The Solutions

Puzzles GaloreA Collection of Mathematical

and Logic Puzzles

Collected by Ian R Humphreys

1. The Flight Around The World

A group of aeroplanes is based on a small island. The tank of each plane holds just enough fuel to take it half-way around the world. Any desired amount of fuel may be transferred from the tank of one plane to the tank of another whilst the planes are in flight. The only source of fuel is on the island, and for the purposes of this problem it is assumed that there is no time lost in refuelling either in the air or on the ground.

What is the smallest number of planes that will ensure the flight, non-stop, of one plane around the world on a great circle assuming that each of the planes has the same constant ground speed and rate of fuel consumption, and that all planes return safely to their island base?

2. Who Is The Engineer?Smith, Jones and Robinson are engineer, brakeman and fireman on a train but

not necessarily in that order. Also riding the train are three passengers with the same three surnames, to be identified in the following premises with a “Mr.” before their names.

• Mr. Robinson lives in Los Angeles.• The brakeman lives in Omaha.• Mr. Jones long ago forgot all the algebra he ever learnt at High School.• The passenger with the same name as the brakeman lives in Chicago.• The brakeman and one of the passengers, a distinguished mathematical physicist,

attend the same church.• Smith beat the fireman at billiards.

The Solutions

What is the name of the engineer?

3. The Coloured HatsThree men, Alan, Brian and Colin are blindfolded and are told

that either a red hat or a green hat will be placed on each of their heads. After this is done, the blindfolds are removed and the men are asked to do the following:

• To raise a hand if they see a red hat.• To leave the room as soon as they are sure of the colour of

their own hat.

All the hats happen to be red, so all the men raise a hand. Several minutes elapse until Colin, who is more astute than the others, leaves the room. How did he deduce the colour of his hat?

4. The Coloured Hats AgainAt a party, four people played a game. Three of them sat one

behind the other, so that Arthur saw Bill and Colin, and Bill saw only Colin who sat in front and saw nobody.

Dave had five hats which were showed to his three friends. Three of the hats were blue and two were red. Now Dave placed a hat on the head of each of his three friends, putting aside the remaining two hats.

He now asked Arthur what colour his hat was. Arthur said he couldn’t tell. Bill, asked the colour of his own hat didn’t know for certain either. Colin, however, although he couldn’t see any hat at all, gave the correct answer when asked what the colour of his hat was! What was the colour of Colin’s hat and how did he deduce it?

5. The Professor On The EscalatorWhen Professor Stanislaw Slapernarski, the Polish mathematician walked

very slowly down the down-moving escalator, he reached the bottom after taking 50 steps. As an experiment, he then ran up the same escalator one step at a time, reaching the top after taking 125 steps.

Assuming that the Professor went up five times as fast as he went down (i.e. he took five steps to every one before), and that he made each trip at a separate constant speed, how many steps would be visible if the escalator stopped running?

6. The Encounter On The BridgeOn a foggy night, a passenger car and a truck meet on a bridge which is so

narrow that the two vehicles can neither pass nor turn. The car has proceeded twice as far onto the bridge as the truck, but the truck has required twice as much time as the car to reach this point. Each vehicle has only half its forward speed when in

The Solutions

reverse. Which of the two vehicles should back up to allow both of them to cross the bridge in the minimum time?

7. The Early CommuterA commuter is in the habit of arriving at his suburban station each evening at

exactly five o’clock. His wife always meets the train and drives him home. One day he takes an earlier train and arrives at his station at four o’clock. The weather is pleasant so instead of telephoning home, he starts walking along the route always taken by his wife. They meet somewhere along the way.

He gets into the car and they drive home arriving ten minutes earlier than they usually do. Assuming that the wife always drives with a constant speed, and on this particular occasion she left just in time to meet the five o’clock train, can you determine how long the husband walked before he was picked up?

8. The Two FerryboatsTwo ferryboats start at the same instant from opposite sides of the river

travelling across the water en route at right angles to the shores. Each travels at a constant speed but one is faster than the other. They pass at a point 720 metres from the nearest shore. Both boats remain at their slips for ten minutes before starting back. On the return trip, they pass 400 metres from the other shore. How wide is the river?

9. The Weights And The ScalesA shopkeeper has a beam balance and requires

to weigh any article of a whole number of grams from 1 to 40 inclusive. What is the least number of weights necessary to do this?

10. The Efficient ElectricianAn electrician is faced with the following annoying dilemma. In the basement

of a three-storey house he finds, bunched together in a hole in the wall, the exposed ends of eleven wires, but he has no way of knowing which end above belongs to which end below. His problem is to match ends.

To accomplish his task, he may do one of two things:

• Short circuit the wires at either end by twisting the ends together in any manner he chooses.

The Solutions

• Test for continuity of a closed circuit by means of a “continuity tester”, consisting of a battery and a bell. The bell rings when the instrument is connected to two ends of a continuous unbroken circuit.

Not wishing to exhaust himself with needless stair climbing and having a passionate interest in operations research, the electrician sat down on the top floor with a paper and pencil and soon devised the most efficient method of labelling the wires. What was his method?

11. How Old Are You?Mr. Jones said to Mr. Smith: “When I was as old as you are now, you were

twice as old as you were when I was three times as old as you.” How old is Mr. Jones if Mr. Smith is 28?

12. Ann And MaryThe present sum of the ages of the two girls is 44. Mary is twice as old as Ann

was when Mary was half as old as Ann will be when she is three times as old as Mary was when she was three times as old as Ann. How old are Ann and Mary?

13. The Tribulations Of A Father Of FiveA father rowing a boat, has to transport his five sons across a river in a

minimum number of one-way crossings, such that finally all the children have had an identical number of crossings.

The boat will hold the father (who does all the rowing) and not more than two children. No pair of children of immediately neighbouring ages can be left together in the absence of their father or they will start to fight. But children more separated in age will content themselves with more peaceful occupations. What is more, the father, whilst handling the oars, is not able to control his children at the same time.

How does he get the five children safely across the river, in the minimum number of one-way crossings, so that each child has the same number of such trips?

14. Three Wives & Their Jealous Husbands

Three jealous husbands with their wives, having to cross a river, find a boat without a boatman. But the boat is so small that it can contain no more than two of them at a time. How can these six persons cross the river so that none of the women shall be left in the company of any of the men unless her husband is present? (Women to row when absolutely necessary to prevent a violation of the above conditions).

The Solutions

15. Four Wives & Their Jealous HusbandsFour couples have to cross a river with the aid of one small boat with the

capacity of only two people. Nowhere, either on land or in the boat, is any woman allowed to be left in the company of any of the men unless her husband is present. How can the transfer be executed if there is an island in the middle of the river where the boat may land and leave some of its cargo?

16. Getting The Nuggets AcrossDuring the Alaska gold-rush, three prospectors who had struck it rich had to

cross the Yukon. But they could find only a small row-boat with the capacity of either two men or one man and a bag of gold nuggets. Each of them owned a bag of the precious metal but the contents of each of the three bags of nuggets were not the same. Smith’s gold was worth $8000, Jones’ $5000 and Brown’s $3000.

None of them trusted the other two, and so after some argument they agreed that the passage should be arranged so that none of them, either on shore or in the boat should be in the presence of nuggets worth more than he owned. How did the men proceed to get themselves and their properties across the Yukon in the least number of crossings consistent with the restrictions?

17. Missionaries and CannibalsThree missionaries and three cannibals have to cross a river with the aid of a

two-seater row-boat. Since the man-eating savages are ravenously hungry, it is imperative that on neither shore they are ever in the majority. The priests’ difficulties are further aggravated by the fact that, of the three savages only one knows how to row, and of the three priests only one knows how to row. How does this uneasy company get across the river?

18. Road Safety Addition H A L T A T M A J O R R O A D A H E A D

The figure represents an addition sum where the digits have been replaced by letters. There are ten letters (A, D, E, H, J, L, M, O, R, T) each representing one of the digits 0 through 9 but not necessarily in that order. What values can you give the letters to make the sum total correctly?

19. The Lonesome Eight x x 8 x x

The Solutions

x x x ) x x x x x x x x x x x x x x x x x x x x x x x x x x

The figure represents a long division, Xs representing unknown digits. Fill in these unknown digits making the division correct. There is only one possible solution.

Note: This problem is a lot easier that it at first appears, yielding readily to a few elementary insights.

20. The Farmer And The Hundred DollarsA farmer went to market to buy some animals. Cows, sheep and pigs must all

be represented in his purchase. Cows sold at $5 each, sheep at $1 each and pigs at $1 for 20. The farmer has $100 to spend and he wants to buy 100 animals. How many of each did he buy?

21. The Knight’s Tour (1)Place a knight on a 6 by 6 chess board (i.e. an ordinary

chess board which is 2 squares short each way). The knight may be placed on any square. The problem is to visit each square once and once only by taking ordinary knight’s chess moves.

22. The Knight’s Tour (2)On an ordinary 8 by 8 chess board, a knight

stands in the top left-hand corner square. The problem is to visit each square once and once only using the knight’s chess move.

The Solutions

23. Alice In Wonderland“Oh, dear,” said Alice mournfully, “I wish I didn’t have to go into the Wabe,

for that is where the Jub-Jub lives.”“Nonsense!”, exclaimed the Mad Hatter. “It lives in a Tum-Tum tree.” And

addressing the March Hare, “Doesn’t it?”“Of course it does,” agreed the March Hare. “Everybody knows that, don’t

you agree?” (Here he prodded the Dormouse who was asleep on his shoulder).The Dormouse opened one sleepy eye and wearily muttered, “Yes, yes of

course,” and promptly went back to sleep again.“Talking of birds,” said the March Hare, “the Bandersnatch is exceptionally

shy this year.”“But I’m sure the Bandersnatch is an animal,” said Alice.“My, you are ignorant,” said the Mad Hatter, “of course it’s a bird.”The Dormouse, who appeared to be sleeping was consulted once more and he

wearily announced, “I agree with Alice.” Then he returned to snoring once again.“The Jabberwock,” announced the Mad Hatter, “is an interesting animal. It

has four wings and feeds on Borogoves.”“It has four wings,” agreed Alice, “but surely it feeds on Raths.”“Poppycock!” snorted the March Hare, disturbing the poor Dormouse, who

woke up suddenly and appeared to be at least half-listening for a change. “It has five wings and feeds on Toves.”

“Five wings and feeds on Raths,” corrected the Dormouse, who found staying awake much too much effort, especially when no-one seemed to know what he was talking about.

At this point in the proceedings, the Cheshire Cat appeared, sitting on the branch of a near-by Tum-Tum tree. He grinned and informed them that of the four zoological facts that each had volunteered, two each were true and two each were false. Whereupon he disappeared until only the grin remained, and this slowly faded, starting from the corners and ending in the middle. Which still leaves the following unanswered: Where does the Jub-Jub live? What is a Bandersnatch? And how many wings has a Jabberwock and what does it feed on?

