3
“A Goodly Struggle”: Problems, Puzzles,
and Challenges
“BECAUSE IT’S THERE” IS A GOOD REASON FOR CLIMBING A MOUNTAIN—SO they say—and it seems a good reason for doing a mathematical problem, too. Mathematicians involved in new research had set each other challenges and gone head to head in competitions since at least the sixteenth century, but competitive puzzling at an amateur level boomed early in the eighteenth century with The Ladies’ Diary, the source of our first extract.
The Diary, like The Athenian Mercury which we will meet in Chapters 5 and 10, was one of the “new media” of its day, offering readers a chance to become writers, so to speak: to see their mathematics in print, and presumably to become still more enthusiastic buyers of the publication as a result. It was, when it worked, a genuinely interactive mode of publication that seems to have caught the imagination of its middle-class British target audience, and the Diary ran for well over a century.
Part of its success was surely its strategy of aiming mathematics at a new audience—women—and the genre of mathematical puzzle writing continued to be marked by a desire to seek out new audiences and to draw new groups of people into mathematical activity. This restless search for new ways to bring mathematics to interested people—and new people to bring it to—is on show in this chapter. Young girls and schoolboys were among the audiences of our later extracts, while in the twentieth century the competitive puzzle for schoolchildren became institutionalized in mathematical olympiads. All of this made for tremendous variation in the difficulty of the problems that were published. The simple conundrums of The Girl’s Own Book (1835) and of Hirschberg’s charming Can You Solve It? (1926) bear witness to the fact that, in mathematics more than in most subjects, one person’s triviality is another’s “goodly struggle.”
The Ladies’ Diary
A long-running publishing phenomenon of eighteenth-century Britain was The Ladies’ Diary, an almanac aimed at women which contained, among other things, a section of mathematical puzzles. The puzzles became very popular, attracting solutions from all over England and beyond (some of them from women). The Diary ran from 1704 to 1841, generating a number of imitators (the inevitable Gentleman’s Diary, for one) and spin-off publications (The Diarian Miscellany). Here we present selections from near the end of the eighteenth century, when the editor was the well-known mathematician and mathematical educator Charles Hutton, whose textbooks were the source for some of the questions asked.
Charles Hutton (1737–1823), The Ladies’ Diary: or Woman’s Almanack 
Containing New Improvements in Arts and Sciences, And many Entertaining Particulars: Designed for the Use and Diversion of the Fair-Sex (London, 1798), pp. 46–48; (London, 1799), pp. 33–34, 43–44.
New Questions
I. by Mr. John Hawkes, of Finedon
What two numbers are those whose product, difference of their squares, and the ratio or quotient of their cubes, are all equal to each other?
II. by Miss Sarah Cowen
Asking lately the contents of a particular field, in form of a trapezium, I was answered that the contents were forgotten, but the dimensions in chains° were as in the annexed figure (3.1); but as I cannot from hence compute the contents, shall be obliged to my friend, Lady Di
, to do it for me.
III. by Mr W
m Newby, Barningham
A gentleman has a field in form of a trapezoid, the area of which is two acres, three roods;° the longest side, or base, AB, which is one of the two parallel sides, is 1432 links; also the angle at A is 14° 17′, and that at B 54° 18′. It is required to find what the four sides of the field will cost fencing, at 6 pence a rod.
Figure 3.1. Sarah Cowen’s field, in the form of a trapezium.
V. by Mr Henry Armstrong, Bewcastle
There is a vessel in the form of a frustum of a cone,° standing on its lesser base, whose 
is 8.67 feet, 
depth 21 inches, its greater base diameter to that of the lesser, as 7 to 5. Into 
a globe had accidentally been put, whose 
was 2
times the measure of its surface. Required, the lineal diameter of the above vessel and globe, and how many gallons of wine would be required just to cover the latter within the former.
VI. by Mr James Sparrow, of Norwich
In a plane triangle, given the angle at the vertex 60°, the length of the line bisecting it and dividing the base into two parts 
5 to 4, equal to 16, to find the sides and area.
