ and 25 acres, and the rest is not disposed of by the will at all. Split equally between the brothers, that would give Alfred $54\tfrac16$ and Benjamin $45\tfrac56$ acres. Which is fair is a question for a lawyer; the word “proportion” in the will supports Dudeney.

Answer Alfred $57\tfrac17$ acres, Benjamin $42\tfrac67$ acres (in the ratio $4:3$)

The Torn Number

Write the number as $100x+y$, where $x$ and $y$ are the two halves, and let $s=x+y$. The condition $s^2=100x+y$ becomes, after subtracting $s=x+y$, $$s(s-1)=99x .$$ That is the crux: 99 must divide the product of two consecutive numbers. Since $s$ and $s-1$ have no common factor, 9 must divide one of them and 11 must divide one of them. For a four-figure square, $32\le s\le99$. The possibilities are: $s$ divisible by 99, giving $s=99$; $s$ divisible by 9 with $s-1$ divisible by 11, giving $s=45$; and $s$ divisible by 11 with $s-1$ divisible by 9, giving $s=55$ (both $s-1$ divisible by 99 and the next values of each case, $144$ and $154$, are out of range). So the only such numbers are $$45^2=2025,\qquad 55^2=3025,\qquad 99^2=9801,$$ and indeed $20+25=45$, $30+25=55$ and $98+01=99$. Of these, 2025 repeats a figure, so the only new number with four different figures is 9,801, as Dudeney says. He also notes 2025 and excludes it for the same reason.

Answer 9,801 (the only other, 2,025, repeats a figure)

Curious Numbers

Write $n+1=a^2$ and $\tfrac n2+1=b^2$. Eliminating $n$ gives $a^2=2b^2-1$, so the crux is to find every pair of whole numbers with $$a^2-2b^2=-1 .$$ The smallest is $(1,1)$, giving $n=0$, and the next $(7,5)$, giving 48. New pairs come from old ones by the rule $$(a,b)\ \longmapsto\ (3a+4b,\ 2a+3b),$$ because $(3a+4b)^2-2(2a+3b)^2=a^2-2b^2$, as expanding both sides shows. From $(7,5)$ it gives $(41,29)$, then $(239,169)$, then $(1393,985)$, so the next three numbers are $$\begin{gather*} n=41^2-1=1{,}680,\qquad 239^2-1=57{,}120,\\ 1393^2-1=1{,}940{,}448, \end{gather*}$$ with halves plus one equal to $29^2=841$, $169^2=28{,}561$ and $985^2=970{,}225$.

Nothing has been skipped. The reverse rule $(a,b)\mapsto(3a-4b,\ 3b-2a)$ also keeps $a^2-2b^2$ fixed, and for any solution with $b>1$ it gives a smaller one in positive whole numbers: $a^2=2b^2-1$ puts $a$ between $b$ and $\tfrac32b$, so $0<3b-2a<b$; and $3a-4b>0$ because $9a^2=18b^2-9>16b^2$ once $b\ge3$ (and $b=2$ gives no solution). Stepping down repeatedly must end at $(1,1)$, so every solution is reached by stepping up from $(1,1)$, and the list above is complete in order.

Dudeney’s three numbers are correct.

Answer 1,680, 57,120 and 1,940,448

A Printer’s Error

The answer is $2^5\times9^2=32\times81=2{,}592$. Here there is no crux in the usual sense and no tidy argument for uniqueness: the exponents make the equation too irregular for algebra to grip. The honest route is the exhaustive one. There are only 9,000 four-figure numbers, and computing $a^b\times c^d$ for each settles the matter.

Dudeney says 2592 is the only solution, and he is right.

Answer $2^5\times9^2=2592$, the only solution

The Converted Miser

As Dudeney points out, the year is not given, so the least sum comes from the most favourable year. A year has 365 or 366 days and 52 or 53 Sundays. The sum $N$ must be a multiple of the number of days and leave 1 over when divided by the number of weekdays.