24. The Spider And The Fly

In a rectangular room 30 metres by 12 metres by 12 metres, a spider is in the middle of one end wall and one metre from the ceiling.

A fly is in the middle of the opposite end wall and one metre from the floor. It is too paralysed with fear to move.

Fly

Spider12 m

12 m

30 m

The Solutions

What is the shortest distance that the spider must crawl across the surfaces of the room in order to reach the fly?

25. Another Escalator PuzzleThere are very long escalators in some New York subway stations. You don’t

have to climb them since the moving steps will do the job for you. However, two brothers have to get to a baseball game and are in a hurry, and so they run up the moving steps, adding their speed to that of the escalator.

The taller boy climbs three times as quickly as his little brother and whilst he runs up he counts 75 steps. The little one counts only 50 steps. How many steps has the visible part of this New York escalator?

26. The Ladder Against The Wall

AB is a wall at right angles to the ground BC. DFEB is a square box 4 feet on a side. AFC is a ladder 20 feet long whose extreme ends touch the wall and the ground, and whose length rests on the corner of the box. How far up the wall does the ladder reach? (i.e. What is the length of AB?)

27. A Coin Puzzle

A

FD

BE C

Figure 1

Figure 2

The Solutions

The problem is to change the arrangement of twenty-cent and ten-cent coins shown in Figure 1 to that shown in Figure 2, in the smallest number of moves.

The only move allowed consists of sliding a pair of touching coins, which must be one twenty-cent coin and one ten-cent coin, from one point on the imaginary base line (which of course is the same in both drawings) to another. The coins must remain in contact throughout the move and may not be rotated during the move to reverse the order of the coins moved. Gaps are allowed at the end of every move except the last.

28. The Dilemma Of A Racecourse OwnerA racecourse owner wished to install photoelectric timing

devices on his racecourse, such that any distance between 1000 metres and 31000 metres (in multiples of 1000 metres) could be timed. The photoelectric devices were extremely expensive and so he wanted to purchase the fewest possible.

It must be possible, by choosing any two sets of these devices, to time one of the desired distances between them. What is least number that he requires, and how would they be set out? Assume

for the purposes of this puzzle that the racecourse is a circular one.

29. The Scrambled Box Tops

Imagine that you have three boxes, one containing two black marbles, one containing two white marbles, and one containing one black marble and one white marble together. The boxes are labelled according to their contents BB, WW, and BW. But someone has switched the lids so that each box is incorrectly labelled. You are allowed to take one marble out of one box of your choosing, and by noting the colour of this marble, you can tell the contents of each of the three boxes. Which box must you draw the marble from and why?

30. The Seventeen SuitorsPrincess Carlotta, the daughter of King Frederik of a certain European

country, was eligible to be wed and her father was inundated with requests from no less than seventeen different suitors for her hand in marriage.

In order that his choice should be impartial, he devised the following scheme whereby his daughter’s future husband was to be chosen. The suitors were to form a circle around the princess and one of them was to be given a gold cross so that each suitor could now be identified by his position in the circle relative to the one who held the cross.

WW BW BB

The Solutions

When all was ready, the king instructed his daughter to start counting the suitors in a clockwise direction, counting the man who held the cross as Number One. When she reached three, that suitor must leave the circle and was no longer eligible. The next suitor round would then be Number One and she would continue to count clockwise, dismissing every third suitor until there was only one left. He would be her future husband.

Now it so happened that one of the suitors was the Princess’ secret lover and Princess Carlotta dearly wanted to marry him, but she dare not defy her father. She could see no immediate way of being certain that her lover would be the last one left. So she started counting round beginning with the man with the cross as directed by her father, and dismissing every third suitor. As she was counting, her eyes went ahead of her hand and whilst there was still a good number of men left in the ring, she could see that her lover would be the next one dismissed. So when the previous suitor had left the ring, she stopped and her father asked her what was wrong.

“I’m dizzy from counting round this way, Father, might I not count round the other way instead?” King Frederik agreed to this seemingly reasonable request, so instead of counting 1,2,3 in a clockwise direction and dismissing her lover as number three, she counted back the way she had already counted. She continued counting in an anticlockwise direction, dismissing every third suitor until there was only one man left. To her surprise it was Rudolf, the man she secretly loved! Can you tell his position in the ring at the start of the count?

31. The Counterfeit CoinsYou have ten stacks of coins,

each consisting of ten dollar coins. One entire stack is known to be counterfeit but you do not know which one. You do know the weight of a genuine dollar coin and you are told that a counterfeit coin weighs exactly one gram more than it should.You are supplied with a direct-

reading pointer balance with a single scale pan and a scale calibrated in grams. What is the

smallest number of weighings required to be certain of identifying the counterfeit pile?

NOTE: A weighing is regarded as a separate one if a different reading is obtained.

32. How Many Children?“I hear some youngsters playing in the garden,” said Jones, a graduate student

in Mathematics. “Are they all yours?”“Heavens, no!” exclaimed Professor Smith, the eminent number theorist, “My

children are playing with children from three other families in the neighbourhood,

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although my family happens to be the largest. The Browns have a smaller number of children, the Greens a still smaller number, and the Blacks the smallest of all.”

“How many children are there altogether?” asked Jones.“Let me put it this way,” said Smith. “There are fewer than 18 children and

the product of the numbers of children in each of the four families happens to be my house number which you saw when you arrived.”

Jones took a notebook and pencil from his pocket and started scribbling. A few moments later he looked up and said, “I need more information. Is there more than one child in the Black family?”

As soon as Smith had replied, Jones smiled and correctly stated the number of children in each family. Knowing the house number and whether the Blacks had more than one child, Jones found the problem trivial. It is a remarkable fact, however, that the number of children can be determined solely from the information given above!

33. A Question Of Cards“Examine your hand carefully”. That, they tell me, is the first precept of

Bridge. So I did. Not that it threw much light on the question of what to play. There was the usual collection of Jacks and Deuces - I always get Jacks and Deuces which serve to remind me that I am a permanent victim of knavish and diabolic devices. So I wasn’t surprised to find that I had more Deuces than Trumps and more Jacks than any other rank in my hand.

As for the rest, I had just got as far as calculating that I had more Diamonds than Spades, more Spades than Kings, more Kings than Clubs and more Clubs than Aces, when someone barked impatiently in my ear, “It’s you to lead, you know.”

Whereupon I tried hard to make up my mind whether to play an Ace or a Queen, but they looked so pretty that I thought it a pity to part with them so early in the game. So I played eenie-meenie-minie-mo with the Sevens that I held in my hand and defiantly hurled a card onto the table. Can anyone please tell me what my original hand consisted of? Each card fully identified by suit and rank. And what were Trumps?

34. The Dovetailed BlockA cubical wooden block is made out of two

pieces of wood, a dark piece and a light piece. They are dovetailed together as shown in the drawing. The other two sides look exactly similar to the two shown. It is completely solid. How is it constructed and fitted together?

The Solutions

35. The Twelve Matches

Assuming one matchstick to be of unit length, it is possible to place twelve matches on a plane surface in various ways to form polygons of integral areas. The drawing show two such polygons of the regular variety. A square with an area of NINE square units and a cross with an area of FIVE square units.

The problem is this. Use all twelve matches (the entire length of each match must be used and must lie on the periphery of the figure) to form, in similar fashion, a polygon with an area of exactly FOUR square units.

36. Conundra Zoo

On taking charge of Conundra Zoo, the new curator was disturbed to find that, by some mischance, the animals were all accommodated in the wrong cages. A notice above each cage indicated where each animal should be, but the Tiger was in the Lion’s cage; the Gorilla in the Rhino’s; the Rhino in the Bear’s; the Lion in the Gorilla’s; and the Bear in the Tiger’s.

He ordered the keeper to move the animals immediately into their proper cages, but, since they were all ferocious beasts, it was unthinkable that any two should be allowed to occupy the same cage at the same time. What is the least

LION RHINO

BEAR GORILLA TIGER

ENCLOSURE

The Solutions

number of moves the keeper would have to make in order to carry out the curator’s instructions and in what sequence would he have to move the beasts?

37. Little Pigley Farm, 1935

ACROSS1. Area in square yards of Dog’s Mead.5. Age of Farmer Dunk’s daughter, Martha.6. Difference in yards between the length and breath of Dog’s Mead.7. Number of roods in Dog’s Mead multiplied by 9 Down.8. Date (A.D.) when the Dunk family originally bought Little Pigley.10. Farmer Dunk’s age.11. The year of birth of Mary, Farmer Dunk’s youngest.14. The perimeter in yards of Dog’s Mead.15. The cube of Farmer Dunk’s walking speed in miles per hour.16. 15 Across minus 9 Down.

DOWN1. The value in shillings per acre of Dog’s Mead.2. The square of Mrs. Grooby’s (Farmer Dunk’s mother-in-law’s) age.3. Age of Mary (see 11 Across).4. Value of Dog’s Mead in pounds sterling.6. The age of Farmer Dunk’s first-born, Ted, who will be twice as old as Mary next year.7. The square of the number of yards in the breadth of Dog’s Mead.8. The number of minutes Farmer Dunk takes to walk one and a third times around Dog’s Mead.9. See 10 Down.10. 10 Across times 9 Down.12. One more than the sum of the digits in the row marked by the arrows.13. The length of tenure in years of Little Pigley by the Dunk family.

NOTE: 1 Rood = ¼ acre = 1210 sq. yds.

1 2 3 4

5 6

7

8 9

10 11 12 13

14

15 16

The Solutions

38. Potton vs BarfordThis year, Potton went to Barford for their annual cricket match, and were well satisfied by their total of 213 runs. Great disappointment was caused however, when their star batsman was dismissed after the third wicket partnership had put on only five runs, but Potton felt that their total was large enough to carry the day, and so it proved. Unfortunately the scorer, unduly elated by his side’s success, lost the scorebook on the way home. Can you help him by telling him the totals at which each of the Potton wickets fell?

ACROSS1. The ninth wicket fell at this total.3. The third wicket fell at this total.5. The number of extras in the innings.7. An odd number.8. The square of the total at which the third wicket fell.10. The second wicket fell at this total.11. The fourth wicket fell at this total.13. The square of 10 Across.16. Half of 17 Across.17. The first wicket fell at this total.19. The same as 10 Across.20. 3 Down reversed.

DOWN1. 5 Across times 2.2. The eighth wicket fell at this total.3. The seventh wicket fell at this total.4. 5 Across times 8.6. The same as 11 Across.7. The sixth wicket fell at this total.9. The square of the total at which the fifth wicket fell.11. This was the score after the fourth wicket partnership had put on 26 runs.12. One greater than 10 Across.13. The same as 10 Across.14. Eight times 10 Across.15. The same as 5 Across.18. Only three partnerships put on more than this number of runs.

