
"Let me think," said Mrs. Allgood. "Yes—yes—that is correct."
"Very well, then. As there are only nine hundred and ninetynine thousand nine hundred and ninetynine different ways of bearing hair, it is clear that the millionth person must repeat one of those ways. Do you see?"
"Yes; I see that—at least I think I see it."
"Therefore two persons at least must have the same number of hairs on their heads; and as the number of people on the earth so greatly exceeds the number of hairs on any one person's head, there must, of course, be an immense number of these repetitions."
"But, Mr. Filkins," said little Willie Allgood, "why could not the millionth man have, say, ten thousand hairs and a half?"
"That is mere hairsplitting, Willie, and does not come into the question."
"Here is a curious paradox," said George. "If a thousand soldiers are drawn up in battle array on a plane"—they understood him to mean "plain"—"only one man will stand upright."
Nobody could see why. But George explained that, according to Euclid, a plane can touch a sphere only at one point, and that person only who stands at that point, with respect to the centre of the earth, will stand upright.
"In the same way," he remarked, "if a billiardtable were quite level—that is, a perfect plane—the balls ought to roll to the centre."
Though he tried to explain this by placing a visitingcard on an orange and expounding the law of gravitation, Mrs. Allgood declined to accept the statement. She could not see that the top of a true billiardtable must, theoretically, be spherical, just like a portion of the orangepeel that George cut out. Of course, the table is so small in proportion to the surface of the earth that the curvature is not appreciable, but it is nevertheless true in theory. A surface that we call level is not the same as our idea of a true geometrical plane.
"Uncle John," broke in Willie Allgood, "there is a certain island situated between England and France, and yet that island is farther from France than England is. What is the island?"
"That seems absurd, my boy; because if I place this tumbler, to represent the island, between these two plates, it seems impossible that the tumbler can be farther from either of the plates than they are from each other."
"But isn't Guernsey between England and France?" asked Willie.
"Yes, certainly."
"Well, then, I think you will find, uncle, that Guernsey is about twentysix miles from France, and England is only twentyone miles from France, between Calais and Dover."
"My mathematical master," said George, "has been trying to induce me to accept the axiom that 'if equals be multiplied by equals the products are equal.'"
"It is selfevident," pointed out Mr. Filkins. "For example, if 3 feet equal 1 yard, then twice 3 feet will equal 2 yards. Do you see?"
"But, Mr. Filkins," asked George, "is this tumbler half full of water equal to a similar glass half empty?"
"Certainly, George."
"Then it follows from the axiom that a glass full must equal a glass empty. Is that correct?"
"No, clearly not. I never thought of it in that light."
"Perhaps," suggested Mr. Allgood, "the rule does not apply to liquids."
"Just what I was thinking, Allgood. It would seem that we must make an exception in the case of liquids."
"But it would be awkward," said George, with a smile, "if we also had to except the case of solids. For instance, let us take the solid earth. One mile square equals one square mile. Therefore two miles square must equal two square miles. Is this so?"
"Well, let me see! No, of course not," Mr. Filkins replied, "because two miles square is four square miles."
"Then," said George, "if the axiom is not true in these cases, when is it true?"
Mr. Filkins promised to look into the matter, and perhaps the reader will also like to give it consideration at leisure.
"Look here, George," said his cousin Reginald Woolley: "by what fractional part does fourfourths exceed threefourths?"
"By onefourth!" shouted everybody at once.
"Try another one," George suggested.
"With pleasure, when you have answered that one correctly," was Reginald's reply.
"Do you mean to say that it isn't onefourth?"
"Certainly I do."
Several members of the company failed to see that the correct answer is "onethird," although Reginald tried to explain that three of anything, if increased by onethird, becomes four.
"Uncle John, how do you pronounce 'too'?" asked Willie.
"'Too," my boy."
"And how do you pronounce 'two'?"
"That is also 'too.'"
"Then how do you pronounce the second day of the week?"
"Well, that I should pronounce 'Tuesday,' not 'Toosday.'"
"Would you really? I should pronounce it 'Monday.'"
"If you go on like this, Willie," said Uncle John, with mock severity, "you will soon be without a friend in the world."
"Can any of you write down quickly in figures 'twelve thousand twelve hundred and twelve pounds'?" asked Mr. Allgood.
His eldest daughter, Miss Mildred, was the only person who happened to have a pencil at hand.
"It can't be done," she declared, after making an attempt on the white tablecloth; but Mr. Allgood showed her that it should be written, "L13,212."
"Now it is my turn," said Mildred. "I have been waiting to ask you all a question. In the Massacre of the Innocents under Herod, a number of poor little children were buried in the sand with only their feet sticking out. How might you distinguish the boys from the girls?"
"I suppose," said Mrs. Allgood, "it is a conundrum—something to do with their poor little 'souls.'"
But after everybody had given it up, Mildred reminded the company that only boys were put to death.
"Once upon a time," began George, "Achilles had a race with a tortoise—"
"Stop, George!" interposed Mr. Allgood. "We won't have that one. I knew two men in my youth who were once the best of friends, but they quarrelled over that infernal thing of Zeno's, and they never spoke to one another again for the rest of their lives. I draw the line at that, and the other stupid thing by Zeno about the flying arrow. I don't believe anybody understands them, because I could never do so myself."
"Oh, very well, then, father. Here is another. The PostOffice people were about to erect a line of telegraphposts over a high hill from Turmitville to Wurzleton; but as it was found that a railway company was making a deep level cutting in the same direction, they arranged to put up the posts beside the line. Now, the posts were to be a hundred yards apart, the length of the road over the hill being five miles, and the length of the level cutting only four and a half miles. How many posts did they save by erecting them on the level?"
"That is a very simple matter of calculation," said Mr. Filkins. "Find how many times one hundred yards will go in five miles, and how many times in four and a half miles. Then deduct one from the other, and you have the number of posts saved by the shorter route."
"Quite right," confirmed Mr. Allgood. "Nothing could be easier."
"That is just what the PostOffice people said," replied George, "but it is quite wrong. If you look at this sketch that I have just made, you will see that there is no difference whatever. If the posts are a hundred yards apart, just the same number will be required on the level as over the surface of the hill."
"Surely you must be wrong, George," said Mrs. Allgood, "for if the posts are a hundred yards apart and it is half a mile farther over the hill, you have to put up posts on that extra halfmile."
"Look at the diagram, mother. You will see that the distance from post to post is not the distance from base to base measured along the ground. I am just the same distance from you if I stand on this spot on the carpet or stand immediately above it on the chair."
But Mrs. Allgood was not convinced.
Mr. Smoothly, the curate, at the end of the table, said at this point that he had a little question to ask.
"Suppose the earth were a perfect sphere with a smooth surface, and a girdle of steel were placed round the Equator so that it touched at every point."
"'I'll put a girdle round about the earth in forty minutes,'" muttered George, quoting the words of Puck in A Midsummer Night's Dream.
"Now, if six yards were added to the length of the girdle, what would then be the distance between the girdle and the earth, supposing that distance to be equal all round?"
"In such a great length," said Mr. Allgood, "I do not suppose the distance would be worth mentioning."
"What do you say, George?" asked Mr. Smoothly.
"Well, without calculating I should imagine it would be a very minute fraction of an inch."
Reginald and Mr. Filkins were of the same opinion.