The crux is that a leap year with 52 Sundays is impossible and the other cases reduce to one small congruence each. In that leap year $N$ is a multiple of 366, so even, while $N=314k+1$ is odd. In an ordinary year beginning on a Sunday there are 53 Sundays and 312 weekdays; $N=365k$ with $365k\equiv53k\equiv1\pmod{312}$. Since $53^2=2{,}809=9\times312+1$, the least $k$ is 53, and $$N=365\times53=19{,}345=62\times312+1 .$$ He gives £53 a day all year, or £62 each weekday with one pound over. The other two kinds of year need $52k\equiv1\pmod{313}$ (ordinary year, 313 weekdays), whose least solution is $k=307$ since $52\times307=51\times313+1$, and $53k\equiv1\pmod{313}$ (leap year, 53 Sundays), least $k=189$ since $53\times189=32\times313+1$. These give £112,055 and £69,174, both far larger.

So the least sum is £19,345, in an ordinary year starting on a Sunday. Dudeney’s table has the same four figures, and his remark that the last such year before he wrote was 1911 is correct: 1 January 1911 was a Sunday.

Answer £19,345

A Fence Problem

Let the side be $s$ poles. The fence runs $4s$ poles with fourteen rails to the pole, $56s$ rails, and the field holds $s^2$ square poles, or $s^2/160$ acres. The crux is that rails grow with the side and acres with its square, so they balance at just one size: $$\frac{s^2}{160}=56s\qquad\Longrightarrow\qquad s=8{,}960\ \text{poles}=28\ \text{miles}.$$ The field has $8{,}960^2/160=501{,}760$ acres, and the fence $56\times8{,}960=501{,}760$ rails. A smaller field has more rails than acres, a larger one fewer. Dudeney’s answer is the same, and so is his general rule: with $r$ rails in the height and half-pole rails, the balance $s^2/160=8rs$ gives $s=1{,}280r$ poles, which is $4r$ miles.

Answer A square 28 miles on a side: 501,760 acres and 501,760 rails

Circling the Squares

Number the squares round the ring and write $x_i$ for the number in square $i$, so that square $i + 5$ is opposite square $i$. The crux is to look at the difference across each diameter, $d_i = x_i^2 - x_{i+5}^2$. The condition for neighbours $i$ and $i+1$ is $$x_i^2 + x_{i+1}^2 = x_{i+5}^2 + x_{i+6}^2, \qquad\text{that is}\qquad d_i = -d_{i+1} .$$ So the differences across the five diameters are all the same size and alternate in sign. The given numbers fix the size: $16^2 - 8^2 = 192$ and $2^2 - 14^2 = -192$. Every diameter therefore holds two numbers whose squares differ by 192, the larger at C, E and A and the smaller at B and D, as the signs alternate.

Now $a^2 - b^2 = (a-b)(a+b) = 192$. The two factors have the same parity and their product is even, so both are even; writing them as $2m$ and $2n$ with $mn = 48$ and $m < n$ gives $a = m + n$ and $b = n - m$. The five factorisations $1 \times 48$, $2 \times 24$, $3 \times 16$, $4 \times 12$ and $6 \times 8$ give exactly five pairs: $$(49, 47), \quad (26, 22), \quad (19, 13), \quad (16, 8), \quad (14, 2).$$ The last two are the given diameters, and the other three pairs must fill the diameters C–H, D–I and E–K. Any order will do, provided each pair is turned the right way round: larger number at C and E, smaller at D. That is $3! = 6$ ways, all using ten different numbers, which is Dudeney’s count. His arrangement, reading clockwise from A, is 16, 2, 49, 22, 19, 8, 14, 47, 26, 13:

The same argument proves Dudeney’s closing remark, that a ring of this kind must have $4n + 6$ squares. With $D$ diameters, square $i + D$ is opposite square $i$, so by definition $d_{i+D} = x_{i+D}^2 - x_i^2 = -d_i$. But the signs alternate from one square to the next, so $d_{i+D} = (-1)^D d_i$. Since $d_i$ is not zero (opposite numbers are different), $D$ must be odd, and the ring has $2D = 6, 10, 14, \dots$ squares.