1 2 3 4

5 6

8 9

10

11 12

13 14

15 16 17 18

19 20

7

The Solutions

39. Another Coin Puzzle

Figure 1 Figure 2

Six twenty-cent pieces are arranged on a flat surface as shown in Figure 1. The problem is to move them into the formation shown in Figure 2 in the smallest number of moves. Each move consists of sliding a twenty-cent piece, without disturbing any of the other coins, to a new position in which it touches two coins. The coins must remain flat on the surface at all times.

40. Calculari’s ConcertsDuring his last series of four concerts, Calculari, the

famous pianist performed works by Beethoven, Brahms, Mozart, Liszt and Chopin. In each concert he played four works, each of the four works being by a different composer. However no two concerts in the series included the same four composers in the same order.

He never played Mozart and Beethoven in the same concert, but if he omitted Beethoven, he always played a Mozart immediately followed by a Chopin. On the other

hand, if Mozart were omitted, Calculari always finished his concert with a work by Brahms. The pianist always followed his invariable rule of commencing a concert with Liszt if Brahms happened to be anywhere in the programme.

The first three of Calculari’s concerts finished with a work by the same composer. What composers did he play in his fourth and final concert, and in which order were they played?

41. Raising Poultry (1)A farmer’s wife is asked whether her poultry-raising business is flourishing.

“Oh, yes,” she replies, being as good at arithmetic as at raising chickens. “Last year I started with 25 chickens and ducks together. Today I have eight times as many. The chickens have multiplied three times as fast as the ducks.”

With how many chickens and how many ducks did this woman start?

1

2

3

4

5

6

The Solutions

42. Raising Poultry (2)At the start there were 17 hens, ducks and geese together. There were more

hens than ducks and more ducks than geese. Now, a year later, there are 248 birds.Hens multiply 7 times as fast as geese and ducks multiply 4 times as fast as

geese. How many hens, how many ducks, and how many geese were there to start with?

43. The Spinning WheelMr. Lucker made his fortune out of

cotton spinning and when he died, he left his six mills to his six nephews, Art, Brad, Charles, Dave, Eddy and Frank. To avoid any appearance of favouritism, he did not specify the mill to be allocated to each nephew, but declared that the decision should be made, as had his money, by spinning!

He provided a contrivance consisting of an hexagonal plate bearing the names of his six nephews and, pivoted to the centre of this

plate, a disk bearing the numbers 0 to 5.Mr. Lucker’s instructions were that the disk should be spun five times and the

scores obtained by each boy on each spin should be totalled up. The nephew gaining the highest total for the five spins was to have first choice of a mill, the nephew with the second highest total second choice and so on.

The diagram shows the result after the first spin. On the second spin Dave forged ahead with his total to take the lead but it was Art who after five spins obtained a total higher than anyone else’s, and therefore had first choice. What were the final totals for each of the six nephews?

44. Walking OutIf a girl takes three steps to a man’s two steps and they both start out on the

left foot, how many steps do they have to take before they are both stepping out on the right foot together?

45. Running To TimeFarmer Root’s property was square in

shape and covered an area of exactly two square miles. It was divided as shown in the diagram into four paddocks. The large shaded paddock was exactly one mile square.

Cottage

WellHouse

Stile

5 4

3 0

1 2

Frank

Eddy

Dave

Charles

Brad

Art

The Solutions

The farmhouse was situated at one corner of the smaller square paddock which paddock also had at its corners a well, an empty cottage and a stile.

Each morning, young Tom the farmer’s son kept himself in training by running from the house to the stile and back again. On one occasion he started off from the house, just as Farmer Root left the house to visit the cottage, walking by way of the well. On his way back from the stile Tom suddenly remembered the cottage key which his father had forgotten. On reaching the house therefore, instead of stopping, Tom continued to run following the same path that his father had taken. He caught up with his father, handed him the key and doubled back the way he had come.

Tom arrived back at the house at exactly the same time that his father reached the cottage. Assuming that Tom ran and his father walked at separate constant speeds all the time and that no time was lost, either in Tom’s turning at the stile or in his handing over the key, how far had Tom run altogether?

46. A Railway Shunting Problem

Two trains find themselves in the above predicament. On a single line track, a train of three carriages and a train of four carriages meet. There is a siding as shown which will only hold one engine or one carriage. How do the two trains pass each other in the least number of moves?

Main Track

C B A

D E F G

Siding

The Solutions

47. Another Shunting ProblemOnly the engine can pass under

the bridge and the problem is to reverse the positions of the truck containing the sheep and the truck containing the cattle, returning the engine to its original position on C.

48. A Third Shunting ProblemTrucks A and B and the Engine are in the positions shown. Siding I will take one truck at a time but not the engine. How can the Trucks A and B be interchanged without cutting the engine off from the main track?

49. A Swastika SquareDuring the Second World War, German prisoners-of-war in Stalag Mathematika XIII were given the chance to escape the rigours of normal camp life and the inevitable torture which followed even the slightest deviation from the very strict camp rules, and be transferred to a special section which allowed them to sit around all day and do mathematical puzzles. All they had to do was to fill the square on the left with the integers from 1 to 25 inclusive so that each row, each column and

each long diagonal added up to the same number. In other words they were to produce a 5 by 5 magic square. But this alone was not enough to earn them a life of

SheepCattle

Bridge

Engine

A

B

C

Siding I

Truck A II

Truck B III

Engine

IV

The Solutions

comparative luxury, the additional condition that all the prime numbers must fall on the shaded “swastika” squares must also be met. Could you have qualified?

50. Weighing The CoinsYou are given twelve coins, one of which is

known to be lighter or heavier than the others. How can this odd coin be located in three weighings along with its relative weight? (i.e. is it lighter or heavier?) You are supplied with a simple beam balance but no weights.

51. The Trees In The Orchard“I’ve just been looking around the orchard,” said Peter, “to find how many

rows of three trees we have - three in a dead straight line I mean. Actually there are eighteen.”

“But we’ve only got twelve trees in the orchard,” I replied, thinking my brother had made a mistake.

“I know that, but all the same there are eighteen rows of three,” repeated Peter. Can you draw a possible layout of trees in our orchard?

52. The Pawns’ Revolt

Now that the black pawns have revolted and overthrown their aristocratic court, they face the question of land distribution. Justice demands that their half of the board should be distributed among the pawnatariat in such a way that each pawn receives a plot containing the same number of squares and of exactly the same shape as all the others. But a serious problem has arisen. No pawn is prepared to move from the square it at present occupies. How did they divide their land?

The Solutions

53. SquarizonaSquarizona is one of those American States whose

boundaries ignore such things as mountains and rivers. In fact it is an exact square, 95 miles by 95 miles, and has six neighbours. Somewhere in there is the State Gaol and in it a mass breakout of long-term prisoners is being hatched. Its leader has been told by an indiscreet warder that it is an exact whole number of miles from the gaol to the nearest part of the six neighbouring States.

Can you tell the escapees how far it is to the nearest State boundary?

54. Staked Out“Yes, it is a bit expensive,” admitted Farmer Roots, “but neater than those old

barbed wire fences. The cattle like it too. Plenty of nice rough sticks to scratch themselves against.”

“The way that cow is scratching herself, I should think that she would push the whole lot over before long,” said I.

“Oh, no. The pens are quite strongly built. The stakes are driven into the ground at one-foot intervals, and then those ash and willow branches are woven in and out between the stakes to form the fence itself. This square pen you are looking at is the first we built, and all the other pens are the same area, but I found it more convenient to build them with one side of the rectangle longer than the other.”

He nodded to the next pen a little further down the paddock. “Now there is one that should interest you,” he said. “The number of stakes in its fence is exactly the same as the square yards they enclose. Now work out the size of that one.”

“Not enough data,” I objected. “There could be half a dozen or more possible answers to that sort of problem.”

Farmer Roots shook his head. “No, just one - provided you really think about it and take everything I’ve told you into account.”

55. Air Miles

ADDON

FIGGAH

MHYNNIS

pOWAH

ADDON 225 540 1296FIGGAH 225 585 1521

MHYNNIS 540 585POWAH 1296 1521

Amazing, isn’t it, how it always seems to be the very thing one is seeking that has managed to get itself lost or mislaid? It was something of this sort that Henshaw

1

2

3

4 5

6

The Solutions

found so frustrating when planning to fly from Mhynnis to Powah during his recent holiday in New Anglia. He had torn off an uninteresting corner of his travel brochure for the purpose of lighting his pipe, only to discover later that he had thereby destroyed the one part of the air-mileage table (printed on the other side) that he really wanted.

Now, sheer hard mental slog has no appeal for Henshaw, so that he was pleased to notice that the table contained certain peculiarities which enabled him to calculate the distance from Mhynnis to Powah with a minimum of effort. What is that distance?

56. Buying CheeseA customer comes to an Italian grocer known for his excellent cheese and asks

for two cheeses which together weigh 37 pounds. They must weigh exactly 37 pounds the customer insists, because today is his wife’s 37th birthday. This, Mr Marinacci thinks is rather a peculiar request, but since this an old and faithful customer, he wants to satisfy him.

So he takes down all his five huge cheeses that he keeps on the shelf and puts first one pair and then another pair on the scales to find two which will satisy his customer. It is all in vain however, because every possible combination of two of the cheeses gave only the following weights: 20, 24, 30, 35, 36, 40, 41, 45, and 51 pounds. So, unfortunately, the customer’s wife never got this pungent 37-pound birthday present. You however, have all you need to figure out how much each of Mr. Marinacci’s five cheeses weighed.

57. A Motorcycle Just In TimeTwo men want to catch a train at a rural railway station 9 miles away. They

do not own a car and they cannot walk faster than 3 miles-per-hour. Fortunately, just as they are about to start on their journey, a friend of theirs comes along on a motorcycle. They tell him that their train leaves in 2½ hours whilst they are afraid that it will take them 3 hours to walk to the station.

The friend offers to help them out. He will take one of them along for part of the journey. Then, whilst this man continues his trip on foot, the motorcyclist will return until he meets the other man who meanwhile will have started his journey on foot. Then the second man will go along as passenger all the way to the station.

The motorcyclist’s speed is 18 miles-per-hour. Where should he stop to let the first man dismount and to return for the second man if he wants both men to reach their goal simultaneously? How long will the whole journey take?

58. Short CircuitWhen a short circuit occurred, Ethel searched her drawer for candles and

found two of equal length. One was a little thicker than the other and would burn for five hours whilst the thinner one would burn for only four. However the electrical trouble did not last as long as the life expectancy of the candles. When it was all over and Ethel was able to blow the candles out, she noticed that the stump of one of the candles was four times as long as that of the other. How long had the power been off?