"I think it will surprise you all," said the curate, "to learn that those extra six yards would make the distance from the earth all round the girdle very nearly a yard!"
"Very nearly a yard!" everybody exclaimed, with astonishment; but Mr. Smoothly was quite correct. The increase is independent of the original length of the girdle, which may be round the earth or round an orange; in any case the additional six yards will give a distance of nearly a yard all round. This is apt to surprise the nonmathematical mind.
"Did you hear the story of the extraordinary precocity of Mrs. Perkins's baby that died last week?" asked Mrs. Allgood. "It was only three months old, and lying at the point of death, when the griefstricken mother asked the doctor if nothing could save it. 'Absolutely nothing!' said the doctor. Then the infant looked up pitifully into its mother's face and said—absolutely nothing!"
"Impossible!" insisted Mildred. "And only three months old!"
"There have been extraordinary cases of infantile precocity," said Mr. Filkins, "the truth of which has often been carefully attested. But are you sure this really happened, Mrs. Allgood?"
"Positive," replied the lady. "But do you really think it astonishing that a child of three months should say absolutely nothing? What would you expect it to say?"
"Speaking of death," said Mr. Smoothly, solemnly, "I knew two men, father and son, who died in the same battle during the South African War. They were both named Andrew Johnson and buried side by side, but there was some difficulty in distinguishing them on the headstones. What would you have done?"
"Quite simple," said Mr. Allgood. "They should have described one as 'Andrew Johnson, Senior,' and the other as 'Andrew Johnson, Junior.'"
"But I forgot to tell you that the father died first."
"What difference can that make?"
"Well, you see, they wanted to be absolutely exact, and that was the difficulty."
"But I don't see any difficulty," said Mr. Allgood, nor could anybody else.
"Well," explained Mr. Smoothly, "it is like this. If the father died first, the son was then no longer 'Junior.' Is that so?"
"To be strictly exact, yes."
"That is just what they wanted—to be strictly exact. Now, if he was no longer 'Junior,' then he did not die 'Junior." Consequently it must be incorrect so to describe him on the headstone. Do you see the point?"
"Here is a rather curious thing," said Mr. Filkins, "that I have just remembered. A man wrote to me the other day that he had recently discovered two old coins while digging in his garden. One was dated '51 B.C.,' and the other one marked 'George I.' How do I know that he was not writing the truth?"
"Perhaps you know the man to be addicted to lying," said Reginald.
"But that would be no proof that he was not telling the truth in this instance."
"Perhaps," suggested Mildred, "you know that there were no coins made at those dates.
"On the contrary, they were made at both periods."
"Were they silver or copper coins?" asked Willie.
"My friend did not state, and I really cannot see, Willie, that it makes any difference."
"I see it!" shouted Reginald. "The letters 'B.C.' would never be used on a coin made before the birth of Christ. They never anticipated the event in that way. The letters were only adopted later to denote dates previous to those which we call 'A.D.' That is very good; but I cannot see why the other statement could not be correct."
"Reginald is quite right," said Mr. Filkins, "about the first coin. The second one could not exist, because the first George would never be described in his lifetime as 'George I.'"
"Why not?" asked Mrs. Allgood. "He was George I."
"Yes; but they would not know it until there was a George II."
"Then there was no George II. until George III. came to the throne?"
"That does not follow. The second George becomes 'George II.' on account of there having been a 'George I.'"
"Then the first George was 'George I.' on account of there having been no king of that name before him."
"Don't you see, mother," said George Allgood, "we did not call Queen Victoria 'Victoria I.;' but if there is ever a 'Victoria II.,' then she will be known that way."
"But there have been several Georges, and therefore he was 'George I.' There haven't been several Victorias, so the two cases are not similar."
They gave up the attempt to convince Mrs. Allgood, but the reader will, of course, see the point clearly.
"Here is a question," said Mildred Allgood, "that I should like some of you to settle for me. I am accustomed to buy from our greengrocer bundles of asparagus, each 12 inches in circumference. I always put a tape measure round them to make sure I am getting the full quantity. The other day the man had no large bundles in stock, but handed me instead two small ones, each 6 inches in circumference. 'That is the same thing,' I said, 'and, of course, the price will be the same;' but he insisted that the two bundles together contained more than the large one, and charged me a few pence extra. Now, what I want to know is, which of us was correct? Would the two small bundles contain the same quantity as the large one? Or would they contain more?"
"That is the ancient puzzle," said Reginald, laughing, "of the sack of corn that Sempronius borrowed from Caius, which your greengrocer, perhaps, had been reading about somewhere. He caught you beautifully."
"Then they were equal?"
"On the contrary, you were both wrong, and you were badly cheated. You only got half the quantity that would have been contained in a large bundle, and therefore ought to have been charged half the original price, instead of more."
Yes, it was a bad swindle, undoubtedly. A circle with a circumference half that of another must have its area a quarter that of the other. Therefore the two small bundles contained together only half as much asparagus as a large one.
"Mr. Filkins, can you answer this?" asked Willie. "There is a man in the next village who eats two eggs for breakfast every morning."
"Nothing very extraordinary in that," George broke in. "If you told us that the two eggs ate the man it would be interesting."
"Don't interrupt the boy, George," said his mother.
"Well," Willie continued, "this man neither buys, borrows, barters, begs, steals, nor finds the eggs. He doesn't keep hens, and the eggs are not given to him. How does he get the eggs?"
"Does he take them in exchange for something else?" asked Mildred.
"That would be bartering them," Willie replied.
"Perhaps some friend sends them to him," suggested Mrs. Allgood.
"I said that they were not given to him."
"I know," said George, with confidence. "A strange hen comes into his place and lays them."
"But that would be finding them, wouldn't it?"
"Does he hire them?" asked Reginald.
"If so, he could not return them after they were eaten, so that would be stealing them."
"Perhaps it is a pun on the word 'lay,'" Mr. Filkins said. "Does he lay them on the table?"
"He would have to get them first, wouldn't he? The question was, How does he get them?"
"Give it up!" said everybody. Then little Willie crept round to the protection of his mother, for George was apt to be rough on such occasions.
"The man keeps ducks!" he cried, "and his servant collects the eggs every morning."
"But you said he doesn't keep birds!" George protested.
"I didn't, did I, Mr. Filkins? I said he doesn't keep hens."
"But he finds them," said Reginald.
"No; I said his servant finds them."
"Well, then," Mildred interposed, "his servant gives them to him."
"You cannot give a man his own property, can you?"
All agreed that Willie's answer was quite satisfactory. Then Uncle John produced a little fallacy that "brought the proceedings to a close," as the newspapers say.
413.—A CHESSBOARD FALLACY.
"Here is a diagram of a chessboard," he said. "You see there are sixtyfour squares—eight by eight. Now I draw a straight line from the top lefthand corner, where the first and second squares meet, to the bottom righthand corner. I cut along this line with the scissors, slide up the piece that I have marked B, and then clip off the little corner C by a cut along the first upright line. This little piece will exactly fit into its place at the top, and we now have an oblong with seven squares on one side and nine squares on the other. There are, therefore, now only sixtythree squares, because seven multiplied by nine makes sixtythree. Where on earth does that lost square go to? I have tried over and over again to catch the little beggar, but he always eludes me. For the life of me I cannot discover where he hides himself."
"It seems to be like the other old chessboard fallacy, and perhaps the explanation is the same," said Reginald—"that the pieces do not exactly fit."