Answer Six ways, e.g. C 49, D 22, E 19, H 47, I 26, K 13

Rackbrane’s Little Loss

Work backwards from the end, where every game can be undone: the two winners each had half of what they hold, and the loser had the rest. Let each player end with $x$. Before the third game, which Mr Potts lost, Mrs Potts and the professor had $\tfrac12 x$ each and Mr Potts $2x$. Before the second, lost by Mrs Potts, Mr Potts had $x$, the professor $\tfrac14 x$ and Mrs Potts $\tfrac74 x$. Before the first, lost by the professor, Mr Potts had $\tfrac12 x$, Mrs Potts $\tfrac78 x$ and the professor $\tfrac{13}{8} x$.

The professor went from $\tfrac{13}{8}x$ to $x$, a loss of $\tfrac58 x$, and this is five shillings. So $x$ is eight shillings, and the players sat down with 13s., 4s. and 7s. The equation is linear, so this is the only answer, as Dudeney says.

Answer 13s. (Mr Potts 4s., Mrs Potts 7s.)

The Farmer and His Sheep

Factor the difference of squares and the puzzle falls apart. If the parts are $x$ and $y$, then $x^2 - y^2 = (x-y)(x+y)$, and since the parts are different, dividing by $x - y$ gives $x + y = 1$. Whatever the split, the flock is exactly one sheep. No split of one sheep into two whole numbers of sheep is possible, so the parts are fractions of a sheep, such as $\tfrac23$ and $\tfrac13$: they differ by $\tfrac13$, and so do their squares, $\tfrac49$ and $\tfrac19$. Any two parts adding up to one will do. This is Dudeney’s answer.

Answer One sheep

Heads or Tails

Each toss multiplies the purse by a fixed factor, so only the number of wins and losses matters, never their order. A win turns a purse into $\tfrac32$ of itself and a loss into $\tfrac12$. A win and a loss together multiply it by $\tfrac32 \times \tfrac12 = \tfrac34$, so after $n$ wins and $n$ losses, in any order, Crooks holds $(\tfrac34)^n$ of what he started with. He loses, and the longer he plays the more he loses: a quarter after two tosses, nearly half after four. The answer agrees with Dudeney.

Answer He loses; after $n$ wins and $n$ losses he keeps $(\tfrac34)^n$ of his money

The See-Saw Puzzle

The two balances are the principle of the lever used twice, and multiplying them makes the unknown lengths cancel. First the brick: if a brick is $b$ pounds, then $b = \tfrac34 b + \tfrac34$, so $b = 3$. Let the boy weigh $w$ and let the two arms of the plank be $L$ (long) and $S$ (short). Bricks on the short arm put the boy on the long one, and bricks on the long arm put him on the short one: $$wL = 48S, \qquad wS = 33L .$$ Multiplying, $w^2 LS = 48 \times 33\, LS$, so $w^2 = 1584$ and $w = 12\sqrt{11} \approx 39.80$ pounds, the geometric mean of the two brick loads. Dudeney gives “about 39.79 lbs”, which is the decimal cut off rather than rounded. The answer depends on ignoring the plank’s own weight, as he does; a real plank pivoted off centre would add a term to each balance.

Answer $12\sqrt{11} \approx 39.8$ pounds

A Legal Difficulty

The will never says what to do with twins, so the question is what it tells us about the testator’s wishes. Its two clauses agree in their ratios: a son was to have twice the mother’s share, and the mother twice a daughter’s. Keeping both ratios, the shares are $4:2:1$, so the son takes $\tfrac47$, the mother $\tfrac27$ and the daughter $\tfrac17$. Those are the only shares that respect both clauses, and this is Dudeney’s answer. It is an interpretation rather than a theorem: a court might equally read the will as having failed for want of a provision for twins.

Answer Son $\tfrac47$, mother $\tfrac27$, daughter $\tfrac17$

A Question of Definition

The two phrases name the same area but not the same shape. A mile square is a square whose sides are a mile long, so it is one square mile of a particular shape. A square mile is a unit of area, and a property of one square mile might be a long thin strip or any other figure. So there is no difference in area, but there may be a great difference in shape, which is Dudeney’s answer. The puzzle is verbal, so there is no program for it.