The Solutions

59. Just Distribution Of MarblesSeven boys owned a certain total number of marbles but there was

considerable difference of opinion as to how many belonged to each. The father of one of the boys, a mathematics teacher, decided to settle the matter once and for all and to everybody’s satisfaction.

So he asked each of the children to get a basket. Then he distributed the marbles in a peculiar way which looked far from just. Finally he asked the first child to dip into the marbles in his basket and to put into each of the others as many marbles as were already in it. Then he asked the second boy to do exactly the same thing, and so on up to the seventh boy. When he asked each of them to count the number of marbles now in his basket, they found that each of them had 128. How many marbles had the father put into each of the seven baskets originally?

60. The Wine Dealer’s TestamentA wine dealer left to his three sons, 24 wine casks, five of which were full,

eleven only half-full and eight empty. His last will stipulated that each of the sons should inherit the same quantity of wine and the same number of casks. Moreover, since each cask contained wine of a different vintage, mixing and decanting was out of the question. How could the wine dealer’s last will be fulfilled?

61. Engineering Drawings

The plan and front elevation of a certain object are shown in Figure 1 and Figure 2 respectively. Draw the side elevation.

62. The Tank And The DrainpipesA large tank is receiving water at a constant rate from a large supply pipe.

The tank is equipped with ten identical drain pipes. If the tank is full then:At the end of two and a half hours it will be empty if all ten of the outlets are

open, but if only six of the outlets are left open it will take five and a half hours to empty.

How long will it take to empty if only three of the outlets are left open?

Figure 1PLAN

Figure 2FRONT ELEVATION

The Solutions

63. Joe NitwitIn his hurry to get on to more pleasant tasks, Joe Nitwit turned on both intake

pipes but forgot to close the drainpipe. The water tank was half full when his boss came by and noticing the drainpipe open, closed it before going about his most urgent task - firing Joe!

Assuming that it takes one of the intake pipes 10 hours to fill the tank, the second intake pipe 8 hours to fill the tank, and the drainpipe only six hours to empty the tank if both intake pipes are closed, how long did it take to fill the tank on this occasion?

64. Filling The BathTwo taps are filling a bath. The first tap, by itself can fill the bath in one hour

if the plug is in. The second tap can fill the bath in 30 minutes if the plug is in. With the plug out and both taps turned off, a full bath will empty in 45 minutes. How long does it take to fill an empty bath with both taps running and the plug out?

65. The Way To The VillageA logician vacationing in the South Seas, finds himself on an island inhabited

by two tribes, one tribe always tells the truth whilst the other tribe invariably lies. He is walking one day and comes to a fork in the road. He knows that one road leads to the village, but which one?

Luckily there is a native nearby but the logician cannot tell from his appearance whether he is from the tribe of truth-tellers or from the tribe of liars. He thinks for a moment and then asks the native just ONE question, and from the native’s reply he knows which road to take. What is his question?

66. Who Owns The Zebra?

There is a terrace of five houses, each of a different color, and inhabited by men of different nationalities, with different pets, and preferences for different drinks and cigarettes. The Englishman lives in the red house. The Spaniard owns the dog. Coffee is drunk in the green house. The Ukranian drinks tea. The green house is immediately to the right (your right) of the ivory house. The Old Gold smoker owns snails. Kools are smoked in the yellow house. Milk is drunk in the middle house. The Norwegian lives in the first house on the left. The man who smokes Chesterfields lives in the house next to the man who owns the fox. Kools are smoked in the house next to the house where the horse is kept. The Lucky Strike smoker

The Solutions

drinks orange juice. The Japanese smokes Parliaments. The Norwegian lives next to the blue house. The problem is, who drinks water and who owns the zebra?

67. A Digit-Placing ProblemPlace the digits from 1 through 8 in the eight

circles with the proviso that no two digits directly adjacent to each other in serial order may go in circles that are directly connected by a line.

For example, if 5 is placed in the top circle, neither 4 nor 6 may be placed in any of the three circles that form a horizontal row beneath it, because each of these circles is joined directly to the top circle by a straight line.

There is only one solution not counting rotations and reflections.

68. A Tennis MatchMiranda beat Rosemary in a set of tennis, winning six games to

Rosemary’s three. Five games were won by the player who did not serve. Who served first?

69. How Far Did The Smith’s Travel?At 10.00 one morning, Mr. Smith and his wife left their house in Connecticut to drive to the home of Mrs. Smith’s parents in Pennsylvania. They planned to stop along the way for lunch at Patricia Murphy’s Candlelight Restaurant at Westchester.

The prospective visit with his in-laws, combined with business worries, put Mr. Smith in a sullen, uncommunicative mood. It was not until 11.00 that Mrs. Smith ventured to ask: “How far have we gone, dear?”

Mr. Smith glanced at the odometer. “Half as far as the distance to Patricia Murphy’s,” he snapped.

They arrived at the restaurant at noon, enjoyed a leisurely lunch, then continued on their way. Not until 5.00 in the afternoon, when they were 200 miles from where Mrs. Smith had asked her first question, did she ask a second one. “How far have we to go, dear?”

“Half as far,” he grunted, “as the distance from here to Patricia Murphy’s.”They arrived at their destination at 7.00 that evening. Because of traffic

conditions, Mr. Smith had driven at widely differing speeds. Nevertheless, it is quite simple to determine (and this is the problem), exactly how far the Smiths travelled from one house to the other.

The Solutions

70. A Pair Of CryptarithmsIn most cryptarithms, a different letter is substituted for each digit in a simple

arithmetical problem. The two remarkable cryptarithms shown below are unorthodox in their departure from this practice, but each is easily solved by logical reasoning and each has a unique answer. (Fitch Cheyney, University of Hartford)

E E O P P P O O P P E O E O P P P P E O O _ P P P P__ O O O O O P P P P P

In the multiplication problem at the left, each E stands for an even digit (zero is even) and each O stands for an odd digit. The fact that every even digit is represented by E does not mean, of course, that all the even digits are the same. For example, one E may stand for 2, another for 4 and so on.

In the multiplication problem at the right, each P stands for a prime digit (2, 3, 5 or 7) (Joseph Ellis Trevor, Cornell University)

71. Square CardsRemove the spades and hearts from a deck of playing cards. Put the spades face up in a row in serial order, Ace at the left, King at the right. Now place a heart beneath each spade so that the sum of each spade-heart pair is a square number. For the purposes of this problem, Ace = 1, Jack = 11, Queen = 12 and King = 13. Prove that the solution is unique.

Jokers

A magician places five cards in a row as shown above. All card backs are either black or red. “Are all the cards with red backs Jokers?” he asks.

The problem is not to answer the question but to determine the minimum number of cards which must be turned over in order to answer it. In other words, assuming all possible variations of the hidden card sides (each Joker may have a red or black back, the card with the visible red back may or may not be a joker and so on), how many cards must you turn over before you can answer the question: “Are all the cards with red backs Jokers?”

It is a confusing problem and one that calls for careful reasoning. There is a surprise in the solution which is closely related to an old joke about three professors on a train in Scotland. Through the window they see a black sheep.

The Solutions

“How interesting,” says the astronomer. “All sheep in Scotland are black.”“A totally unwarranted inference,” the physicist replies. “We can conclude

only that some sheep in Scotland are black.”“Correction,” says the logician. “At least one sheep in Scotland is black on at

least one side.”

The Solutions1. The Flight Around The World

Three aeroplanes are sufficient to ensure the flight of one of them around the world. There are many ways in which this may be done, but the following seems to be the most efficient.

Planes A and B and C take off together. After going one-eighth of the way, C transfers ¼ tank to A and ¼ tank to B. This gives A and B full tanks and leaves C with a ¼ tank sufficient just to get back to base.

Planes A and B continue a further one-eighth of the way when B transfers ¼ tank to A. B now has ½ tank left, sufficient just to get back to base.

Plane A meanwhile, with a full tank continues until he runs out of fuel ¼ way from base (¾ way round the world). Here A is met by C which has been refuelled at base. C transfers ¼ tank to A and both planes head for home.

A and C run out of fuel one-eighth way from base where they are met by refuelled plane B. B transfers ¼ tank to A and ¼ tank to C and the three planes then have just enough fuel to reach base with empty tanks.

2. Who Is The Engineer?• From 2 we see that Mr. Robinson lives in Los Angeles.• From 3 and 6 we can deduce that the distinguished mathematical physicist lives in Omaha.• From 4 we see that Mr. Jones cannot live in Omaha (6) or Los Angeles (2) so he must live in

Chicago.• From 5 we now know that the brakeman’s name is Jones.From Smith cannot be the fireman; he cannot be the brakeman (5) so he must be the engineer.

3. The Coloured HatsC asks himself, “Can my hat be green?” If so then A will know immediately that he has a red hat, for only a red hat on his head would cause B to lift his hand. A would have therefore left the room. B would reason in the same way and also leave. Since neither has left the room, C deduces that his own hat must be red.

4. The Coloured Hats AgainIf Bill and Colin are wearing red hats, Arthur would have known his hat was blue, because there were only two red hats. Since Arthur didn’t know the correct answer, Colin concluded that there remained only two possiblities for himself and Bill. Either both had blue hats, or one had a blue hat and the other a red hat. If he himself had a red hat, Colin reasoned, Bill would have concluded that he, Bill, had a blue hat, because otherwise Arthur would have known that he, Arthur, must have a blue hat. So Bill, because he was not able to tell the colour of his own hat, involuntarily betrayed to Colin that his, Colin’s hat was not red. Therefore Colin concluded that his hat was blue.

The Solutions

5. The Professor On The EscalatorLet n be the number of steps visible when the escalator is not moving, and let a unit of time be the time it takes the professor to walk down one step. If he walks down the down-moving escalator in 50 steps, then n - 50 steps have gone out of sight in 50 units of time. It takes him 125 steps to run up the same escalator taking five steps to every one before. In this time 125 - n steps have disappeared in 125 5 units of time. Since the escalator can be assumed to run at a constant speed, we have the following linear equation that readily yields n = 100 steps:

n - 50 125 - n = 50 25

6. The Encounter On The BridgeIf we call the length of the bridge 3d, the two vehicles will meet when the car is 2d over the bridge and the truck, d. To reach this point, the car has required t seconds and the truck 2t seconds. To back up the car would need 2t seconds for the distance 2d; the truck 4t seconds for the distance d. Now let us consider the possiblities:

Case 1 The car backs up. It requires 2t seconds, meanwhile the truck proceeds. It needs 4t seconds to finish the journey across the bridge so the car has to wait 2t seconds until it can start across again. It requires 1½t seconds to drive over the bridge. Thus 5½t seconds are required for the operation.