"But they do fit," said Uncle John. "Try it, and you will see."
Later in the evening Reginald and George, were seen in a corner with their heads together, trying to catch that elusive little square, and it is only fair to record that before they retired for the night they succeeded in securing their prey, though some others of the company failed to see it when captured. Can the reader solve the little mystery?
UNCLASSIFIED PROBLEMS.
"A snapper up of unconsidered trifles." Winter's Tale, iv. 2.
414.—WHO WAS FIRST?
Anderson, Biggs, and Carpenter were staying together at a place by the seaside. One day they went out in a boat and were a mile at sea when a rifle was fired on shore in their direction. Why or by whom the shot was fired fortunately does not concern us, as no information on these points is obtainable, but from the facts I picked up we can get material for a curious little puzzle for the novice.
It seems that Anderson only heard the report of the gun, Biggs only saw the smoke, and Carpenter merely saw the bullet strike the water near them. Now, the question arises: Which of them first knew of the discharge of the rifle?
415.—A WONDERFUL VILLAGE.
There is a certain village in Japan, situated in a very low valley, and yet the sun is nearer to the inhabitants every noon, by 3,000 miles and upwards, than when he either rises or sets to these people. In what part of the country is the village situated?
416.—A CALENDAR PUZZLE.
If the end of the world should come on the first day of a new century, can you say what are the chances that it will happen on a Sunday?
417.—THE TIRING IRONS.
The illustration represents one of the most ancient of all mechanical puzzles. Its origin is unknown. Cardan, the mathematician, wrote about it in 1550, and Wallis in 1693; while it is said still to be found in obscure English villages (sometimes deposited in strange places, such as a church belfry), made of iron, and appropriately called "tiringirons," and to be used by the Norwegians today as a lock for boxes and bags. In the toyshops it is sometimes called the "Chinese rings," though there seems to be no authority for the description, and it more frequently goes by the unsatisfactory name of "the puzzling rings." The French call it "Baguenaudier."
The puzzle will be seen to consist of a simple loop of wire fixed in a handle to be held in the left hand, and a certain number of rings secured by wires which pass through holes in the bar and are kept there by their blunted ends. The wires work freely in the bar, but cannot come apart from it, nor can the wires be removed from the rings. The general puzzle is to detach the loop completely from all the rings, and then to put them all on again.
Now, it will be seen at a glance that the first ring (to the right) can be taken off at any time by sliding it over the end and dropping it through the loop; or it may be put on by reversing the operation. With this exception, the only ring that can ever be removed is the one that happens to be a contiguous second on the loop at the righthand end. Thus, with all the rings on, the second can be dropped at once; with the first ring down, you cannot drop the second, but may remove the third; with the first three rings down, you cannot drop the fourth, but may remove the fifth; and so on. It will be found that the first and second rings can be dropped together or put on together; but to prevent confusion we will throughout disallow this exceptional double move, and say that only one ring may be put on or removed at a time.
We can thus take off one ring in 1 move; two rings in 2 moves; three rings in 5 moves; four rings in 10 moves; five rings in 21 moves; and if we keep on doubling (and adding one where the number of rings is odd) we may easily ascertain the number of moves for completely removing any number of rings. To get off all the seven rings requires 85 moves. Let us look at the five moves made in removing the first three rings, the circles above the line standing for rings on the loop and those under for rings off the loop.
Drop the first ring; drop the third; put up the first; drop the second; and drop the first—5 moves, as shown clearly in the diagrams. The dark circles show at each stage, from the starting position to the finish, which rings it is possible to drop. After move 2 it will be noticed that no ring can be dropped until one has been put on, because the first and second rings from the right now on the loop are not together. After the fifth move, if we wish to remove all seven rings we must now drop the fifth. But before we can then remove the fourth it is necessary to put on the first three and remove the first two. We shall then have 7, 6, 4, 3 on the loop, and may therefore drop the fourth. When we have put on 2 and 1 and removed 3, 2, 1, we may drop the seventh ring. The next operation then will be to get 6, 5, 4, 3, 2, 1 on the loop and remove 4, 3, 2, 1, when 6 will come off; then get 5, 4, 3, 2, 1 on the loop, and remove 3, 2, 1, when 5 will come off; then get 4, 3, 2, 1 on the loop and remove 2, 1, when 4 will come off; then get 3, 2, 1 on the loop and remove 1, when 3 will come off; then get 2, 1 on the loop, when 2 will come off; and 1 will fall through on the 85th move, leaving the loop quite free. The reader should now be able to understand the puzzle, whether or not he has it in his hand in a practical form.
The particular problem I propose is simply this. Suppose there are altogether fourteen rings on the tiringirons, and we proceed to take them all off in the correct way so as not to waste any moves. What will be the position of the rings after the 9,999th move has been made?
418.—SUCH A GETTING UPSTAIRS.
In a suburban villa there is a small staircase with eight steps, not counting the landing. The little puzzle with which Tommy Smart perplexed his family is this. You are required to start from the bottom and land twice on the floor above (stopping there at the finish), having returned once to the ground floor. But you must be careful to use every tread the same number of times. In how few steps can you make the ascent? It seems a very simple matter, but it is more than likely that at your first attempt you will make a great many more steps than are necessary. Of course you must not go more than one riser at a time.
Tommy knows the trick, and has shown it to his father, who professes to have a contempt for such things; but when the children are in bed the pater will often take friends out into the hall and enjoy a good laugh at their bewilderment. And yet it is all so very simple when you know how it is done.
419.—THE FIVE PENNIES.
Here is a really hard puzzle, and yet its conditions are so absurdly simple. Every reader knows how to place four pennies so that they are equidistant from each other. All you have to do is to arrange three of them flat on the table so that they touch one another in the form of a triangle, and lay the fourth penny on top in the centre. Then, as every penny touches every other penny, they are all at equal distances from one another. Now try to do the same thing with five pennies—place them so that every penny shall touch every other penny—and you will find it a different matter altogether.
420.—THE INDUSTRIOUS BOOKWORM.
Our friend Professor Rackbrane is seen in the illustration to be propounding another of his little posers. He is explaining that since he last had occasion to take down those three volumes of a learned book from their place on his shelves a bookworm has actually bored a hole straight through from the first page to the last. He says that the leaves are together three inches thick in each volume, and that every cover is exactly oneeighth of an inch thick, and he asks how long a tunnel had the industrious worm to bore in preparing his new tube railway. Can you tell him?
421.—A CHAIN PUZZLE.
This is a puzzle based on a pretty little idea first dealt with by the late Mr. Sam Loyd. A man had nine pieces of chain, as shown in the illustration. He wanted to join these fifty links into one endless chain. It will cost a penny to open any link and twopence to weld a link together again, but he could buy a new endless chain of the same character and quality for 2s. 2d. What was the cheapest course for him to adopt? Unless the reader is cunning he may find himself a good way out in his answer.
422.—THE SABBATH PUZZLE.
I have come across the following little poser in an old book. I wonder how many readers will see the author's intended solution to the riddle.
Christians the week's first day for Sabbath hold; The Jews the seventh, as they did of old; The Turks the sixth, as we have oft been told. How can these three, in the same place and day, Have each his own true Sabbath? tell, I pray.
423.—THE RUBY BROOCH.