VII. by Mr Wm Burden, of Acaster Malbis
How long will the sun be in rising out of the horizon, the 12th of October, 1798, to those inhabitants who have the duration of the twilight at that time the shortest possible?
VIII. by Mr T. Molineux, Macclesfield, Cheshire
Required, a general theorem for determining, by means of a rain gauge, the height of water which falls upon the ground; the weight of a cubic foot of water, the area of the aperture of the rain-gauge, and the weight of the water caught in the vessel being given.
X. by James Glenie, Esq.
On a given 
line as a base,° to constitute a triangle, such that the ratio of the square on the other two sides, to the ratio of the sum of their cubes to the difference of their cubes, shall be equal to the ratio of 171 to 140, whilst the area of the triangle has to the square of the base a given ratio, suppose that of 1 to 2.
XI. by Mr Tho
Coulthend, of Frosterly
Being in the head of a dale, about seven miles from Kendal, remarkable for the height of the hills on each side, I observed that the road on which I was going intersected at right angles a line joining the middle of the bases of two amazing high, rocky hills; the angle of elevation of the one was 50°, and of the other 60°, taken at the point of intersection. I then proceeded forward in the same direction, up a gentle declivity, rising at an angle of 10°, for the space of 200 yards, and then the angle of altitude of the first was 48° 10′, and that of the other 58°. I desire to know the perpendicular height of the hills from the bottom of the dale, the distance between their summits, and how far I was from the top of each at both stations.
XII. by Mr John Rutherford, Schoolmaster, Lanchester
On Lammas day,° in latitude 54° 40′ north, I observed a tree, which I knew to be 20 yards in height, to cast its shadow along the declivity of a hill. Now, admitting the declivity to be a plane of an indefinite length, and to incline regularly from the east towards the west at an angle of 20°, also the tree to stand perpendicular to the horizon, and exactly at the top of the said declivity, I desire to know how long the shadow was, supposing the time to be an hour before noon?
XIII. by Mr John Ryley, of Leeds
I should like to see a scientific solution to question 69, page 164, of Dr Hutton’s Conic Sections and Select Exercises, which is this: “If a glass tube, 36 inches long, closed at top, be sunk perpendicularly into water, till its lower or open end be 30 inches below the surface of the water, how high will the water rise within the tube, the quicksilver in the common barometer at the same time standing at 29 inches?” The answer being only there given, but not the solution.
XIV. by Mr Thomas Milner, Lartington Free School
I have seen a sheep leap from a bridge very high, into water, and swim out. Now, if a globe, whose weight is 112 pounds, and one foot in diameter, fall from an eminence ten yards high, how deep must the water be, just to destroy all the globe’s velocity, supposing the density of air, water, and the globe to be 
respectively.
XV. or Prize Question, by Amicus
To construct a triangle, having two of the sides given, and such that, a perpendicular to the third side being drawn from its opposite angle, that perpendicular shall be the height of a prism whose base is the triangle, and 
.
I. Question, answered by Mr Wm Davis, Schoolmaster, of Crowan
Put x = the greater number, and y = the less. Then xy = x2 − y2 and xy = x3 ÷ y3, or x2 = y4, or x = y2; then by substitution, etc., we have y2 − y = 1. By completing the square, etc., we find y = + 
= 1.61803. Consequently x = 2.61803.
The Same, by Mr John Eadon, Junior, Sheffield
Let x = the greater, and y = the less number. Then, by the question, xy = x2 − y2, and xy = x3 ÷ y3; therefore y4x = x3, and y4 = x2, and y2 = x. Put y2 for x in the first equation, and we get y3 = y4 − y2, or y2 − y = 1. Hence y = + 
, and then x = y2 = 1 + , which are the two numbers sought.
For proof: xy = 2 + , and x2 − y2 = 2 + , and x3 ÷ y3 = 2 + .
The same, by Mr John Ramsay, London
Suppose x the greater number, and y the less. 
xy = x2 − y2 = x3 ÷ y3. By equating the two first quantities is got x = y × (1±), and by equating the first and third x = y2; hence y = ± = 1.618, etc., or −0.618, etc., and x = y2 = 
± = 2.618, etc., or 0.382, etc.