Case 2 The truck backs up. It requires 4t seconds. Simultaneously the car proceeds though only with one-eighth its speed because of the truck’s snail pace. They arrive at the end of the bridge together, then the truck drives over the bridge which takes 6t seconds. Thus in this case 10t seconds are need to complete the operation.

Therefore, Case 1 is the solution most favourable to both vehicles.

7. The Early CommuterThe commuter walked for 55 minutes before his wife picked him up.Since they arrive home 10 minutes earlier than usual, this means that the wife had chopped 10 minutes off her usual travel time to and from the station. It follows that she must have met her husband five minutes before his usual pick-up time of five o’clock or at 4:55 pm. He started walking at 4pm and was therefore walking for 55 minutes.

8. The Two Ferry BoatsWhen the two ferry boats meet for the first time, the combined distance travelled by the boats is equal to the width of the river. When they reach the opposite shores, the combined distance travelled is twice the width of the river and when they meet for the second time, the combined distance is three times the width of the river.

Since both boats are moving at a constant speed for the same period of time, it follows that each boat has gone three times as far as when they first met (after they had travelled a combined distance of one river’s width). Since the white boat, (slow boat), had travelled 720 metres before the first meeting, its total distance at the time of the second meeting must be 3 x 720 or 2160 metres. This is 400 metres more than the width of the river, so we must subtract 400 from 2160 to obtain 1760 metres as the width of the river.

9. The Weights And The ScalesFour weights are quite sufficient for this purpose. 1 gram, 3 grams, 9 grams and 27 grams.

To weight two grams, the 3-gram weight is put on one pan and the 1-gram weight on the other, making a difference of two grams.

The SolutionsTo weight five grams, the 9-gram weight is put on one pan and the and the 3-gram weight and the 1-gram weight are put on the other, making a difference of five grams.

By using this method of adding and subtracting weights, the four weights will weigh any number of whole grams from one to forty grams inclusive.

10. The Efficient ElectricianOn the top floor, the electrician shorted five pairs of wires by twisting them together, leaving the last wire free. Then he walked to the basement and identified the lower ends of the shorted pairs by means of his continuity tester. He labelled the bottom ends of the first shorted pair A1 and A2 respectively, the ends of the second shorted pair B1 and B2 respectively, the ends of the third shorted pair C1 and C2 respectively, the ends of the fourth shorted pair D1 and D2 respectively, the ends of the fifth shorted pair E1 and E2 respectively and the unpaired end he labelled F. He shorted the bottom ends of the wires by leaving A1 free, connecting A2 to B1, B2 to C1, C2 to D1, D2 to E1, and E2 to F.

Back on the top floor, he removed all the shorts but left the pairs twisted together at insulated portions so that the pairs were still identifiable. He then checked for continuity between the free wire, which he knew to be the upper end of F and some other wire. When he found the other wire he was at once able to label it E2 and identify its mate as E1. He next tested for contintuity between E1 and another end which when found could be marked D2 and its mate D1. Continuing in this fashion, the remaining ends were easily identified.

11. How Old Are You?If when Jones was 28, Smith was 2x years old, then he was x years old when Jones was 3x years old. Whatever their respective ages, Jones is always the same number of years older than Smith. Age difference when Smith is x and Jones is 3x is 2x years.

Age difference when Jones is 28 and Smith 2x is 28 - 2x.

Now the age differences are equal, therefore 2x = 28 - 2x or 4x = 28 or x = 7.Age difference is 2x or 14 years. Therefore when Jones was 28 years old, Smith was 14 years old.Smith is 28 so Jones is now 28 + 14 = 42 years old.

12. Ann And MaryWork from the end backwards. Let Ann be x years old, then Mary was 3x when she was three times as old as Ann. Age difference is 2x. Ann will be three times 3x when she is 9x years old. Mary will be 4½x when she is half as old as 9x. Ann will be 4½x - 2x or 2½x when Mary is 4½x. Therefore Mary now being twice as old as 2½x is now 5x. Ann is 2x years younger or 3x years old. But we are told that 5x + 3x = 44. And from this x = 5½.

Mary is 5x or 27½ years old whilst Ann is 3x or 16½ years old. Easy isn’t it?

13. The Tribulations Of A Father Of FiveLet A,B,C,D,E be the five children in descending order of age.

Left Bank Crossing Right BankABCDE - -

ACE BD -ACE B DBE AC DBE AD CBD AE CBD CE ABD CE A

The SolutionsBD - ACE

- BD ACE- - ABCDE

The minimum number of crossings is nine with three crossings per child. Note that if C and E have only one crossing each, the total number of crossings is reduced from nine to seven.

The Solutions

14. Three Wives & Their Jealous HusbandsLet the three husbands be A,B and C and their wives a,b and c respectively. The following moves are typical of a minimum solution with men rowing where possible.

Left Bank Crossing Right BankABCabc - -

ACac Bb -ACac B bABC ac bABC a bcAa BC bcAa Bb Ccab AB Ccab c ABCb ac ABCb B ACac- Bb ACac- - ABCabc

Note that only a, B, and c need be able to row.

15. Four Wives & Their Jealous HusbandsLet the husbands be ABCD and their wives abcd respectively. Then the following crossings are necessary:

Left bank River Island River Right bankABCDabcd - - - -ABCDcd ab - - -ABCDcd b a - -ABCDd bc a - -ABCDd c ab - -CDcd AB ab - -CDcd - AB ab -CDcd - AB b aCDcd - b AB aCDcd - b B AaCDcd B b - AaBCD cd b - AaBCD d bc - AaDd BC bc - AaDd - bc BC AaDd - bc a ABCDd - c ab ABCDd - c C ABabDd C c - ABabd CD c - ABabd - c CD ABabd - c b ABCDad - - bc ABCDad - - c ABCDab

The Solutionsd c - - ABCDab- cd - - ABCDab- - - cd ABCDab- - - - ABCDabcd

16. Getting The Nuggets AcrossLet Smith, Jones and Brown be designated by the numbers 3, 2 and 1 respectively and their respective properties by (3), (2) and (1) . In any group on either bank the total of the numbers in parentheses must never exceed the total of the plain numbers. One solution is as follows:

Left bank River Right bank123(1)(2)(3) - -13(1)(3) 2(2) -13(1)(3) 2 (2)12(3) 3(1) (2)12(3) 3 (1)(2)3(3) 12 (1)(2)3(3) 2(2) 1(1)2(2) 3(3) 1(1)2(2) 1(1) 3(3)(1)(2) 12 3(3)(1)(2) 3 12(3)(1) 3(2) 12(3)(1) 1 23(2)(3)- 1(1) 23(2)(3)- - 123(1)(2)(3)

A minimum of thirteen trips.

17. Missionaries and CannibalsM = misionary C = cannibal Bracketed letters are the ones who can row.

Left bank River Right bankMM(M)CC(C) - -

MMC(C) (M)C -MMC(C) (M) CMM(M) (C)C CMM(M) (C) CC

M(C) (M)M CCM(C) (M)C MCMC (M)(C) MCMC (M)C M(C)CC (M)M M(C)CC (C) MM(M)C (C)C MM(M)C (C) MM(M)C- (C)C MM(M)C- - MM(M)CC(C)

A total of thirteen crossings.

The Solutions

18. Road Safety AdditionThere are seven basic solutions, values for D and H are the two remaining digits in each case. These two digits can be put in either order making 14 solutions altogether. If zero cannot be the first digit, #4 has only one, not two solutions (H not equal 0) making only 13 solutions in total.

R T A M L O J E1 2 9 7 6 1 0 5 32 2 9 7 6 0 1 4 33 2 9 7 6 0 1 5 44 2 9 6 5 4 8 7 35 2 4 7 6 3 9 0 86 6 7 3 2 0 5 9 87 6 7 3 2 4 1 5 0

19. The Lonesome EightWhen two digits are brought down instead of one, there must be a zero in the quotient. This occurs twice, so we know at once that the quotient must be x080x.

When the divisor is multiplied by the quotient’s last digit, the produxt is a four-digit number. But we see that 8 times the divisor is a three-digit number. So the quotient’s last digit must be nine.

The divisor must be less than 125 because 8 125 = 1000, a four-digit number.

We can now deduce that the quotient’s first digit must be more than 7, for 7 times a divisor less than 125 would give a product which would be more than 100 less than 1000, the lowest four-digit number, and therefore would give a difference of more than 100 (a three digit number) when subtracted from the first four digits of the dividend, instead of the two-digit difference it does, in fact, give.

The first digit of the quotient cannot be nine because nine times the divisor is a four figure number, so it must be 8. This makes the full quotient 80809.

The divisor must be more than 123 because 123 80809 is a seven-digit number and our dividend has eight digits. Therefore it must be 124.

The final result is now easily found.

20. The Farmer And The Hundred DollarsTo buy 100 animals with $100, the number of anomals bought must equal the number of dollars spent. Since sheep sell at the price of one sheep for $1, they can be ignored until we have bought a number of cows and pigs such that the number of cows and pigs bought, equals the number of dollars paid for them. The rest of the $100 can be made up of sheep, which will not affect the balance of animals bought against dollars spent.

For every cow bought, we have spent $4 more than we need to, to make the number of cows bought equal to dollars spent. For every pig bought, we have spent 95 cents less than necessary to make the number of pigs bought equal to the number of dollars spent.

Thus if we buy cows and pigs in the inverse ratio of this excess and insufficiency, (i.e. 19 cows for every 80 pigs) the number of animals bought will be equal to the number of dollars spent.

But 80 + 19 = 99 so we can only buy one such balanced set for under $100. The other $1 can be used to buy a sheep. Therefore the farmer bought 19 cows, 1 sheep and 80 pigs.

The Solutions

21. The Knight’s Tour (1)1 10 35 28 3 1218 27 2 11 34 299 36 17 30 13 426 19 8 33 22 317 16 21 24 5 1420 25 6 15 32 23

There are numerous solutions to this problem. The solution given here is just one of them to prove to those who gave up that it is possible. The numbers indicate the order of the moves starting at 1 and finishing at 36. Note that the solution given here is a circular tour where the knight finishes on a cell exactly a knight’s move from where it started.

22. The Knight’s Tour (2)1 14 63 40 3 24 57 4262 39 2 13 60 41 4 2315 64 61 36 25 56 43 5838 33 12 55 52 59 22 511 16 37 26 35 54 51 4432 29 34 53 50 47 6 2117 10 27 30 19 8 45 4828 31 18 9 46 49 20 7

The numbers show the 63 moves in numerical order. There are many more solutions to this problem. The solution shown here is a circular tour where the knight finishes on a cell exactly a knight’s move away from where it started.