The annals of Scotland Yard contain some remarkable cases of jewel robberies, but one of the most perplexing was the theft of Lady Littlewood's rubies. There have, of course, been many greater robberies in point of value, but few so artfully conceived. Lady Littlewood, of Romley Manor, had a beautiful but rather eccentric heirloom in the form of a ruby brooch. While staying at her town house early in the eighties she took the jewel to a shop in Brompton for some slight repairs.
"A fine collection of rubies, madam," said the shopkeeper, to whom her ladyship was a stranger.
"Yes," she replied; "but curiously enough I have never actually counted them. My mother once pointed out to me that if you start from the centre and count up one line, along the outside and down the next line, there are always eight rubies. So I should always know if a stone were missing."
Six months later a brother of Lady Littlewood's, who had returned from his regiment in India, noticed that his sister was wearing the ruby brooch one night at a county ball, and on their return home asked to look at it more closely. He immediately detected the fact that four of the stones were gone.
"How can that possibly be?" said Lady Littlewood. "If you count up one line from the centre, along the edge, and down the next line, in any direction, there are always eight stones. This was always so and is so now. How, therefore, would it be possible to remove a stone without my detecting it?"
"Nothing could be simpler," replied the brother. "I know the brooch well. It originally contained fortyfive stones, and there are now only fortyone. Somebody has stolen four rubies, and then reset as small a number of the others as possible in such a way that there shall always be eight in any of the directions you have mentioned."
There was not the slightest doubt that the Brompton jeweller was the thief, and the matter was placed in the hands of the police. But the man was wanted for other robberies, and had left the neighbourhood some time before. To this day he has never been found.
The interesting little point that at first baffled the police, and which forms the subject of our puzzle, is this: How were the fortyfive rubies originally arranged on the brooch? The illustration shows exactly how the fortyone were arranged after it came back from the jeweller; but although they count eight correctly in any of the directions mentioned, there are four stones missing.
424.—THE DOVETAILED BLOCK.
Here is a curious mechanical puzzle that was given to me some years ago, but I cannot say who first invented it. It consists of two solid blocks of wood securely dovetailed together. On the other two vertical sides that are not visible the appearance is precisely the same as on those shown. How were the pieces put together? When I published this little puzzle in a London newspaper I received (though they were unsolicited) quite a stack of models, in oak, in teak, in mahogany, rosewood, satinwood, elm, and deal; some half a foot in length, and others varying in size right down to a delicate little model about half an inch square. It seemed to create considerable interest.
425.—JACK AND THE BEANSTALK.
The illustration, by a British artist, is a sketch of Jack climbing the beanstalk. Now, the artist has made a serious blunder in this drawing. Can you find out what it is?
426.—THE HYMNBOARD POSER.
The worthy vicar of Chumpley St. Winifred is in great distress. A little church difficulty has arisen that all the combined intelligence of the parish seems unable to surmount. What this difficulty is I will state hereafter, but it may add to the interest of the problem if I first give a short account of the curious position that has been brought about. It all has to do with the church hymnboards, the plates of which have become so damaged that they have ceased to fulfil the purpose for which they were devised. A generous parishioner has promised to pay for a new set of plates at a certain rate of cost; but strange as it may seem, no agreement can be come to as to what that cost should be. The proposed maker of the plates has named a price which the donor declares to be absurd. The good vicar thinks they are both wrong, so he asks the schoolmaster to work out the little sum. But this individual declares that he can find no rule bearing on the subject in any of his arithmetic books. An application having been made to the local medical practitioner, as a man of more than average intellect at Chumpley, he has assured the vicar that his practice is so heavy that he has not had time even to look at it, though his assistant whispers that the doctor has been sitting up unusually late for several nights past. Widow Wilson has a smart son, who is reputed to have once won a prize for puzzlesolving. He asserts that as he cannot find any solution to the problem it must have something to do with the squaring of the circle, the duplication of the cube, or the trisection of an angle; at any rate, he has never before seen a puzzle on the principle, and he gives it up.
This was the state of affairs when the assistant curate (who, I should say, had frankly confessed from the first that a profound study of theology had knocked out of his head all the knowledge of mathematics he ever possessed) kindly sent me the puzzle.
A church has three hymnboards, each to indicate the numbers of five different hymns to be sung at a service. All the boards are in use at the same service. The hymnbook contains 700 hymns. A new set of numbers is required, and a kind parishioner offers to present a set painted on metal plates, but stipulates that only the smallest number of plates necessary shall be purchased. The cost of each plate is to be 6d., and for the painting of each plate the charges are to be: For one plate, 1s.; for two plates alike, 113/4d. each; for three plates alike, 111/2d. each, and so on, the charge being one farthing less per plate for each similarly painted plate. Now, what should be the lowest cost?
Readers will note that they are required to use every legitimate and practical method of economy. The illustration will make clear the nature of the three hymnboards and plates. The five hymns are here indicated by means of twelve plates. These plates slide in separately at the back, and in the illustration there is room, of course, for three more plates.
427.—PHEASANTSHOOTING.
A Cockney friend, who is very apt to draw the long bow, and is evidently less of a sportsman than he pretends to be, relates to me the following not very credible yarn:—
"I've just been pheasantshooting with my friend the duke. We had splendid sport, and I made some wonderful shots. What do you think of this, for instance? Perhaps you can twist it into a puzzle. The duke and I were crossing a field when suddenly twentyfour pheasants rose on the wing right in front of us. I fired, and twothirds of them dropped dead at my feet. Then the duke had a shot at what were left, and brought down threetwentyfourths of them, wounded in the wing. Now, out of those twentyfour birds, how many still remained?"
It seems a simple enough question, but can the reader give a correct answer?
428.—THE GARDENER AND THE COOK.
A correspondent, signing himself "Simple Simon," suggested that I should give a special catch puzzle in the issue of The Weekly Dispatch for All Fools' Day, 1900. So I gave the following, and it caused considerable amusement; for out of a very large body of competitors, many quite expert, not a single person solved it, though it ran for nearly a month.
"The illustration is a fancy sketch of my correspondent, 'Simple Simon,' in the act of trying to solve the following innocent little arithmetical puzzle. A race between a man and a woman that I happened to witness one All Fools' Day has fixed itself indelibly on my memory. It happened at a countryhouse, where the gardener and the cook decided to run a race to a point 100 feet straight away and return. I found that the gardener ran 3 feet at every bound and the cook only 2 feet, but then she made three bounds to his two. Now, what was the result of the race?"
A fortnight after publication I added the following note: "It has been suggested that perhaps there is a catch in the 'return,' but there is not. The race is to a point 100 feet away and home again—that is, a distance of 200 feet. One correspondent asks whether they take exactly the same time in turning, to which I reply that they do. Another seems to suspect that it is really a conundrum, and that the answer is that 'the result of the race was a (matrimonial) tie.' But I had no such intention. The puzzle is an arithmetical one, as it purports to be."
429.—PLACING HALFPENNIES.
Here is an interesting little puzzle suggested to me by Mr. W. T. Whyte. Mark off on a sheet of paper a rectangular space 5 inches by 3 inches, and then find the greatest number of halfpennies that can be placed within the enclosure under the following conditions. A halfpenny is exactly an inch in diameter. Place your first halfpenny where you like, then place your second coin at exactly the distance of an inch from the first, the third an inch distance from the second, and so on. No halfpenny may touch another halfpenny or cross the boundary. Our illustration will make the matter perfectly clear. No. 2 coin is an inch from No. 1; No. 3 an inch from No. 2; No. 4 an inch from No. 3; but after No. 10 is placed we can go no further in this attempt. Yet several more halfpennies might have been got in. How many can the reader place?