II Question, answered by Mr Tho Coultherd, Frosterly
Let CE be drawn, and produce CA to F, letting fall the perpendicular EF. Then AF = (BE2 − BA2 − AE2) ÷ 2AB = 2.7727, and 
= 
= 17.9125; also 
= CE = 24.5394, and AC × FE = 134.3437, the area of the triangle ACE. Again, in the triangle CDE, having the three sides given, by a like process is easily found the perpendicular GE = 15.469589, and thence CD × EG = 92.8175, the area. Consequently the sum of these two areas gives 227.162 square chains, or 22 Acres, 2 Roods, 34 Perches for the area required.
The same, by Mr J. Gee, Elswick, near Newcastle
In the triangle ABE, the three sides are given, 
find the angle A = 95° 39′. Hence, if the diagonal CE be drawn, we shall have two sides and the included angle of the triangle ACE, find the said diagonal = 24.539 chains. Then in each of the triangles ACE, DCE, the three sides are known, whence the sum of their areas is easily found = 22 Acres, 2 Roods, 35 Perches = the content required.
The same, by Mr R
d Oliver, Assistant to the Rev. Mr Cursham, Sutton, near Mansfield
In the triangle ABE, all the three sides are given, 
find the angle ABE = 54° 30′, the supplement of which is 125° 30′ = the angle EBC. If CE be drawn, we then have the sides EB, BC, and the included angle, whence CE is easily found = 24.5. Hence we have the sides of all the triangles AEB, EBC, ECD, from which their areas may be found, the sum of which comes out 22 Acres, 3 Roods nearly.
XIII. answered by Mr J. Gough, Kendal
The density of the air is 
its spring,° which in the open tube is equal to a column of mercury of the same base and 29 inches high; but in the immersed tube this weight is increased by a column of water 3 − x inches high, x denoting the height of the water in the tube. But 13600 : 1000 = 20 − x : 2.205 − 0.0735x = a column of mercury of the same weight, and the whole pressure = 29.5 + 2.205 − 0.0735x = 31.705 − 0.0735x. But when the matter is given, the magnitude is inversely the density, or pressure in the present case. Therefore 31.705 − 0.0753x : 29.5 = 36 : 36 − x; hence x2 − 467.36x = −1080, and x = 2.33 inches, as required.
The same, by Miss Maria Middleton, Eden, near Durham
Let l = 36 inches the length of the tube, b = 30 inches the part immersed, x = the height of water in the tube, and f = 413 inches, the height of a column of water equal to the pressure of the atmosphere, when the quicksilver stands at 29 inches. Then, since the spaces occupied by the same quantity of air are reciprocally as the compressing forces, it will be as l − x : l = f : 
= force of the air in l − x; hence + x = b + f, and x = 2.2654115 inches.
Notes
chains: a unit of length equal to 66 feet or 100 links.
roods: a unit of area equal to a square perch, 272
square feet.
frustum of a cone: a cone whose point has been cut off, leaving a figure with two parallel circular faces of different sizes, and one curved face.
On a given straight line . . . : this obscure question asks the reader to construct a triangle with sides of lengths a, b, c, where a is given and b and c are to be found, subject to the constraint that the triangle’s area should equal 2a2 and
Approached using algebra the problem is intractable, but it is possible to solve it quite elegantly using a geometric approach.
Lammas day: 1 August.
spring: pressure.
The Girl’s Own Book
First published in Boston in 1834 by Lydia Marie Child—also known as a writer of fiction and an activist for women’s rights and the abolition of slavery—The Girl’s Own Book describes a whole range of pastimes and activities, moved by the thought that “every mind should seek to improve itself to the utmost.” It went through many editions on both sides of the Atlantic, and a recent reprinting has proved a minor hit, too. The arithmetical puzzles it contains are pitched at the very lowest level of mathematical knowledge, and there is no denying they seem disappointingly trivial: compare those in The Boy’s Own Magazine, which follow.