23. Alice In WonderlandThe Jub-Jub lives in a Tum-Tum tree, the Bandersnatch is an animal, and the Jabberwock has four wings and feeds on Toves. This problem yields readily to a little trial and error.

24. The Spider And The Fly

Ceiling Spider

FloorFly

24m

32m

40m

The SolutionsThe easiest way to solve this puzzle is to imagine the room as a carboard box and to open it out flat as shown in the diagram above. The shortest walking path for the fly to be reached is exactly 40 metres. The spider crawls along the straight dotted line labelled 40m in the diagram of the unfolded room above.

25. Another Escalator PuzzleLet the number of visible steps be n and the little boy move up one step in unit time. Therefore n - 50 steps disappear in 50 units of time. As the bigger boy travels three times the speed of his little brother, n - 75 steps disappear in 75 3 units of time, or 25 units of time.

Therefore: n - 50 = n - 75 50 25

From this equation, n is readily deduced as 100. Therefore the escalator had 100 steps.

26. The Ladder Against The WallCall AD = a and EC = b.

Since triangle ADF and triangle FEC are similar then:

a = 4 or ab = 16...............................................................................................(1)4 b

AB = a + 4 BC = b + 4.

Therefore by Pythgoras’ Theorem:

AB2 + BC2 = AC2 or (a + 4)2 + (b + 4)2 = 400.

Expanding and collecting terms together:a2 + b2 + 8(a + b) + 32 = 400..............................................................................(2)

Now from (1) 32 = 2ab

Substituting in (2):a2 + b2 + 8(a + b) + 2ab = 400...........................................................................(3)

Now: a2 + 2ab + b2 = (a + b)2

So substituting in (3) we get:

(a + b)2 + 8(a + b) = 400

Add 16 to both sides:(a + b)2 + 8(a + b) + 16 = 416

Or: (a + b + 4)2 = 416So: a + b + 4 = 416..........................................................................................(4)

From (1) b = 16 aSubstituting in (4):a + 16 + 4 = 416 aBy using the formula for solution of quadratic equations, a is found to be 15.36 metres or 1.04 metres. This makes AB = 19.36 metres or 5.04 metres.

The Solutions

27. A Coin Puzzle

The rearrangement from Figure 1 to Figure 2 can be effected in four moves.

1. Move C and D to the right leaving room for another 20c and another 10c between E and C.2. Move A and B to the right so that A touches D.3. Move D and A into the gap between E and C.4. Move E and D into the gap between C and B.

28. The Dilemma Of A Racecourse Owner

29. The Scrambled Box TopsThe key to this solution is your knowledge that all three labels on the boxes are incorrect. If this is the case, then the box labelled BW cannot contain a black and a white marble. It must contain either two black marbles or two white marbles.

So if we choose the box marked BW and draw one marble from it without looking inside, we can tell by the colour of this marble whether two black marbles or two white marbles are in this box. If it is (say) black, the box marked BW must contain two black marbles. The box marked WW we know cannot contain two white marbles because it is incorrectly labelled. We know also that it cannot contain two black marbles because we have already identified this box. Therefore the box marked WW must contain one black marble and one white marble. Therefore the BB box contains two white marbles. Similar reasoning will give us the contents of the boxes if the marble drawn from the BW box is white.

30. The Seventeen SuitorsThe Princess Carlotta’s secret lover was the man who was given the gold cross to hold.

A B C D E

A BC DE

Figure 1

Figure 2

The racecourse owner needs six sets of equipment set out as in the diagram. A race between any two sets will be a different multiple of 1000 metres.

5000 metres

1000 metres

2000 metres

7000 metres

4000 metres

12000 metres

The Solutions

31. The Countefeit CoinsThe counterfeit stack can be identified by a single weighing of coins. Take one coin from the first stack, two from the second stack, three from the third stack and so on, up to all ten from the tenth stack. Weight the whole sample collection, and the excess weight of this collection, in grams, indicates the counterfeit pile. For example, if the excess were four grams, then the counterfeit pile would be the pile from which we removed four coins.

32. How Many Children?When Jones began to work on the Professor’s problem, he knew that each of the four families had a different number of children and that they totalled less than 18. He futher knew that the product of the four numbers gave the Professor’s house number. Therfore his obvious first step was to factorise into four different numbers which together summed less than 18.

If there had been only one way to do this, he would have immediately solved the problem. Since he needed more information, we conclude that he must have found more than one way of factorising the house number. Our next step is to write down all possible combinations of four different numbers which totla less than 18, and obtain the products of each group. We find there are many cases where more than one combination gives the same product. How do we decide which product is the house number? The clue lies in the fact that Jones asked if there was more than one child in the smallest family. The question is meaningful only if the house number is 120 which can be factorised as1 3 5 8; 1 4 5 6; 2 3 4 5. Had Smith answered “No”, the problem would remain unsolved. Since Jones did solve it, we know the answer to be “Yes”. The families therefore contained 2, 3, 4 and 5 children.

33. A Question Of CardsThe hand was a follows:

J K, J A, K, Q, J, 7, 2 K, J, 7, 2

Hearts were trumps.

34. The Dovetailed Block

The cube is constructed from two blocks of wood with dovetails running at forty-five degrees to the faces of the block. The two pieces just slide together along the line of the dovetails.

The Solutions

35. The Twelve Matches

Figure 1 Figure 2

Twelve matches can be used to form a right triangle with sides of 3, 4 and 5 units as shown in Figure 1. This triangle has an area of 6 square units. By altering the position of three matches as shown in Figure 2, we remove two square units leaving a polygon with an area of exactly FOUR square units. Who said anything about the figure having to be a regular polygon?

36. Conundra ZooIn full, the moves are:Rhino to enclosure; Gorilla to Bear’s cage; Tiger to Rhino’s cage; Rhino to Lion’s cage; Gorilla to enclosure; Lion to Bear’s cage; Bear to Gorilla’s cage; Gorilla to Tiger’s cage; Lion to enclosure; Tiger to Bear’s cage; Rhino to Rhino’s cage; Lion to Lion’s cage; Tiger to enclosure; Bear to Bear’s cage; Gorilla to Gorilla’s cage; Tiger to Tiger’s cage. Sixteen moves in all. There is no other sequence of moves which will achieve the object so economically. Notice that the moves fall into four successive cycles of four moves each, one cycle to get each animal into its proper cyclic order relative to one other animal.

37. Little Pigley Farm, 1935

The two clues having four-figure year numbers (8 Across and 11 Across) must begin with a 1 and so 8 Down ends with the same number that starts 15 Across (see 10 Down). 15 Across must either be 27 or 64 and we see that if Farmer Dunk’s walking speed is 4 miles-per-hour he must take 16 minutes to walk one-and-a-third times round Dog’s Mead. This would make the perimeter 1408 yards which is

3 8 7 2 0 155

3 2 4 49 3 5 2

1 6 1 027

92 7

1 9 1 37

1 69 2

5

The Solutionsimpossible as 14 Across has only three digits. Farmer Dunk’s walking speed must therefore be 3 miles-per-hour and we can fill in 15 Across as 27; 8 Down as 12; 9 Down as 11; 14 Across as 792; 16 Across as 16.

The sum of the numbers in the row indicated by the arrows cannot be 29 and therefore 12 Down must be 19. The most difficult step now is to find a square number of 5 digits ending in 796, and bearing in mind that the difference in length and breadth of Dog’s Mead is a 2-digit number. The only possible solution is 1762 = 30976. The length follows as 220 yards and the rest of the puzzle fits into place. There is slight ambiguity about Farmer Dunk’s age, but since he has a son of 45 it seems fairly certain that his age is 72.

38. Potton vs Barford

Figure 1 Figure 2

Considering 13 Across together with 13 Down, we see that each of these numbers must begin with a 1; so that 19 Across also starts with 1 and the number of extras (15 Down) must be 11, since 8 times this (4 Down) is two digits only. Since the third wicket put on five runs only, and 3 Across ends with 8, it must begin with 1.

Therefore, every wicket up to and including the seventh, (3 Down) fell at a total smaller than 200. 14 Down begins with 8 or 9 and since it is not possible for a square (13 Across) to end with an 8, this can be eliminated. The square can now be partially completed as in Figure 1.

Since 8 times 19 Across is less than 1000, and since its square ends with 9, the only possiblities are: 113, 117, 123 and the only one of these which will fit with 12 Down is 123; and 1232 = 15129. We know now that the third wicket fell at 128 and filling in as much of the puzzle as this information allows us to do, we discover from 6 Down that the fourth wicket fell at 161, and from 7 Down that the sixth wicket fell at 183. 11 Down is therefore 154 which gives (16 Across) that the first wicket fell at 48.

It follows from 2 Down that the eighth wicket was taken at 211 and from 1 Across, the ninth fell at 212. Since 7 Across is an odd number, the seventh wicket (3 Down) must have fallen at 194 and it only remains to find a number whose square is a number of five digits beginning with 32 and ending with 61. This can only be 181 whose square is 32761. The remaining clue is 18 Down and from our complete score card we see this must be 31.

The scores were, then, as follows:1st wicket, 48; 2nd at 123; 3rd at 128; 4th at 161; 5th at 181; 6th at 183; 7th at 194; 8th at 211; 9th at 212; 10th at 213.

22 1 1

1

11 8

8

1 11 9

111

2 1 2 1914836

1112

3844292151

1617

321

882

211

4 9 13

The Solutions

39. Another Coin PuzzleThe following moves are typical of many solutions:

1. Move 1 to touch 2 and 42. Move 4 to touch 5 and 63. Move 5 to touch 1 and 2 below4. Move 1 to touch 4 and 5.

40. Calculari’s ConcertsFor his last concert, Calculari played: Liszt, Brahms, Mozart and Chopin in that order.

Since Beethoven and Mozart were never played in the same concert, one of these must always have been excluded. Consequently every concert must have included Brahms, Chopin and Liszt and (because of the inclusion of Brahms) Liszt must have started each concert. Then, if Beethoven were omitted, since in such cases Mozart must have been followed immediately by Chopin, these two composers can fill only (a) second and third or (b) third and fourth places respectively. And since Brahms must fill the other place, these are the only two alternatives in a non-Beethoven concert.

Similarly, if Mozart were omitted (Brahms consequently finishing the programme), Beethoven and Chopin can fill only (c) second and third or (d) third and second places respectively. Only four arrangements are therefore possible and three of these viz. (a), (c) and (d) result in Brahms finishing the concert. These three cases must therefore have accounted for Calculari’s first three concerts. (b) must have been his last concert.

41. Raising Poultry (1)Let d be the number of ducks she starts with and c be the number of chickens.Then d + c = 25.

Let x be the number of times the ducks multiply, then 3x will be the number of times chickens multiply.So now, a year later, she has:

25 + 3xc + xd birds. This is equal to 200.