430.—FIND THE MAN'S WIFE.
One summer day in 1903 I was loitering on the Brighton front, watching the people strolling about on the beach, when the friend who was with me suddenly drew my attention to an individual who was standing alone, and said, "Can you point out that man's wife? They are stopping at the same hotel as I am, and the lady is one of those in view." After a few minutes' observation, I was successful in indicating the lady correctly. My friend was curious to know by what method of reasoning I had arrived at the result. This was my answer:—
"We may at once exclude that Sister of Mercy and the girl in the short frock; also the woman selling oranges. It cannot be the lady in widows' weeds. It is not the lady in the bath chair, because she is not staying at your hotel, for I happened to see her come out of a private house this morning assisted by her maid. The two ladies in red breakfasted at my hotel this morning, and as they were not wearing outdoor dress I conclude they are staying there. It therefore rests between the lady in blue and the one with the green parasol. But the left hand that holds the parasol is, you see, ungloved and bears no weddingring. Consequently I am driven to the conclusion that the lady in blue is the man's wife—and you say this is correct."
Now, as my friend was an artist, and as I thought an amusing puzzle might be devised on the lines of his question, I asked him to make me a drawing according to some directions that I gave him, and I have pleasure in presenting his production to my readers. It will be seen that the picture shows six men and six ladies: Nos. 1, 3, 5, 7, 9, and 11 are ladies, and Nos. 2, 4, 6, 8, 10, and 12 are men. These twelve individuals represent six married couples, all strangers to one another, who, in walking aimlessly about, have got mixed up. But we are only concerned with the man that is wearing a straw hat—Number 10. The puzzle is to find this man's wife. Examine the six ladies carefully, and see if you can determine which one of them it is.
I showed the picture at the time to a few friends, and they expressed very different opinions on the matter. One said, "I don't believe he would marry a girl like Number 7." Another said, "I am sure a nice girl like Number 3 would not marry such a fellow!" Another said, "It must be Number 1, because she has got as far away as possible from the brute!" It was suggested, again, that it must be Number 11, because "he seems to be looking towards her;" but a cynic retorted, "For that very reason, if he is really looking at her, I should say that she is not his wife!"
I now leave the question in the hands of my readers. Which is really Number 10's wife?
The illustration is of necessity considerably reduced from the large scale on which it originally appeared in The Weekly Dispatch (24th May 1903), but it is hoped that the details will be sufficiently clear to allow the reader to derive entertainment from its examination. In any case the solution given will enable him to follow the points with interest.
SOLUTIONS.
1.—A POSTOFFICE PERPLEXITY.
The young lady supplied 5 twopenny stamps, 30 penny stamps, and 8 twopencehalfpenny stamps, which delivery exactly fulfils the conditions and represents a cost of five shillings.
2.—YOUTHFUL PRECOCITY.
The price of the banana must have been one penny farthing. Thus, 960 bananas would cost L5, and 480 sixpences would buy 2,304 bananas.
3.—AT A CATTLE MARKET.
Jakes must have taken 7 animals to market, Hodge must have taken 11, and Durrant must have taken 21. There were thus 39 animals altogether.
4.—THE BEANFEAST PUZZLE.
The cobblers spent 35s., the tailors spent also 35s., the hatters spent 42s., and the glovers spent 21s. Thus, they spent altogether L6,13s., while it will be found that the five cobblers spent as much as four tailors, twelve tailors as much as nine hatters, and six hatters as much as eight glovers.
5.—A QUEER COINCIDENCE.
Puzzles of this class are generally solved in the old books by the tedious process of "working backwards." But a simple general solution is as follows: If there are n players, the amount held by every player at the end will be m(2^n), the last winner must have held m(n + 1) at the start, the next m(2n + 1), the next m(4n + 1), the next m(8n + 1), and so on to the first player, who must have held m(2^{n  1}n + 1).
Thus, in this case, n = 7, and the amount held by every player at the end was 2^7 farthings. Therefore m = 1, and G started with 8 farthings, F with 15, E with 29, D with 57, C with 113, B with 225, and A with 449 farthings.
6.—A CHARITABLE BEQUEST.
There are seven different ways in which the money may be distributed: 5 women and 19 men, 10 women and 16 men, 15 women and 13 men, 20 women and 10 men, 25 women and 7 men, 30 women and 4 men, and 35 women and 1 man. But the last case must not be counted, because the condition was that there should be "men," and a single man is not men. Therefore the answer is six years.
7.—THE WIDOW'S LEGACY.
The widow's share of the legacy must be L205, 2s. 6d. and 10/13 of a penny.
8.—INDISCRIMINATE CHARITY
The gentleman must have had 3s. 6d. in his pocket when he set out for home.
9.—THE TWO AEROPLANES.
The man must have paid L500 and L750 for the two machines, making together L1,250; but as he sold them for only L1,200, he lost L50 by the transaction.
10.—BUYING PRESENTS.
Jorkins had originally L19, 18s. in his pocket, and spent L9, 19s.
11.—THE CYCLISTS' FEAST.
There were ten cyclists at the feast. They should have paid 8s. each; but, owing to the departure of two persons, the remaining eight would pay 10s. each.
12.—A QUEER THING IN MONEY.
The answer is as follows: L44,444, 4s. 4d. = 28, and, reduced to pence, 10,666,612=28.
It is a curious little coincidence that in the answer 10,666,612 the four central figures indicate the only other answer, L66, 6s. 6d.
13.—A NEW MONEY PUZZLE.
The smallest sum of money, in pounds, shillings, pence, and farthings, containing all the nine digits once, and once only, is L2,567, 18s. 93/4d.
14.—SQUARE MONEY.
The answer is 11/2d. and 3d. Added together they make 41/2d., and 11/2d. multiplied by 3 is also 41/2d.
15.—POCKET MONEY.
The largest possible sum is 15s. 9d., composed of a crown and a halfcrown (or three halfcrowns), four florins, and a threepenny piece.
16.—THE MILLIONAIRE'S PERPLEXITY.
The answer to this quite easy puzzle may, of course, be readily obtained by trial, deducting the largest power of 7 that is contained in one million dollars, then the next largest power from the remainder, and so on. But the little problem is intended to illustrate a simple direct method. The answer is given at once by converting 1,000,000 to the septenary scale, and it is on this subject of scales of notation that I propose to write a few words for the benefit of those who have never sufficiently considered the matter.
Our manner of figuring is a sort of perfected arithmetical shorthand, a system devised to enable us to manipulate numbers as rapidly and correctly as possible by means of symbols. If we write the number 2,341 to represent two thousand three hundred and fortyone dollars, we wish to imply 1 dollar, added to four times 10 dollars, added to three times 100 dollars, added to two times 1,000 dollars. From the number in the units place on the right, every figure to the left is understood to represent a multiple of the particular power of 10 that its position indicates, while a cipher (0) must be inserted where necessary in order to prevent confusion, for if instead of 207 we wrote 27 it would be obviously misleading. We thus only require ten figures, because directly a number exceeds 9 we put a second figure to the left, directly it exceeds 99 we put a third figure to the left, and so on. It will be seen that this is a purely arbitrary method. It is working in the denary (or ten) scale of notation, a system undoubtedly derived from the fact that our forefathers who devised it had ten fingers upon which they were accustomed to count, like our children of today. It is unnecessary for us ordinarily to state that we are using the denary scale, because this is always understood in the common affairs of life.