Answer Equal in area; only the mile square must be a square

The Miners’ Holiday

Bill’s extra three shillings are shared across the seven when the average is taken, so the six others’ total must fall short of it by six shares of three shillings. The six spent 60 shillings between them. If the average is $m$, Bill spent $m + 3$, and $7m = 60 + m + 3$, so $6m = 63$ and $m = 10\tfrac12$ shillings. Bill spent 13s. 6d.; the average, 10s. 6d., is half a guinea. The answer is unique and agrees with Dudeney.

Answer 13s. 6d.

Simple Multiplication

Moving the leading 3 to the back is the same as asking for the repeating block of a fraction, here $\tfrac{6}{17}$. Suppose the number has $k$ digits, so it is $3 \times 10^{k-1} + r$ with $r < 10^{k-1}$. Moving the 3 gives $10r + 3$, and we need $2(10r + 3) = 3(3 \times 10^{k-1} + r)$, that is, $$17r = 9 \times 10^{k-1} - 6 .$$ So $9 \times 10^{k-1} \equiv 6 \pmod{17}$. Since $9 \times 2 = 18 \equiv 1$, this says $10^{k-1} \equiv 12 \pmod{17}$. The powers of 10 modulo 17 run $10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12, \dots$, and the first 12 is $10^{15}$. So the shortest row has $k = 16$ cards, and $$\begin{align*} N &= 3 \times 10^{15} + \frac{9 \times 10^{15} - 6}{17}\\ &= \frac{6(10^{16} - 1)}{17} = 3529411764705882 . \end{align*}$$ The last expression shows that $N$ is the repeating block of $\tfrac{6}{17} = 0.\overline{3529411764705882}$. Since the powers of 10 repeat with period 16 modulo 17, the longer answers are this block written twice, three times and so on, just as Dudeney says. His answer agrees.

Answer 3529411764705882

Simple Division

If two numbers leave the same remainder on division by $d$, then $d$ divides their difference. So $d$ must divide $1059 - 701 = 358$, $1417 - 1059 = 358$ and $2312 - 1417 = 895$. Now $358 = 2 \times 179$ and $895 = 5 \times 179$, and 179 is prime, so the largest candidate is 179. It works: every number leaves remainder 164 (for instance $701 = 3 \times 179 + 164$). Apart from 1, it is the only such divisor. This agrees with Dudeney.

Answer 179, with remainder 164

A Problem in Squares

Work in inches, where five square feet is 720 square inches, and look for whole sides $a < b < c$ with $b^2 - a^2 = c^2 - b^2 = 720$. Dudeney’s boards have sides 31, 41 and 49 inches: $41^2 - 31^2 = 1681 - 961 = 720$ and $49^2 - 41^2 = 2401 - 1681 = 720$. Among whole numbers of inches they are the only ones.

The crux of the second half is that three whole squares in arithmetic progression always have a common difference divisible by 24, which is why 7 and 13 need fractions. Let the squares be $a^2, a^2 + d, a^2 + 2d$.

• Squares leave remainder 0 or 1 on division by 3. If 3 did not divide $d$, the three terms would leave all three remainders 0, 1 and 2, so 3 divides $d$.

• Squares leave remainder 0 or 1 on division by 4. The first and third terms differ by $2d$; if $d$ were odd they would differ by 2 modulo 4, so $d$ is even. Then the three sides all have the same parity. If they are all odd, each square leaves 1 on division by 8, so 8 divides $d$. If they are all even, halve every side: the squares shrink by 4 and so does $d$, and we repeat until the sides are odd; at that point 8 divides the reduced difference, so 8 divides $d$ too.

So $24 \mid d$, and 720 $= 24 \times 30$ passes, as it must. For 7 and 13 the sides must be fractions, and Dudeney’s triples check exactly: $\bigl(\tfrac{113}{120}\bigr)^2, \bigl(\tfrac{337}{120}\bigr)^2, \bigl(\tfrac{463}{120}\bigr)^2$ step by 7, and $\bigl(\tfrac{80929}{19380}\bigr)^2, \bigl(\tfrac{106921}{19380}\bigr)^2, \bigl(\tfrac{127729}{19380}\bigr)^2$ step by 13. He leaves the difference 23 as a hard nut for the reader, and it is left here too.