Lydia Marie Child (1802–1880), The Girl’s Own Book (London, 1835), pp. 169–171, 179.
Arithmetical Puzzles
1. How can you take away one from nineteen, and have twenty remain?
2. What is the difference between twice twenty-five and twice five and twenty?
3. If you can buy a herring and a half for three halfpence, how many herrings can you buy for eleven pence?
4. A and B made a bet concerning which could eat the most eggs. A ate ninety-nine; B ate one hundred, and won. How many more did B eat than A?
5. Place four nines together so as to make exactly one hundred. In the same way four may be made from three threes, three may be made from three twos, etc.
6. If a person hold in his hands a piece of silver, and a piece of gold, you can ascertain in which hand is the silver, and in which the gold, by the following simple process. The gold must be named some even number, say eight; the silver must be named an odd number, say three. Then tell the person to multiply the number in his right hand by an even number, and that in his left hand by an odd number, and make known the amount of the two added together. If the whole sum be odd, the gold is in his right hand; if it be even, the silver is in the right hand. For the sake of concealing the artifice better, you need not know the amount of the product, but simply ask if it can be halved without a remainder; if it can, the sum is, of course, an even one.
7. The figure 9 has one remarkable characteristic, which belongs to no other number. Multiply it by any figure you will, the product added together will still be nine. Thus, twice 9 are 18; 8 and 1 are 9. Three times 9 are 27; 7 and 2 are 9. Eight times 9 are 72; 7 and 2 are nine, etc.
If you multiply it by any figures larger than 12, the result will differ only by there being a plurality of nines.
8. When first the marriage knot was tied
Between my wife and me,
My age exceeded hers as much
As three times three does three.
But when ten years and half ten years
We man and wife had been,
Her age approached as near to mine.
As eight is to sixteen.
Question. How old were they when they married?
9. A room with four corners had a cat in each corner: three cats before each cat, and a cat on every cat’s tail. How many cats were there?
10. If you cut thirty yards of cloth into one yard pieces, and cut one yard every day, how long will it take you?
11. How can you show that seven is the half of twelve?
Keys to Puzzles, Conundrums, etc.
1. XIX. XX.
2. Twice twenty-five is fifty; twice five, and twenty, is thirty.
3. If a herring and a half are three halfpence, of course each herring is a penny a piece.
4. Those who hear you will think you say one.
8. The bride was 15, and the bridegroom 45.
9. Four cats.
10. Twenty-nine days.
11. XII.
12. Draw a line through the middle of XII, the upper half will be VII.
The Boy’s Own Magazine
Beginning in 1855, The Boy’s Own Magazine, “An Illustrated Journal of Fact, Fiction, History and Adventure,” contained a mix of stories, recreations, and nonfiction somewhat in the vein of the later and better known Boy’s Own Paper (or for that matter The Dangerous Book for Boys of 2006). The mathematical questions presented here illustrate not just the sturdy manliness of the publication (rockets and all that) but also the (albeit modest) level of mathematical sophistication expected of its readers. Solutions were not given.
“F.L.J., B.A.,” “Mathematical Questions,” The Boy’s Own Magazine, issue 11 (London, 1855), p. 352.
1. C spent two 11ths of his life as a child at home, three years more than twice that time as a boy at school, three 11ths as a student at home or at college, and four 55ths in a public appointment. How long was he a schoolboy?
2. At 25 years of age, C was 4lbs. 8oz. more than 18 times the average weight of a newly-born infant, but 9lbs. 8oz. added made him just 20 times as heavy. What was his weight?
3. T’s expenditure fell short of his income for the first half of the present year by a fifth, but having since increased, by change of residence, he finds that he will spend .
more than his income for the second half, and have £9, 5 shillings and 8 pence left at the year’s end. Find his yearly income.
4. A walked from Bedford to Cardington in three-quarters of an hour, and, at a speed two-thirds of a mile per hour less, took as long to reach the top of Hillfoot Hill, after which he reached Warden in an hour, at a rate one-sixth of a mile 
slower still. Had he walked uniformly as at first, he would have reached Warden at 24 minutes before 1, but at his last rate not till 14 minutes after. What is the distance from Bedford to Cardington and Warden?