But if a man said that he had 6,553 dollars in the septenary (or seven) scale of notation, you will find that this is precisely the same amount as 2,341 in our ordinary denary scale. Instead of using powers of ten, he uses powers of 7, so that he never needs any figure higher than 6, and 6,553 really stands for 3, added to five times 7, added to five times 49, added to six times 343 (in the ordinary notation), or 2,341. To reverse the operation, and convert 2,341 from the denary to the septenary scale, we divide it by 7, and get 334 and remainder 3; divide 334 by 7, and get 47 and remainder 5; and so keep on dividing by 7 as long as there is anything to divide. The remainders, read backwards, 6, 5, 5, 3, give us the answer, 6,553.
Now, as I have said, our puzzle may be solved at once by merely converting 1,000,000 dollars to the septenary scale. Keep on dividing this number by 7 until there is nothing more left to divide, and the remainders will be found to be 11333311 which is 1,000,000 expressed in the septenary scale. Therefore, 1 gift of 1 dollar, 1 gift of 7 dollars, 3 gifts of 49 dollars, 3 gifts of 343 dollars, 3 gifts of 2,401 dollars, 3 gifts of 16,807 dollars, 1 gift of 117,649 dollars, and one substantial gift of 823,543 dollars, satisfactorily solves our problem. And it is the only possible solution. It is thus seen that no "trials" are necessary; by converting to the septenary scale of notation we go direct to the answer.
17.—THE PUZZLING MONEY BOXES.
The correct answer to this puzzle is as follows: John put into his moneybox two double florins (8s.), William a halfsovereign and a florin (12s.), Charles a crown (5s.), and Thomas a sovereign (20s.). There are six coins in all, of a total value of 45s. If John had 2s. more, William 2s. less, Charles twice as much, and Thomas half as much as they really possessed, they would each have had exactly 10s.
18.—THE MARKET WOMEN.
The price received was in every case 105 farthings. Therefore the greatest number of women is eight, as the goods could only be sold at the following rates: 105 lbs. at 1 farthing, 35 at 3, 21 at 5, 15 at 7, 7 at 15, 5 at 21, 3 at 35, and 1 lb. at 105 farthings.
19.—THE NEW YEAR'S EVE SUPPERS.
The company present on the occasion must have consisted of seven pairs, ten single men, and one single lady. Thus, there were twentyfive persons in all, and at the prices stated they would pay exactly L5 together.
20.—BEEF AND SAUSAGES.
The lady bought 48 lbs. of beef at 2s., and the same quantity of sausages at 1s. 6d., thus spending L8, 8s. Had she bought 42 lbs. of beef and 56 lbs. of sausages she would have spent L4, 4s. on each, and have obtained 98 lbs. instead of 96 lbs.—a gain in weight of 2 lbs.
21.—A DEAL IN APPLES.
I was first offered sixteen apples for my shilling, which would be at the rate of ninepence a dozen. The two extra apples gave me eighteen for a shilling, which is at the rate of eightpence a dozen, or one penny a dozen less than the first price asked.
22.—A DEAL IN EGGS.
The man must have bought ten eggs at fivepence, ten eggs at one penny, and eighty eggs at a halfpenny. He would then have one hundred eggs at a cost of eight shillings and fourpence, and the same number of eggs of two of the qualities.
23.—THE CHRISTMASBOXES.
The distribution took place "some years ago," when the fourpennypiece was in circulation. Nineteen persons must each have received nineteen pence. There are five different ways in which this sum may have been paid in silver coins. We need only use two of these ways. Thus if fourteen men each received four fourpennypieces and one threepennypiece, and five men each received five threepennypieces and one fourpennypiece, each man would receive nineteen pence, and there would be exactly one hundred coins of a total value of L1, 10s. 1d.
24.—A SHOPPING PERPLEXITY.
The first purchase amounted to 1s. 53/4d., the second to 1s. 111/2d., and together they make 3s. 51/4d. Not one of these three amounts can be paid in fewer than six current coins of the realm.
25.—CHINESE MONEY.
As a chingchang is worth twopence and fourfifteenths of a chingchang, the remaining elevenfifteenths of a chingchang must be worth twopence. Therefore eleven chingchangs are worth exactly thirty pence, or half a crown. Now, the exchange must be made with seven roundholed coins and one squareholed coin. Thus it will be seen that 7 roundholed coins are worth sevenelevenths of 15 chingchangs, and 1 squareholed coin is worth oneeleventh of 16 chingchangs—that is, 77 rounds equal 105 chingchangs and 11 squares equal 16 chingchangs. Therefore 77 rounds added to 11 squares equal 121 chingchangs; or 7 rounds and 1 square equal 11 chingchangs, or its equivalent, half a crown. This is more simple in practice than it looks here.
26.—THE JUNIOR CLERKS' PUZZLE.
Although Snoggs's reason for wishing to take his rise at L2, 10s. halfyearly did not concern our puzzle, the fact that he was duping his employer into paying him more than was intended did concern it. Many readers will be surprised to find that, although Moggs only received L350 in five years, the artful Snoggs actually obtained L362, 10s. in the same time. The rest is simplicity itself. It is evident that if Moggs saved L87, 10s. and Snoggs L181, 5s., the latter would be saving twice as great a proportion of his salary as the former (namely, onehalf as against onequarter), and the two sums added together make L268, 15s.
27.—GIVING CHANGE.
The way to help the American tradesman out of his dilemma is this. Describing the coins by the number of cents that they represent, the tradesman puts on the counter 50 and 25; the buyer puts down 100, 3, and 2; the stranger adds his 10, 10, 5, 2, and 1. Now, considering that the cost of the purchase amounted to 34 cents, it is clear that out of this pooled money the tradesman has to receive 109, the buyer 71, and the stranger his 28 cents. Therefore it is obvious at a glance that the 100piece must go to the tradesman, and it then follows that the 50piece must go to the buyer, and then the 25piece can only go to the stranger. Another glance will now make it clear that the two 10cent pieces must go to the buyer, because the tradesman now only wants 9 and the stranger 3. Then it becomes obvious that the buyer must take the 1 cent, that the stranger must take the 3 cents, and the tradesman the 5, 2, and 2. To sum up, the tradesman takes 100, 5, 2, and 2; the buyer, 50, 10, 10, and 1; the stranger, 25 and 3. It will be seen that not one of the three persons retains any one of his own coins.
28.—DEFECTIVE OBSERVATION.
Of course the date on a penny is on the same side as Britannia—the "tail" side. Six pennies may be laid around another penny, all flat on the table, so that every one of them touches the central one. The number of threepennypieces that may be laid on the surface of a halfcrown, so that no piece lies on another or overlaps the edge of the halfcrown, is one. A second threepennypiece will overlap the edge of the larger coin. Few people guess fewer than three, and many persons give an absurdly high number.
29.—THE BROKEN COINS.
If the three broken coins when perfect were worth 253 pence, and are now in their broken condition worth 240 pence, it should be obvious that 13/253 of the original value has been lost. And as the same fraction of each coin has been broken away, each coin has lost 13/253 of its original bulk.