Answer 31, 41 and 49 inches

The Battle of Hastings

We need the smallest whole $y$ with $61y^2 + 1 = x^2$, which is Pell’s equation $x^2 - 61y^2 = 1$.

Trial is hopeless here, and the crux is the classical method: the continued fraction of $\sqrt{61}$. Its convergents $p/q$ are the best rational approximations to $\sqrt{61}$, and a theorem of Lagrange says that the least solution of $x^2 - 61y^2 = 1$ is always among them, at the end of the first period (or the second, when the period is odd, as it is for 61). Running the algorithm gives $$x = 1\,766\,319\,049, \qquad y = 226\,153\,980,$$ and exact arithmetic confirms $x^2 - 61y^2 = 1$. Each of the sixty-one squares has $y^2 = 51\,145\,622\,669\,840\,400$ men, the army is $61y^2 = 3\,119\,882\,982\,860\,264\,400$, and with Harold it makes a square of side $x$. This is Dudeney’s answer. The same method gives the small cases he quotes, $60 \times 4^2 + 1 = 31^2$ and $62 \times 8^2 + 1 = 63^2$, and also his remark that for 97 the least $y$ is 6 377 352. His conclusion follows: over three trillion Saxons is too many, so the chronicle is wrong.

Answer $61 \times 226\,153\,980^2 = 3\,119\,882\,982\,860\,264\,400$ men

The Sculptor’s Problem

Both sides of the equation share the factor $a + b$, and cancelling it turns a cubic into a quadratic. With edges $a$ and $b$ we need $a^3 + b^3 = a + b$, and since $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, this becomes $$a^2 - ab + b^2 = 1 .$$ Whole numbers cannot satisfy it unless $a = b = 1$, since $a^2 - ab + b^2 = (a-b)^2 + ab$ and $ab \ge 2$ for any other pair; the unequal pedestals forbid that, so the edges are fractions. Write them over a common denominator as $p/q$ and $r/q$; then $p^2 - pr + r^2 = q^2$. Trying $q = 2, 3, \dots$ in turn, a short search finds that the smallest denominator that works is $q = 7$, with $8^2 - 8 \cdot 3 + 3^2 = 49$ and $8^2 - 8 \cdot 5 + 5^2 = 49$. So there are two answers in the smallest figures: edges of $\tfrac87$ and $\tfrac37$ feet, and edges of $\tfrac87$ and $\tfrac57$ feet. For the first pair the lengths add to $\tfrac{11}{7}$ and the volumes to $\tfrac{512 + 27}{343} = \tfrac{539}{343} = \tfrac{11}{7}$. Dudeney gives the first pair and mentions the second.

Answer Edges $\tfrac87$ and $\tfrac37$ feet (or $\tfrac87$ and $\tfrac57$)

The Spanish Miser

The outer boxes of an arithmetic progression add up to twice the middle one, so the middle box must be half a square. Call the boxes $a < b < c$. Then $a + c = 2b$ is a square, so $b = 2k^2$; and with $a + b = s^2$ and $b + c = t^2$, adding gives $s^2 + t^2 = 4b = 8k^2$. For each $k$ it is a short search for the ways of writing $8k^2$ as a sum of two squares with $s^2 > 2k^2$.

The smallest total comes out as $482 + 3362 + 6242$, with sums $3844 = 62^2$, $6724 = 82^2$ and $9604 = 98^2$; since the total is $3b$, a search over small $b$ settles it completely. For the question asked, the smallest single box, the search finds $386, 8450, 16514$ (pair sums $94^2$, $130^2$ and $158^2$), and nothing with a smaller box. That is Dudeney’s answer, but it is only as good as the search: a very large middle box could in principle pair with a smaller bottom box, and no elementary argument rules that out. Dudeney states it without proof.

Answer 386 (in the triple 386, 8450, 16514)

The Nine Treasure Boxes

The bound on box A is what makes the search finite in practice: A is 1, 4 or 9, so only a difference $d$ for which $A + d$ and $A + 2d$ are also squares is worth trying, and for each such $d$ we list every progression of three squares with that difference. The common difference $d$ must be a multiple of 24 (see No. 128), and the first that carries three progressions, one starting below twelve, is $d = 3360$: $$2^2, 58^2, 82^2; \qquad 46^2, 74^2, 94^2; \qquad 97^2, 113^2, 127^2,$$ that is, A, B, C $= 4, 3364, 6724$; D, E, F $= 2116, 5476, 8836$; and G, H, I $= 9409, 12769, 16129$. This agrees with Dudeney. His claim that the bound on A makes the answer unique is stronger than he proves: there are infinitely many differences with a progression starting at 4, and a search can only say that no other works up to a large limit.