5. B noticed that a skyrocket at its highest point of ascent formed an angle 45° of elevation with the horizon. He moved 80 yards nearer to the rocket-post, when another rocket, which mounted as high as the former, made an angle of elevation of 60°. How high did the rockets ascend?
6. L saw a boy fishing on the opposite side of a river in a direction forming an angle of 60° with the bank; the boy then moved 42 yards farther down the river, when the angle of his direction with either bank had decreased to 30°. Find the breadth of the river.
7. Standing 5.05 feet from a wall directly facing the sun when the altitude of the latter was 30°, I noticed that my shadow on the wall was just half of my real height. What is my height?
8. From Sandown Castle, Kent (July, ’54) F noticed the Hannibal, man-of-war, and the Prince, transport, anchored in the Downs,° in directions forming angles with the direction of Walmer Castle, the Prince of 60°, the Hannibal of 50°; from Walmer, however, 16 furlongs from Sandown Castle, the direction of the latter made with that of the Hannibal an angle of 70°, with that of the Prince of 40°. How far from the Prince was the Hannibal?
Note
The Downs: an anchorage on the southeast coast of England. Man-of-war and transport are in this context types of ships.
“The Analyst”
Beginning in 1874 and continuing as Annals of Mathematics from 1884 onward, The Analyst appeared monthly, published in Des Moines, Iowa, and was intended as “a suitable medium of communication between a large class of investigators and students in science, comprising the various grades from the students in our high schools and colleges to the college professor.” It carried a range of mathematical articles, both pure and applied, and a regular series of mathematical problems of varying difficulty: on the whole they seem harder than those in The Ladies’ Diary and possibly easier than the Mathematical Challenges in the extract after the next. Those given here appeared in the very first issue.
The Analyst: A monthly journal of pure and applied mathematics, vol. 1, no. 1 (Des Moines, 1874), p. 15.
1. Find the value of x and y in the following equations:
—Communicated by U. Jesse Knisely, President and Professor of Mathematics in Luther College, Newcomerstown, Ohio.
2. Let a regular polygon of 14 sides be described, each of whose equal sides shall be one. Then will the radius of its circumscribing circle, which put = t, be more than two and less than three. Put r = 2 + x; then is x a positive quantity less than one. Let another regular polygon of half the number of sides (7) be inscribed in a circle whose radius is one, and determine one of its equal sides in functions of x expressed in its simplest form.
3. If a line make an angle of 40° with a fixed plane, and a plane embracing this line be perpendicular to the fixed plane, how many degrees from its first position must the plane embracing the line revolve in order that it may make an angle of 45° with the fixed plane?
—Communicated by Prof. A. Schuyler, Berea, Ohio.
4. A cask containing a gallons of wine stands on another containing a gallons of water; they are connected by a pipe through which, when open, the wine can escape into the lower cask at the rate of c gallons per minute, and through a pipe in the lower cask the mixture can escape at the same rate; also, water can be let in through a pipe on the top of the upper cask at a like rate. If all the pipes be opened at the same instant, how much wine will be in the lower cask at the end of t minutes, supposing the fluids to mingle perfectly?
—Communicated by Artemas Martin, Mathematical Editor of Schoolday Magazine, Erie, Pennsylvania.
Can You Solve It?
This collection of puzzles, which ranges across mathematics, natural science, and word games of various kinds, promised the reader “a goodly struggle” and that “feeling of intense pleasure which is aroused by the accomplishment of any difficult intellectual task,” although many of the puzzles are in fact not all that difficult. The author, Arthur Hirschberg, is otherwise completely unknown.
Arthur Hirschberg (dates unknown), Can You Solve It? A Book of Puzzles and Problems (London, 1926), pp. 167–169, 277–278.
Puzzles
1100
Two clerks, A and B, were hired in an office at the same time. A’s salary was £1,000 a year with an increase of £200 a year. B’s salary was also to commence at £1,000 a year, but his increase was £50 every half-year. In each case payments were made half-yearly. Which clerk received the better offer, and why?