30.—TWO QUESTIONS IN PROBABILITIES.
In tossing with the five pennies all at the same time, it is obvious that there are 32 different ways in which the coins may fall, because the first coin may fall in either of two ways, then the second coin may also fall in either of two ways, and so on. Therefore five 2's multiplied together make 32. Now, how are these 32 ways made up? Here they are:—
(a) 5 heads 1 way (b) 5 tails 1 way (c) 4 heads and 1 tail 5 ways (d) 4 tails and 1 head 5 ways (e) 3 heads and 2 tails 10 ways (f) 3 tails and 2 heads 10 ways
Now, it will be seen that the only favourable cases are a, b, c, and d—12 cases. The remaining 20 cases are unfavourable, because they do not give at least four heads or four tails. Therefore the chances are only 12 to 20 in your favour, or (which is the same thing) 3 to 5. Put another way, you have only 3 chances out of 8.
The amount that should be paid for a draw from the bag that contains three sovereigns and one shilling is 15s. 3d. Many persons will say that, as one's chances of drawing a sovereign were 3 out of 4, one should pay threefourths of a pound, or 15s., overlooking the fact that one must draw at least a shilling—there being no blanks.
31.—DOMESTIC ECONOMY.
Without the hint that I gave, my readers would probably have been unanimous in deciding that Mr. Perkins's income must have been L1,710. But this is quite wrong. Mrs. Perkins says, "We have spent a third of his yearly income in rent," etc., etc.—that is, in two years they have spent an amount in rent, etc., equal to onethird of his yearly income. Note that she does not say that they have spent each year this sum, whatever it is, but that during the two years that amount has been spent. The only possible answer, according to the exact reading of her words, is, therefore, that his income was L180 per annum. Thus the amount spent in two years, during which his income has amounted to L360, will be L60 in rent, etc., L90 in domestic expenses, L20 in other ways, leaving the balance of L190 in the bank as stated.
32.—THE EXCURSION TICKET PUZZLE.
Nineteen shillings and ninepence may be paid in 458,908,622 different ways.
I do not propose to give my method of solution. Any such explanation would occupy an amount of space out of proportion to its interest or value. If I could give within reasonable limits a general solution for all money payments, I would strain a point to find room; but such a solution would be extremely complex and cumbersome, and I do not consider it worth the labour of working out.
Just to give an idea of what such a solution would involve, I will merely say that I find that, dealing only with those sums of money that are multiples of threepence, if we only use bronze coins any sum can be paid in (n + 1) squared ways where n always represents the number of pence. If threepennypieces are admitted, there are
2n cubed + 15n squared + 33n —————————— + 1 ways. 18
If sixpences are also used there are
n^{4} + 22n cubed + 159n squared + 414n + 216 ———————————————— 216
ways, when the sum is a multiple of sixpence, and the constant, 216, changes to 324 when the money is not such a multiple. And so the formulas increase in complexity in an accelerating ratio as we go on to the other coins.
I will, however, add an interesting little table of the possible ways of changing our current coins which I believe has never been given in a book before. Change may be given for a
Farthing in 0 way. Halfpenny in 1 way. Penny in 3 ways. Threepennypiece in 16 ways. Sixpence in 66 ways. Shilling in 402 ways. Florin in 3,818 ways. Halfcrown in 8,709 ways. Double florin in 60,239 ways. Crown in 166,651 ways. Halfsovereign in 6,261,622 ways. Sovereign in 500,291,833 ways.
It is a little surprising to find that a sovereign may be changed in over five hundred million different ways. But I have no doubt as to the correctness of my figures.
33.—A PUZZLE IN REVERSALS.
(i) L13. (2) L23, 19s. 11d. The words "the number of pounds exceeds that of the pence" exclude such sums of money as L2, 16s. 2d. and all sums under L1.
34.—THE GROCER AND DRAPER.
The grocer was delayed half a minute and the draper eight minutes and a half (seventeen times as long as the grocer), making together nine minutes. Now, the grocer took twentyfour minutes to weigh out the sugar, and, with the halfminute delay, spent 24 min. 30 sec. over the task; but the draper had only to make fortyseven cuts to divide the roll of cloth, containing fortyeight yards, into yard pieces! This took him 15 min. 40 sec., and when we add the eight minutes and a half delay we get 24 min. 10 sec., from which it is clear that the draper won the race by twenty seconds. The majority of solvers make fortyeight cuts to divide the roll into fortyeight pieces!
35.—JUDKINS'S CATTLE.
As there were five droves with an equal number of animals in each drove, the number must be divisible by 5; and as every one of the eight dealers bought the same number of animals, the number must be divisible by 8. Therefore the number must be a multiple of 40. The highest possible multiple of 40 that will work will be found to be 120, and this number could be made up in one of two ways—1 ox, 23 pigs, and 96 sheep, or 3 oxen, 8 pigs, and 109 sheep. But the first is excluded by the statement that the animals consisted of "oxen, pigs, and sheep," because a single ox is not oxen. Therefore the second grouping is the correct answer.
36.—BUYING APPLES.
As there were the same number of boys as girls, it is clear that the number of children must be even, and, apart from a careful and exact reading of the question, there would be three different answers. There might be two, six, or fourteen children. In the first of these cases there are ten different ways in which the apples could be bought. But we were told there was an equal number of "boys and girls," and one boy and one girl are not boys and girls, so this case has to be excluded. In the case of fourteen children, the only possible distribution is that each child receives one halfpenny apple. But we were told that each child was to receive an equal distribution of "apples," and one apple is not apples, so this case has also to be excluded. We are therefore driven back on our third case, which exactly fits in with all the conditions. Three boys and three girls each receive 1 halfpenny apple and 2 thirdpenny apples. The value of these 3 apples is one penny and onesixth, which multiplied by six makes sevenpence. Consequently, the correct answer is that there were six children—three girls and three boys.
37.—BUYING CHESTNUTS.
In solving this little puzzle we are concerned with the exact interpretation of the words used by the buyer and seller. I will give the question again, this time adding a few words to make the matter more clear. The added words are printed in italics.
"A man went into a shop to buy chestnuts. He said he wanted a pennyworth, and was given five chestnuts. 'It is not enough; I ought to have a sixth of a chestnut more,' he remarked. 'But if I give you one chestnut more,' the shopman replied, 'you will have fivesixths too many.' Now, strange to say, they were both right. How many chestnuts should the buyer receive for half a crown?"
The answer is that the price was 155 chestnuts for half a crown. Divide this number by 30, and we find that the buyer was entitled to 5+1/6 chestnuts in exchange for his penny. He was, therefore, right when he said, after receiving five only, that he still wanted a sixth. And the salesman was also correct in saying that if he gave one chestnut more (that is, six chestnuts in all) he would be giving fivesixths of a chestnut in excess.
38.—THE BICYCLE THIEF.
People give all sorts of absurd answers to this question, and yet it is perfectly simple if one just considers that the salesman cannot possibly have lost more than the cyclist actually stole. The latter rode away with a bicycle which cost the salesman eleven pounds, and the ten pounds "change;" he thus made off with twentyone pounds, in exchange for a worthless bit of paper. This is the exact amount of the salesman's loss, and the other operations of changing the cheque and borrowing from a friend do not affect the question in the slightest. The loss of prospective profit on the sale of the bicycle is, of course, not direct loss of money out of pocket.