Answer A to I: 4, 3364, 6724, 2116, 5476, 8836, 9409, 12769, 16129

The Five Brigands

Eliminate the two awkward shares and the count becomes a count of whole points in a slab. Multiply the second condition by 6 and subtract twice the first: $$70a + 16b + 4c + d = 800, \qquad e = 69a + 15b + 3c - 600 .$$ Here $a, \dots, e$ are the shares of Alfonso to Esteban. Every choice of positive $a, b, c$ with $70a + 16b + 4c < 800$ and $69a + 15b + 3c > 600$ gives exactly one answer. Note also that the halves and thirds come out whole automatically: $3d + 2e = 1200 - 6(12a + 3b + c)$ forces $d$ even and $e$ a multiple of 3. The first inequality alone limits Alfonso to at most 11.

Counting the points gives 6627 answers, split by Alfonso’s share as 1005, 985, 977, 903, 832, 704, 570, 388, 200, 60 and 3 for shares 1 to 11. That is exactly Dudeney’s count and table, so Labosne’s 6639 is twelve too many.

Answer 6627 answers

The Banker’s Puzzle

A pile of sixpences cannot be split into equal piles (other than single coins or one pile) exactly when their number is prime, and the crux is Euler’s remarkable polynomial $n^2 - n + 41$, which is prime for $n = 1, 2, \dots, 40$. The banker puts in 40 sixpences and tells the customer to transfer the square of one less than the number he put in. If the customer put in $n$, the box then holds $$40 + n + (n - 1)^2 = n^2 - n + 41 ,$$ and since the customer may put in at most forty sixpences (a pound), this is prime every time: 41 when $n = 1$, up to $1601$ when $n = 40$. The banker wins with certainty. The limit of a pound matters, because at $n = 41$ the total is $41^2$.

Two small points of reading. The instruction is a rule, not a number, and the customer applies it knowing his own $n$; the banker never learns what the customer put in, so the rule “neither knowing what the other put in” is kept. And when $n = 1$ the rule asks for no sixpences at all, which is taken here to be allowed. The answer agrees with Dudeney.

Answer Put in 40; ask for $(n-1)^2$ when the customer put in $n$

The Stonemason’s Problem

The key is the fact the mason noticed: $1^3 + 2^3 + \dots + n^3 = T_n^2$, where $T_n = \tfrac12 n(n+1)$ is the $n$th triangular number. A run of heaps from $a^3$ to $b^3$ is then a difference of two such squares, $$a^3 + \dots + b^3 = T_b^2 - T_{a-1}^2 ,$$ and the question is when a difference of two squares of triangular numbers is itself a square.

Dudeney’s answer is twelve heaps, $14^3 + 15^3 + \dots + 25^3 = T_{25}^2 - T_{13}^2 = 325^2 - 91^2 = 97{,}344 = 312^2$. A neat check: $(325 - 91)(325 + 91) = 234 \times 416 = 2^6 \cdot 3^2 \cdot 13^2$, a perfect square.

That nothing smaller works is settled by search, and the search is complete. Any run below 97,344 blocks has its largest heap $b^3$ below 97,344, so $b \le 45$, and a program tries every run with $2 \le a \le b \le 45$ of four or more heaps.

It finds none smaller. Dudeney’s two remarks also check: with three heaps allowed, $23^3 + 24^3 + 25^3 = 41{,}616 = 204^2$ would be smaller still, and five heaps $25^3 + \dots + 29^3 = 99{,}225 = 315^2$ come close without beating twelve.