Therefore x(3c + d) = 175 But c + d = 25.

So x(2c + 25) = 175

Now x and c must be whole numbers.

The prime factors of 175 are 5 5 7

So x = 5, 7, 25 or 35.

If x = 35; 2c + 25 = 5 this gives a negative number of chickensIf x = 25; 2c + 25 = 7 this also gives a negative number of chickensIf x = 7; 2c + 25 = 25 this makes the number of chickens zero which cannot be.

So x must be 5 and 2c + 25 = 35. Thus c = 5. In which case d = 20.

This is the only possible solution.

42 Raisning Poultry (2)This solution is arrived at in a similar way to that of the previous problem. Introduction of another variable doesn’t really make our task any more difficult, if we go about the solution logically.

Let the number of hens be h; the number of ducks d; the number of geese g.

The Solutions

Let geese multiply x times a year.Then hens multiply 7x times a yearand ducks multiply 4x times a year.

At the end of the year there are:

17 + 7xh + 4xd + xg = 248 birds. But h + d + g = 17

So x(17 + 3d + 6h) = 231

Now the prime factors of 231 are: 3 7 11So x is one of the following: 3, 7, 11, 21, 33, 77

If x = 77, 33 or 21, h and d would be negative numbers which is impossible.

If x = 11 or 7, h and d would be fractional numbers which again cannot be,

When x = 3 2h + d = 20

This gives the following possibilites:

h = 9 8 7 6 5 4 3 2 1d = 2 4 6 8 10 12 14 16g = 6 5 4 3 2 1 0 -1 -2

The only combination which gives hens more than ducks and ducks more than geese as stipulated by the original problem, is th third set.

This is the unique solution required. Originally there were 7 hens, 6 ducks and 4 geese.

43. The Spinning WheelArthur 15, Barry 14, Charles 13, David 10, Ernest 11, Frank 12.

Each pair of opposite numbers on the disk totals 5. Each pair of opposite boys must have therefore scored a total of five each spin, or 25 for all 5 spins. No more, no less! If Arthur finished with the highest total, then it follows that David finished with the lowest.

The lowest scorer, David, could not have made more than 11 for, had he scored 12, Arthur would have scored only 13 and could not possibly have been the clear winner. The disposal of the numbers on the disk is such that David could have taken the lead on the second spin only by obtaining a 5 giving him a progressive total of 9. Then since 11 is his possible maximum, at least one of the remaining spins must have resulted in a 0. Three spins have therefore been uniquely determined, for which A = 6; B = 7; C = 10; D = 9; E = 8; F = 5. For the remaining two spins, David’s scores could be: 0-0; 0-1; 1-1 or 2-0. But the first and last of these would result in Arthur tying with someone else instead of winning. Again 1-1 for David would give two 5s to Frank making Frank the winner. In the remaining two spins David therefore scored 0-1.

All five positions of the disk are now uniquely determined. From these positions the score for each player may simply be added up.

44. Walking OutThey never step out on their right feet together.

45. Running To TimeTom ran exactly two miles.

The Solutions

One side of the small field is obviously 2 - 1 miles. So M ( the distance the farmer walked) is 2.2 - 2 miles. Call the total distance run by Tom, D miles. Now when Tom caught up with his father, the farmer has walked ½(D + M) miles. Similarly, for the whole distance, the farmer walked M miles, whilst Tom ran D miles. Therefore, since their relative speeds are constant,

D = ½(D + M) or D = D + MM ½(D - M ) M D M

Cross multiplying we get: D(D M) = M(D M) or D2 - DM DM + M2

This means that D2 - 2DM M2. Complete the square by adding M2 to both sides.

D2 - 2DM + M2 = 2M2. Extract the square root of both sides to obtain:

D - M = M2 i.e. D = M2 + M or D = M(2 + 1)

Now we resubstitute for M which gives us

D = (22 - 2)(2 + 1)

The 2 multiplies itself out leaving us simply with D = 2.

46. A Railway Shunting ProblemFirst, the white engine uncouples and shunts onto the siding. The black train then moves forward and couples up with A, pushing the white coaches until G is to the left of the siding. The white engine then moves onto the main tarck. A is then uncoupled from B and the black engine reverses pulling A and pushing its own four coaches until A is on the right of the siding. The black engine then pushes A onto the siding and then hauls its own coaches past the siding. The white engine couples to A and pulls A onto the main track. The same procedure is then repeated for B and C whence the two trains have passed and can proceed on their journeys.

47. Another Shunting ProblemFirst the engine manoeuvres the sheep truck onto C. Then the engine does a complete circuit of the track passing over A and under the bridge and then pushes the cattle truck onto C to couple with the sheep truck. The engine then pulls the two trucks onto B and then pushes them onto A where the sheep truck is uncoupled. The cattle truck is then pulled onto B and pushed onto C. The engine then goes back over B and under the bridge, pushing the sheep truck onto B, where it is then uncoupled. The engine returns to C and couples up to the cattle truck, pulling it onto B and pushing it on to A. The engine then uncouples the cattle truck and returns to C.

48. A Third Shunting ProblemE pushes A onto I and returns to the main track. E then pulls B onto IV and pushes B past the junction of II with IV and uncouples. E returns to III and pulls A onto III and thence pushes them up to III until B is on I. B is uncoupled and A is pulled onto III and uncoupled. E returns to IV and thence to II where it pulls B onto II. E uncouples and returns to IV.

The Solutions

49. A Swastika Square

This is the swastika square the leader’s men finally came up with - 5 by 5 with all lines adding up to 65, and all prime numbers within the swastika shape.

50. Weighing The CoinsNumber the coins 1, 2, 3, 4..., 12.Weight 1, 2, 3, 4 against 6, 5, 7, 8If they balance, the odd one must be amongst 9, 10, 11, 12 and we have two weighings in which to locate it. We also know that 1 to 8 are perfect.

Weigh 9, 10, 1 against 11, 2, 3If they balance, 12 must be the odd one and if it is weighed against a perfect one, we can tell whether it is lighter or heavier.

If they do not balance, note which side is the heavier...............................................................(A)

Now weigh 9 against 10.............................................................................................................(B)If they balance, 11 is the imperfect one and 9 and 10 are perfect. (A) tells us whether it is lighter or heavier.

If 9 and 10 do not balance, 11 must be perfect and we know whether the odd one is heavy or light from (A). The actual coin may now be located from (B).

If the first weighing of 1, 2, 3, 4 against 5, 6, 7, 8 does not balance, we know 9, 10, 11 and 12 are perfect. Note which side is heavier.............................................................................................(C)

Weigh 1, 9, 5 against 6, 7, 2. If they balnce, the odd one is among 3, 4, 8.

Weigh 3 against 4 if they balance, 8 is the odd one and is light or heavy according to (C). If 3 and 4 do not balance, then 8 is perfect and the odd one, and whether it is heavier or lighter can be determined from (C).

If 1, 9, 5 against 6, 7, 2 do not balance, note which is heavier and compare with (C). Depending on this result, we either weigh 5 against 2 or 6 against 7.

Alternative Solution

Label the twelve coins with letters, thus: S, H, I,N ,E, B, A, L, D, T, O, P.Now weigh SLAB against PIED and note the positon of the scales.Now weigh SPOT against HEAD and note the position of the scales.Now weigh PELT against BOND and note the position of the scales.

17 5 13 21 9

4 12 25 8 16

11 24 7 20 3

10 18 1 14 22

23 6 19 2 15

The SolutionsOnly consider the weighings which don’t balance.If, say, SLAB was light and SPOT was light and PELT against BOND balanced, the coin must be light and amongst SLAB and SPOT, and heavy and amongst PIED and HEAD. A letter must be found which is common between SLAB and SPOT or between PIED and HEAD which is not to be found anywhere else. There are three letters which are common: S (in SLAB and SPOT) and E, D (in PIED and HEAD). E cannot be the one because it appears in PELT, which balances. D cannot be the one because it appears in BOND which balances. Therefore it must be S. S is on the light side of both weighings, so the odd coin must be S and it is light.

Another example, suppose BOND and SLAB are heavy and SPOT against HEAD balances. We must find a common letter between heavy BOND and SLAB or between light PELT and PIED which does not appear anywhere else. B is common between BOND and SLAB and P, E are common between PELT and PIED. P appears in SPOT which balances, so it cannot be P. E appears in HEAD which balances, so it cannot be E. B satisfies the conditions, so the odd coin is B and, as it is on the heavy side of the weighings, it is heavy.

Another example, Suppose SLAB is light and SPOT against HEAD and PELT against BOND both balance. Here we must find a letter amongst SLAB and PIED which does not appear anywhere else. The only letter of the eight which does not appear in either of the two balanced weighings, is I. The odd coin is therefore I, and it is heavy as it is on the heavy side of its weighing.

A final example. Suppose all three weighings do not balance. PIED, HEAD and PELT are all heavy. Here we must find a letter which appears in all three of PIED, HEAD and PELT or all three of SLAB, SPOT and BOND. The only letter to fit the bill is E, so E is the odd coin and it is heavy as it appears on the heavy sides of all three weighings.

51. The Trees In The Orchard

The line of four trees has been discounted

52. The Pawns’ Revolt

This is the only possible solution

The Solutions

53. Squarizona

54. Staked Out180 feet by 20 feet.

If the lengths of the sides are x feet and y feet, the perimeter will contain 2(x + y) stakes, and this will equal the area in square yards which is xy/9, from which x = 18y/(y - 18) = 324/(y - 18) + 18. For 324 to be divisible by (y - 18), y (or x) must equal 19, 20, 21, 22, 24, 27, 30 or 36. The last would give dimensions of 36 by 36 making it a square, whereas the pen is not a square. Nevertheless, the pen’s area must be a square number because it is the same as the first pen which is a square. The only other value making xy a square is x = 20.

55. Air Miles

56. Buying CheeseMr. Marinacci got only nine different weights for any combination of any two of his five cheeses. It ought to be 10 because five different quantities can be paired in ten different ways. Therefore two different combinations must have resulted in the same weight. If each cheese is combined with each of the others in turn, each will be weighed four times. Hence the sum of the weights of all 10 combinations must be divisible by four.

91

84 11

6061

95

1 2

6

5 4

3

35

The problem here is to find two right-angled triangles one side of one triangle and a side of the other triangle total 95 in two different instances (see the diagram opposite). The shortest distance to the nearest boundary was 11 miles.

1404

1296 225

585540

P

M

FA

Notice that PA = 1296 miles, and AF = 225 miles, a total distance of 1521 miles. But this is the distance PF which means that PAF lie on a straight line. Again, because 2252 + 5402 = 5852, FAM is a right angle, and since FAP is a straight line, MAP is a right angle, making MP the hypotenuse of a Pythagorean triangle with legs 1296 and 540 miles.