1101
If five cats catch five rats in five minutes, how many cats will it require to catch one hundred rats in one hundred minutes?
1102
A swimming pool which contains 1000 cubic yards of water can be filled by either of two inlet pipes in three and four hours respectively, and can be emptied by either of two outlet pipes in five and six hours respectively. How long does it take to fill the tank if both inlet pipes and the two outlet pipes are open?
1103
A hymn-board in a church has four grooved rows on which the numbers of the four hymns chosen for the service may be placed. The hymn book contains 700 hymns. The numbers of the hymns chosen are displayed on plates, each carrying one digit. What is the smallest number of plates which can be carried in stock so that the numbers of any four hymns can be displayed? What is the smallest number, if an inverted six can be used for a nine?
1104
A man decides to save a penny on the first day of the month, twopence on the second day, fourpence on the third day, eightpence on the fourth day, and so on—doubling the amount to be saved each day. How much does he save in a month of 31 days?
1105
Two clubs were formed. The first had 15 members and it was agreed upon to dine four at a time until all possible combinations have been formed.
The second club consisted of 21 members and dined in groups of three at a time, until all possible combinations had been formed.
Which group served more dinners, and how many more?
1106
In a potato race, each contestant started from a basket, ran 3 yards to the first potato, returned and deposited the potato in the basket. He repeated this with each of the other potatoes, but each potato was 3 yards farther from the basket. If each contestant picked up 24 potatoes in this manner, what was the total distance he ran?
Answers
1100
B is better off. His income is always £50 a year more than A’s. In the first year A received £1,000, and B receives £500 for the first half-year, and £550 for the second half-year, or a total of £1,050. In a similar manner it can be shown that there will always be a difference of £50 in B’s favour.
1101
The same number—that is, 5 cats.
1102
4
hours.
1103
Nine plates, each carrying digits from 1 to 6, inclusive, 
eight plates, each carrying the digits 0, 7, 8, 9, are needed. This makes a total of 86 plates.
If inverted sixes are used for nines, then twelve sixes must be provided, but no nines are needed. This makes a total of 81 plates.
1104
£2,147,483,647.
1105
The first group had to serve 1,365 dinners, and the second group 1,330 dinners; or, 35 more dinners were served by the first group.
1106
1,800 yards.
Mathematical Challenges
We have seen competitive mathematical puzzles from the eighteenth century in The Ladies’ Diary, and from the nineteenth century in The Analyst. In the twentieth century more formal competitions began to take place: mathematical olympiads have run in various countries, and, since 1959, internationally. All such mathematical competitions rely on a plentiful supply of mathematical problems of a suitable kind: typically they need to be hard yet require little factual knowledge (all the same, the explanation of what a prime number is, in the solution to question 2 in this extract, comes as something of a surprise). Here we present selections from one collection of such material, the Mathematical Challenges of the Scottish Mathematical Council. We present two sets of problems and one of solutions, leaving the solutions of the second set as an exercise for the reader.
Mathematical Challenges (Edinburgh, 1989), pp. 30–31, 37, 128–30.
1. Ten married couples sit down for dinner at a circular table. Each man sits opposite his wife. After the meal, a wit in the company points out that if all the ladies get up and move the same number of places to the right then each one will end up sitting on the knee of some gentleman other than her husband. How were the couples arranged around the table?
2. A printer makes mistakes in setting up type and prints 5423 as 5423.
He makes similar mistakes with another product but this time it does not matter. What four-digit number does he print? (Solutions involving the use of a computer will not be accepted.)
3. Show that the equation
x4 + 131 = 3y4
has no solution in integers x and y.
4. In a certain garden there are several beds each planted with flowers. For any two different varieties of plants in the garden, there is exactly one bed containing both. If any bed is considered, there is exactly one other bed in which none of the same varieties of plants is growing.
Show that
(i) every bed has at least two varieties,
(ii) there are at least four varieties in the garden,
(iii) there are at least six flower beds, and
(iv) no bed has more than two varieties.