Answer 97,344 blocks, heaps $14^3$ to $25^3$

The Sultan’s Army

The number of ways to write $N$ as a sum of two squares depends only on its prime factors, and that is the crux. Primes of the form $4k+3$ (3, 7, 11, …) can never be a sum of two squares; primes of the form $4k+1$ (5, 13, 17, 29, 37, …) can, in exactly one way; the factor 2 changes nothing. Fermat stated these facts and Euler proved them, and Jacobi’s two-square theorem turns them into a count. If $N = p_1^{e_1} p_2^{e_2} \cdots$ times a power of 2, the $p_i$ being primes $4k+1$ and any prime $4k+3$ appearing to an even power, and $N$ is not itself a square, then the number of ordered pairs of positive whole numbers with $a^2 + b^2 = N$ is exactly $(e_1+1)(e_2+1)\cdots$. Unordered pairs are half as many, a pair with $a = b$ counting once.

For twelve unordered ways the product $(e_1+1)(e_2+1)\cdots$ must be 24 (or 23, which would need a prime raised to the 22nd power). Spread 24 over the smallest primes $4k+1$ with the largest exponent on the smallest prime: $3 \times 2 \times 2 \times 2$ gives $5^2 \cdot 13 \cdot 17 \cdot 29 = 160{,}225$, while $4 \times 3 \times 2$ gives $5^3 \cdot 13^2 \cdot 17 = 359{,}125$ and the other splits are larger still. So the army is 160,225 men, Dudeney’s number, and a direct count of every pair of squares up to 400,000 confirms it is the smallest with exactly twelve ways, and also the smallest with at least twelve.

The twelve formations are $$\begin{aligned} &400^2+15^2,\ 399^2+32^2,\ 393^2+76^2,\ 392^2+81^2,\\ &384^2+113^2,\ 375^2+140^2,\ 360^2+175^2,\ 356^2+183^2,\\ &337^2+216^2,\ 329^2+228^2,\ 311^2+252^2,\ 300^2+265^2. \end{aligned}$$ Dudeney’s example checks too: $130 = 2 \times 5 \times 13$ has $2 \times 2 / 2 = 2$ ways.

Answer 160,225 men

A Study in Thrift

Two of the four conditions come almost free. Every square is a sum of two triangles, since $T_s + T_{s-1} = \tfrac12 s(s+1) + \tfrac12 (s-1)s = s^2$: a square of counters cut along a diagonal, one side of the cut including the diagonal. Dudeney states without proof that every triangular number above 6 splits into three triangles; for the numbers here, the splits are given below and they check. So the real condition is that the number is both a square and a triangle.

A number $T$ is triangular exactly when $8T + 1$ is a square (for $T = \tfrac12 m(m+1)$ it is $(2m+1)^2$). So $y^2$ is square and triangular when $8y^2 + 1 = x^2$: Pell’s equation $x^2 - 8y^2 = 1$. Its smallest solution is $x = 3$, $y = 1$, and each solution gives the next by $$(x, y) \mapsto (3x + 8y,\ x + 3y),$$ which is multiplication by $3 + \sqrt8$; as with the Battle of Hastings, these are all the solutions. The values of $y$ run 1, 6, 35, 204, 1189, 6930, 40391, and the square triangular numbers are $$1,\ 36,\ 1225,\ 41{,}616,\ 1{,}413{,}721,\ 48{,}024{,}900,\ 1{,}631{,}432{,}881 .$$ After Mrs McAllister’s 36 come five more, so her sixth present needs £1,631,432,881, which is Dudeney’s figure. A search of all squares up to a million finds exactly the first four.

Dudeney’s table of splittings checks in every place but one. For 1225 he gives two triangles of sides 36 and 34, but those hold $666 + 595 = 1261$ counters. The pattern of the other rows, side of the square and one less, gives 35 and 34, which hold $630 + 595 = 1225$; the 36 is a misprint.

Answer 1,631,432,881 sovereigns

The Artillerymen’s Dilemma

A square pyramid with $n$ balls along each side of its base holds $$1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$$ balls, and the question is when this is a perfect square.

Apart from the single ball, the answer is $n = 24$: the pyramid holds $\tfrac16 \times 24 \times 25 \times 49 = 4900 = 70^2$ balls. Dudeney found it by writing out the pyramid numbers until a square appeared, and a search up to $n = 2$ million finds no other.