The SolutionsThe sum of all nine weights mentioned is 322 pounds. To this figure, the missing tenth weight must be added, to get four times the weight of all five cheeses. This means, that among the weights given in the problem, one has to be found which, added to 322 yields a number divisible by four. The only possible figure is 30 (322 + 30) / 4 = 88, and 88 is the sum of the weights of all five cheeses. The weights given in the problem indicate that the two lightest cheeses together weighed 20 pounds and the two heaviest together weighed 51 pounds. The remaining fifth cheese, therefore, weighed 88 = (20 + 51) = 17 pounds. By trial and error you will find in a jiffy that the four remaining cheese weighed 7, 13, 23, and 28 pounds.

57. A Motorcycle Just In TimeThe distance, x, the second man has to walk until he meets the motorcyclist and the distance the first man must walk from the moment he has dismounted until he reaches the station must be equal because both men are supposed to arrive simultaneously and therefore must have walked and ridden for equal lengths of time. The motorcyclist altogether travels (9 - x) + (9 - 2x) + (9 - x) = 27 - 4x miles at 18 miles per hour whilst each of his two friends in the same time travel (9 - x) miles at 18 miles per hour and x miles at 3 miles per hour. So we have the equation:

27 - 4x = 9 - x + x 18 18 3

which yields x = 2.

Therefore the first man has to dismount after a ride of seven miles. The motorcyclist returns and meets the second man at a point 2 miles from the start. The whole journey will take 1 hour and 3-and-a-third minutes.

58. Short CircuitLet the candles be 1 unit long to begin with. And let them burn for t hours. Then the candle with the shorter life will burn t / 4 of its length in t hours. And the candle with the longer life, will burn t / 5 of its length in t hours. Therefore the respective lengths of candle remaining in each case are: 1 - t / 4 and 1 - t / 5.

Now from the problem we know that: 4 (1 - t / 4) = 1 - t / 5 or 4 - t = (5 - t) / 5.

Or 20 - 5t = 5 - tor 4t = 15Therefore t = 3¾ hours.

59. Just Distribution Of MarblesYou do not need algebraic formulae for this one. Just get a piece of papaer and a pencil and start from the end, that is, from the 128 marbles in each basket. That way, you will quickly find that the baskets contained 449, 225, 113, 57, 29, 15, and 8 marbles.

60. The Wine Dealer’s TestamentThree solutions to this problem are easily found with trial and error.

Full Half full EmptyFirst son 3 1 4

Second son 2 3 3Third son 0 7 1

Full Half full EmptyFirst son 2 3 3

Second son 2 3 3Third son 1 5 2

Full Half full Empty

The SolutionsFirst son 3 1 4

Second son 1 5 2Third son 1 5 2

61. Engineering Drawings

62. The Tank And The DrainpipesLet the capacity of the supply pipe be s gallons per hour and the capacity of the one of the drainpipes be d gallons per hour, then:

1. Q + 2½(s) = 2½(10d) from statement a.2. Q + 5½(s) = 5½(6d) from statement b.3. Q + x(s) = x(3d) where x is the number of hours it will take before the tank is empty

if only three of the drainpipes are open.At first glance it would appear that these three equations with their four unknowns will not be sufficient to give the information desired, but let us see. Subtracting equation 1 from equation 2 and then from equation 3 gives:

4. 3s = (33 - 25)d = 8d or s = 8d / 3 and5. (x - 2½)s = (3x - 25)d

Now, substituting for s in equation 5, its value 8d / 3 gives:

((x - 2½)8d) / 3 = (3x - 25)d

or, dividing by d and simplifying:

8x - 20 = 9x - 75or x = 55

The tank will be empty at the end of 55 hours if three of the drainpipes are open and the remainder closed.

63. Joe NitwitPipe 1 fills the tank in 10 hours or in one hour it fills 1/10 th of the tank.Pipe 2 fills the tank in 8 hours or in one hour it fills 1/8 th of the tank.The drainpipe empties the tank in 6 hours or in one hour it empties 1/6 th of the tank.

Therefore, the tank is filling up at a rate of 1 / 10 + 1 / 8 - 1 / 6 per hour.

This equals 7 / 120. Therefore at this rate it will take 120 / 7 hours to fill the tank or 60 / 7 hours to half fill it.

There can be no holes in the object for otherwise there would be dotted lines on the plan or elevation to show their course. The solution is shown at the left.

The Solutions

When the drainpipe is closed, the rate then increases to 1 / 10 + 1/ 8 per hour.

This equals 27 / 120. Therefore it would take 120 / 27 hours to fill the tank at this rate or 60 / 27 hours to half fill it.

Total time to fill tank would be 60 / 7 + 60 / 27 hoursor 60 / 7 + 20 / 9 hoursor (540 + 140) / 63 = 10 50/63 hours.or a little over 10 hours 47 minutes and 37 seconds.

64. Filling The BathTap 1 fills the bath in 1 hour therfore in one hour it fills one bath.Tap 2 fills the bath in ½ an hour, therefore in one hour it fills 2 baths.The drain will empty the bath in ¾ hour, therefore in one hour it empties 4/3 baths.

Therefore, in one hour, with both taps running and the drainpipe open, they will fill:

1 + 2 - 4/3 = 5 / 3 baths.

Therefore, it will take 3 / 5 hours to fill one bath or 36 minutes.

65. The Way To The VillageThe logician points to one of the roads and asks the native: “If I were to ask you if this road leads to the village, would you answer ‘Yes’?” The native is forced to give the right answer, even if he is a liar!

If the road does not lead to the village, the liar would say “No” to the direct question; but as the question is put, he lies and and says he would reply “Yes”. Thus the logician can be certain that the road does lead to the village, whether the respondant is a truth-teller or a liar. If the road does actually not go to the village, the liar is forced in the same way to answer “No” to the question.

66. Who Owns The Zebra?House Yellow Blue Red Ivory Green

Inhabitant Norwegian Ukranian Englishman Spaniard JapanesePet Fox Horse Snails Dog Zebra

Beverage Water Tea Milk Orange Juice CoffeeCigarettes Kool Chesterfield Old Gold Lucky Strike Parliament

The Japanese owns the zebra; the Norwegian drinks water.

67. A Digit-Placing Problem

7

13 4

5 8 6

2

A

D

G

B C

E

F

In the series 1, 2, 3, 4, 5, 6, 7, 8 each digit has two neighbours except 1 and 8. In the diagram at the left, circle C is connected to every circle except H and likewise, circle F is connected to every circle except circle A. It follows that these two circles (C and F) must contain the numbers 1 and 8, although it doesn’t matter which way round they are placed. With 1 in C and 8 in F, only one circle is available for 2 and 7. The remaining four numbers are now easily placed.

The Solutions

68. A Tennis MatchWhoever served first, served five games and the other player served four. Suppose the first server won x games of the games she served and y of the other four games. The total number of games lost by the player who served them is then 5 - x + y. But we are told that 5 games were won by the player who did not serve them, which is another way of saying that 5 games were lost by the player that served them. Therefore 5 - x + y = 5 so that x = y. Therefore the player who served first won a total of x + y or 2x games. Because this must be an even number, and only Miranda won an even number of games, she must have been the first server.

69. How Far Did The Smiths Travel?

Draw a diagram as above and mark on it the information as stated in the problem. Q1 is the point where Mrs. Smith wasked her first question. If they had travelled x miles at this point, from Mr. Smith’s reply it must be 2x miles from here to Patricia Murphy’s. Q2 is the point where Mrs. Smith asked her second question. If they has travelled 2y miles from Patricia Murphy’s, then from Mr. Smith’s reply, it must be y miles to their destination in the state of Pennsylvania. Now we are told that the distance travelled between questions was 200 miles. Therefore we arrive at the equation

2x + 2y = 200 or x + y = 100.

The total distance travelled is x + y + 200 miles or 100 + 200 = 300 miles.

70. A Pair Of Cryptarithms 2 8 5 7 7 5 3 9 3 3 2 5 6 5 2 3 2 5 8 5 5 2 3 2 5 1 1 1 1 5 2 5 5 7 5

2 7

6 3

8 1

5 4

AH

B

C

DE

F

G

Another way of solving this puzzle is to draw a new diagram. Label the eight circles with letters as before and in the new diagram, connect the circles together only if in the old network they were not connected. The original problem then becomes a matter of placing the eight digits in the circles so that a connected path can be traced from 1 to 8 taking the digits in order. The numbers are then transcribed into their corresponding circles in the original network.

H

C Px 2x 2y y

PMQ1 Q2

200 miles

The SolutionsThe second problem, the more difficult of the two is perhaps best approached by searching first for all the three-digit numbers composed of prime digits that yield four prime digits when multiplied by a prime. There are only four.

755 x 3 = 2325 555 x 5 = 2775 755 x 5 = 3775 325 x 7 = 2275

No three-digit number has more than one multiplier, therefore the multiplier in the problem must consist of two identical digits. Thus there are only four possibilities that need to be tested.

71. Square Cards

A 2 3 4 5 6 7 8 9 10 J Q K

8 2 K Q J 10 9 A 7 6 5 4 3

9, 10 and J must be paired with 7, 6, and 5. This establishes 6 pairings. Since the 6s have been used, the 3 pairs only with the K. Since the 5s have been used, the 4 pairs only with the Q. The remaining 3 gaps are filled in only one way, proving the uniqueness of the solution.

72. JokersLetter the five cards from A through E.Obviously D must be turned to see if it is a Joker, and E must be turned to see if it has a red back.. This gives us the possiblities:

1. D is a Joker, E has a black back.2. D is a Joker, E has a red back.3. D is not a Joker, E has a black back.4. D is not a Joker, E has a red back.

For cases 2, 3 and 4, the answer to the question: “Are all the cards with red backs Jokers?” is ‘No’. No more cards need to be turned. For case 1, the answer is ‘Yes’, but it takes more thinking to realise that turning the other three cards cannot contradict this answer.

B is irrelevant because it has a black back. Seeing the back of either Joker is also irrelevant. If a Joker’s back is black, it is not involved in the question. If it is red, the answer is still ‘Yes’. Most people staring at an actual row of cards have such an overwhelming desire to see the backs of the Jokers that they usually answer A, C, D, E.

One might conclude, therefore, that turning D and E is sufficient to answer the question. It is not! Recall the story about the cautious logician who observed a black sheep in Scotland and concluded that at least one sheep in Scotland is black on at least one side. What about card B? Nowhere in the original problem was it stated that the five cards were normal playing cards! When someone thinks he has solved the problem, the magician turns over card B that its other side is a red back! This of course contradicts a ‘Yes’ answer. The correct solution, therefore, is that card B as well as cards D and E must be turned.