1. A wine merchant had twelve containers of the following capacities: 13, 15, 16, 18, 19, 21, 31, 40, 50, 51, 81 and 91 litres. One container contained beer for himself, and the rest, which contained wine, were sold intact. All the wine was sold to two customers. The second customer bought twice the amount of wine bought by the first customer.
How much wine did each customer buy?
2. Prove that the only solution in integers of the equation
x2 − 2xy + 2y2 − 4y3 = 0
is x = 0, y = 0.
3. In the diagram (Figure 3.2), A, B and C are the centres of three circles each of which touches the other two.
Prove that the perimeter of the triangle ABC is equal to the diameter of the circle with centre C.
4. You are given a red, a white and a blue marble, all of the same weight, and also a red, a white and a blue marble again equal in weight to each other but heavier than the first three. The heavier marbles look identical to the lighter ones. Describe a procedure to find the three heavier marbles in two weighings on a balance scale.
Solutions
1. Number the places 0, 1, 2, . . . , 19, starting with 0 for the seat occupied by one of the men and then moving counter-clockwise around the table.
Figure 3.2. Three circles each of which touches the other two.
Then label each seat M or W depending on whether the occupant at the beginning is a man or a woman.
Suppose that all the ladies were to move n places to the right (that is, counter-clockwise around the table). Then 0 < n < 20. After the move each seat labelled M would have two occupants and each seat labelled W would be empty. Thus n places to the right of a seat labelled W is one labelled M. Furthermore a seat n places to the right of one labelled M could not be occupied by a lady after the move and therefore would be empty, hence it is labelled W. Therefore since 0 is an M, it follows that n is a W, the seat with number equal to the remainder on dividing 2n by 20 is an M, that with number equal to the remainder on dividing 3n by 20 is a W and so on.
Suppose now that n is odd. Then 10n leaves remainder 10 on division by 20. Hence seat 10 is an M. But this cannot be true, since seat 10 is opposite seat 0 and so is occupied by the wife of the man in seat 0. Hence n is even. Next suppose that n is a multiple of 4. Then 5n leaves remainder 0 on division by 20. By the statement at the end of the previous paragraph this seat is a W; but we know that it is an M. Hence n is not a multiple of 4.
This shows that the possible values of n are 2, 6, 14 and 18. No matter which of these we take we find that the arrangement must follow the pattern MMWWMMWW. . . .
2. A prime is a positive integer which cannot be expressed as the product of two smaller positive integers. For example, 2, 3, 5 and 7 are primes but 4, 6, 8 and 9 are not. Any positive integer greater than 1 is either a prime or can be expressed as a product of primes (with possible repetitions). We shall use this as a basis for an organised search for solutions of the equation abcd = abcd.
Since a, c are single digit numbers, the primes which feature when abcd is expressed as a product of primes are all less than 10 and so the possibilities are 2, 3, 5 and 7. Moreover the only number less than 10 involving two different primes in a factorization is 6 = 2 × 3 and so abcd can be expressed in one of the following forms:
Here m and n are positive integers.
We check these possibilities in turn, listing all the four digit numbers of the required form. First we have the powers of a prime.
(i) 1024, 2048, 4096, 8192.
(ii) 2187, 6561.
(iii) 3125.
(iv) 2401.
None of these has the required property. Next we consider case (v), listing first three times a power of 2, then 32 times a power of 2 and so on.
(v) 1536, 3072, 6144; 1152, 2304, 4608, 9216; 1728, 3456, 6912; 1296, 2592, 5184, 1944, 3888, 7776; 1458, 2916, 5832; 4374, 8748. Of these, just one number has the required property: 2592 = 2592.
(vi) In this case, each number is divisible by 2 and by 5 and hence is divisible by 10. Therefore if abcd is to equal abcd, the digit d is 0; but then abcd = ab, which requires a to be divisible by both 2 and 5, which is not possible.
In the remaining cases, with the exception of (xi), we use a system similar to that followed in (v). Case (xi) is similar to case (vi).