Dudeney confessed that he had no proof that 4900 is the only answer, though he did not believe another existed. He was right. The problem had been posed by Édouard Lucas, and in 1918, the year after this book appeared, G. N. Watson proved that 1 and 4900 are the only square pyramids that are also squares. His proof used elliptic functions, and no proof within the reach of this book is known, so here the answer rests on Watson’s theorem.

Answer 4900 balls (a pyramid of side 24, a square of side 70)

The Dutchmen’s Wives

Someone who buys $x$ hogs at $x$ shillings pays $x^2$, so each husband and wife bought numbers whose squares differ by 63. Now $h^2 - w^2 = (h - w)(h + w) = 63$, and the factorisations $1 \times 63$, $3 \times 21$ and $7 \times 9$ give the only three pairs: $(32, 31)$, $(12, 9)$ and $(8, 1)$. So the husbands bought 32, 12 and 8 hogs, and the wives 31, 9 and 1.

Hendrick bought 23 more than Katruen: among husbands 32, 12, 8 and wives 31, 9, 1, the only difference of 23 is $32 - 9$. So Hendrick bought 32 and Katruen 9. Elas bought 11 more than Gurtruen, and of the remaining numbers only $12 - 1$ works, so Elas bought 12 and Gurtruen 1. That leaves Cornelius with 8 and Anna with 31. Pairing each husband with the wife whose square is 63 less than his: Hendrick (32) married Anna (31), Elas (12) married Katruen (9) and Cornelius (8) married Gurtruen (1). There is no other solution.

Answer Hendrick and Anna, Elas and Katruen, Cornelius and Gurtruen

Find Ada’s Surname

This is the last puzzle again with a larger number. Each woman spends the square of her number of feet in farthings, and 8s. 5¼d. is 405 farthings, so each mother and daughter bought $m$ and $d$ feet with $m^2 - d^2 = 405 = 3^4 \times 5$. Its five factorisations give exactly five pairs, one for each family: $$(203, 202), \quad (69, 66), \quad (43, 38), \quad (27, 18), \quad (21, 6).$$ The mothers bought 203, 69, 43, 27 and 21 feet. Mrs Robinson spent 6s., which is 288 farthings, more than Mrs Evans, and among the mothers’ squares $41209$, $4761$, $1849$, $729$, $441$, only $729 - 441 = 288$; so Mrs Robinson bought 27 feet and Mrs Evans 21. Four times Mrs Evans’s 441 is 1764, and the nearest of the others is Mrs Jones at 1849 (43 feet). Mrs Smith spent most, 203 feet, and Mrs Brown has the 69.

Mrs Brown bought 21 yards, 63 feet, more than Bessie, so Bessie bought 6 feet and is Bessie Evans. Annie bought 16 yards, 48 feet, more than Mary: of the daughters’ 202, 66, 38, 18, only $66 - 18 = 48$, so Annie Brown and Mary Robinson. Annie spent $66^2 = 4356$ farthings, and £3 0s. 8d. is 2912 farthings, so Emily spent 1444, which is $38^2$: Emily Jones. Ada is the remaining daughter, with 202 feet, and her surname is Smith.

Answer Ada Smith

Saturday Marketing

Everything is in whole shillings, so what is left is a whole number of shilling coins, and sharing it equally among eight needs a multiple of eight. The wives spend 10s. If the husband’s multiplier for the wife who spent $w$ shillings is $k$, the husbands spend the sum of $k \times w$, and pairing the numbers 1 to 4 with 1 to 4 makes this sum between $1 \cdot 4 + 2 \cdot 3 + 3 \cdot 2 + 4 \cdot 1 = 20$ and $1 + 4 + 9 + 16 = 30$. So between 30s. and 40s. is spent and between 0 and 10s. is left. There must be something to share, so 8s. is left, 32s. spent, and the husbands spent 22s.

Which pairing gives 22? Checking the 24 pairings, only one: Ann with 3 (Jones), Mary with 4 (Robinson), Jane with 1 (Smith) and Kate with 2 (Brown), since $3 + 8 + 3 + 8 = 22$. Each person takes home a shilling.

Answer Ann Jones, Mary Robinson, Jane Smith, Kate Brown