(vii) 1792, 3584, 7168; 1568, 3136, 6272; 1372, 2744, 5488; 4802, 9604.
(viii) 1215, 3645; 2025, 6075; 1125, 3375; 1875, 5625; 9375.
(ix) 1701, 5103; 1323, 3969; 1029, 3087, 9261; 7203.
(x) 1715; 1225, 8575; 6125; 4375.
(xi) 2058; 1764; 1512; 9072.
In all these cases, no number listed has the required property. Hence the only number which has the property is 2592, which is equal to 2592.
3. The last digit of the square of an integer is either 0, 1, 4, 5, 6 or 9. A fourth power is the square of a square, and so its last digit is either 0, 1, 5 or 6. Hence the last digit of x4 + 131 is either 1, 2, 6 or 7 whilst the last digit of 3y4 is either 0, 3, 5 or 8. It follows that an integer cannot be both of the form 3y4 and of the form x4 + 131. Therefore the equation x4 + 131 = 3y4 has no integer solutions.
4. There are three rules:
(P) All beds contain flowers.
(Q) For any two varieties of flower there is exactly one bed containing both.
(R) If X is a flower bed there is exactly one bed which contains none of the varieties from bed X.
(i) Suppose there is a bed A which contains just the one variety, α. By rule (R) there is exactly one bed which does not contain α. Call this bed B. B contains flowers (rule (P)). Let β be one of the varieties in B. By rule (Q) there is a bed, C, containing both α and β. Clearly C is not either A or B. Now use rule (R) again to deduce the existence of a bed D containing neither α nor β. Bed D is not the same as bed B. We now know of two beds—B and D—neither of which contains any of the varieties from A. This is contrary to rule (R). Consequently A cannot contain just the one variety.
(ii) Label one of the beds A. In view of (i) A contains at least two varieties, say α, β. By rule (R) there is abed B containing neither α nor β. Because of (i) B must contain two varieties. Call them γ, δ. We now have four distinct varieties.
(iii) Continuing the argument of (ii) we have bed A which contains α, β but not γ, δ and bed B which contains γ, δ but not α, β. By rule (Q) there is a bed C containing α, γ. If C were to contain β or δ we should contradict the “exactly one” part of rule (Q) applied either to varieties α, β or to varieties γ, δ. So C contains α, γ but not β, δ.
Similarly we can deduce the existence of beds D, E, F such that
(a) D contains α, δ but not β, γ.
(b) E contains β, γ but not α, δ.
(c) F contains β, δ but not α, γ.
These six beds A–F are clearly all different, and so we have established (iii).
(iv) Let A be one of the beds and suppose it contains m varieties. By rule (R) there is a bed B containing no varieties from A. Suppose B contains n varieties. We count the number of beds.
It follows from rules (Q) and (R) that all beds other than A and B contain exactly one variety from A and exactly one variety from B. So there are mn beds other than A and B. So there are mn + 2 beds in total.
Let α be a variety from A. It appears once with each variety from B and once with no varieties from B (the latter in bed A). So α appears in n + 1 beds.
Let C be a bed other than A or B, and suppose C contains q varieties. By rule (R) there is a bed D containing no varieties from C. D does however contain a variety β from bed A. It follows from the previous paragraph that B appears in n + 1 beds. However if we repeat the argument of the previous paragraph but with D, C replacing A, B respectively then we see that β appears in q + 1 beds. It follows that q = n. This shows that all beds, with the possible exception of A, contain n varieties. Now repeat the argument of the second paragraph with C, D in place of B, A. This will show that there are n2 + 2 beds. So n2 + 2 = mn + 2. Therefore m = n so all beds have n varieties.
Now we count the number of varieties. There are n2 + 2 beds each containing n varieties. Each variety appears in n + 1 beds. So the number of varieties is
This has to be an integer.
n + 1 and n have no common factor. So if
is to be an integer n + 1 must divide n2 + 2.
n2 + 2 = (n + 1)2 − 2(n + 1) + 3. So if n + 1 divides n2 + 2 we must have n + 1 dividing 3. This will only happen if n = 2.