39.—THE COSTERMONGER'S PUZZLE.
Bill must have paid 8s. per hundred for his oranges—that is, 125 for 10s. At 8s. 4d. per hundred, he would only have received 120 oranges for 10s. This exactly agrees with Bill's statement.
40.—MAMMA'S AGE.
The age of Mamma must have been 29 years 2 months; that of Papa, 35 years; and that of the child, Tommy, 5 years 10 months. Added together, these make seventy years. The father is six times the age of the son, and, after 23 years 4 months have elapsed, their united ages will amount to 140 years, and Tommy will be just half the age of his father.
41.—THEIR AGES.
The gentleman's age must have been 54 years and that of his wife 45 years.
42.—THE FAMILY AGES.
The ages were as follows: Billie, 31/2 years; Gertrude, 13/4 year; Henrietta, 51/4 years; Charlie, 101/2; years; and Janet, 21 years.
43.—MRS. TIMPKINS'S AGE.
The age of the younger at marriage is always the same as the number of years that expire before the elder becomes twice her age, if he was three times as old at marriage. In our case it was eighteen years afterwards; therefore Mrs. Timpkins was eighteen years of age on the weddingday, and her husband fiftyfour.
44.—A CENSUS PUZZLE.
Miss Ada Jorkins must have been twentyfour and her little brother Johnnie three years of age, with thirteen brothers and sisters between. There was a trap for the solver in the words "seven times older than little Johnnie." Of course, "seven times older" is equal to eight times as old. It is surprising how many people hastily assume that it is the same as "seven times as old." Some of the best writers have committed this blunder. Probably many of my readers thought that the ages 241/2 and 31/2 were correct.
45.—MOTHER AND DAUGHTER.
In four and a half years, when the daughter will be sixteen years and a half and the mother fortynine and a half years of age.
46.—MARY AND MARMADUKE.
Marmaduke's age must have been twentynine years and twofifths, and Mary's nineteen years and threefifths. When Marmaduke was aged nineteen and threefifths, Mary was only nine and fourfifths; so Marmaduke was at that time twice her age.
47.—ROVER'S AGE.
Rover's present age is ten years and Mildred's thirty years. Five years ago their respective ages were five and twentyfive. Remember that we said "four times older than the dog," which is the same as "five times as old." (See answer to No. 44.)
48.—CONCERNING TOMMY'S AGE.
Tommy Smart's age must have been nine years and threefifths. Ann's age was sixteen and fourfifths, the mother's thirtyeight and twofifths, and the father's fifty and twofifths.
49.—NEXTDOOR NEIGHBOURS.
Mr. Jupp 39, Mrs. Jupp 34, Julia 14, and Joe 13; Mr. Simkin 42; Mrs. Simkin 40; Sophy 10; and Sammy 8.
50.—THE BAG OF NUTS.
It will be found that when Herbert takes twelve, Robert and Christopher will take nine and fourteen respectively, and that they will have together taken thirtyfive nuts. As 35 is contained in 770 twentytwo times, we have merely to multiply 12, 9, and 14 by 22 to discover that Herbert's share was 264, Robert's 198, and Christopher's 308. Then, as the total of their ages is 171/2 years or half the sum of 12, 9, and 14, their respective ages must be 6, 41/2, and 7 years.
51.—HOW OLD WAS MARY?
The age of Mary to that of Ann must be as 5 to 3. And as the sum of their ages was 44, Mary was 271/2 and Ann 161/2. One is exactly 11 years older than the other. I will now insert in brackets in the original statement the various ages specified: "Mary is (271/2) twice as old as Ann was (133/4) when Mary was half as old (243/4) as Ann will be (491/2) when Ann is three times as old (491/2) as Mary was (161/2) when Mary was (161/2) three times as old as Ann (51/2)." Now, check this backwards. When Mary was three times as old as Ann, Mary was 161/2 and Ann 51/2 (11 years younger). Then we get 491/2 for the age Ann will be when she is three times as old as Mary was then. When Mary was half this she was 243/4. And at that time Ann must have been 133/4 (11 years younger). Therefore Mary is now twice as old—271/2, and Ann 11 years younger—161/2.
52.—QUEER RELATIONSHIPS.
If a man marries a woman, who dies, and he then marries his deceased wife's sister and himself dies, it may be correctly said that he had (previously) married the sister of his widow.
The youth was not the nephew of Jane Brown, because he happened to be her son. Her surname was the same as that of her brother, because she had married a man of the same name as herself.
53.—HEARD ON THE TUBE RAILWAY.
The gentleman was the second lady's uncle.
54.—A FAMILY PARTY.
The party consisted of two little girls and a boy, their father and mother, and their father's father and mother.
55.—A MIXED PEDIGREE.
[Illustration:
Thos. Bloggs m . . . . . + W. Snoggs m Kate Bloggs. . . m Henry Bloggs. Joseph Bloggs m +  Jane John Alf. Mary Bloggs m Snoggs Snoggs m Bloggs
]
The letter m stands for "married." It will be seen that John Snoggs can say to Joseph Bloggs, "You are my father's brotherinlaw, because my father married your sister Kate; you are my brother's fatherinlaw, because my brother Alfred married your daughter Mary; and you are my fatherinlaw's brother, because my wife Jane was your brother Henry's daughter."
56.—WILSON'S POSER.
If there are two men, each of whom marries the mother of the other, and there is a son of each marriage, then each of such sons will be at the same time uncle and nephew of the other. There are other ways in which the relationship may be brought about, but this is the simplest.
57.—WHAT WAS THE TIME?
The time must have been 9.36 p.m. A quarter of the time since noon is 2 hr. 24 min., and a half of the time till noon next day is 7 hr. 12 min. These added together make 9 hr. 36 min.
58.—A TIME PUZZLE.
Twentysix minutes.
59.—A PUZZLING WATCH.
If the 65 minutes be counted on the face of the same watch, then the problem would be impossible: for the hands must coincide every 65+5/11 minutes as shown by its face, and it matters not whether it runs fast or slow; but if it is measured by true time, it gains 5/11 of a minute in 65 minutes, or 60/143 of a minute per hour.
60.—THE WAPSHAW'S WHARF MYSTERY.
There are eleven different times in twelve hours when the hour and minute hands of a clock are exactly one above the other. If we divide 12 hours by 11 we get 1 hr. 5 min. 27+3/11 sec., and this is the time after twelve o'clock when they are first together, and also the time that elapses between one occasion of the hands being together and the next. They are together for the second time at 2 hr. 10 min. 54+6/11 sec. (twice the above time); next at 3 hr. 16 min. 21+9/11 sec.; next at 4 hr. 21 min. 49+1/11 sec. This last is the only occasion on which the two hands are together with the second hand "just past the fortyninth second." This, then, is the time at which the watch must have stopped. Guy Boothby, in the opening sentence of his Across the World for a Wife, says, "It was a cold, dreary winter's afternoon, and by the time the hands of the clock on my mantelpiece joined forces and stood at twenty minutes past four, my chambers were wellnigh as dark as midnight." It is evident that the author here made a slip, for, as we have seen above, he is 1 min. 49+1/11 sec. out in his reckoning.
61.—CHANGING PLACES.
There are thirtysix pairs of times when the hands exactly change places between three p.m. and midnight. The number of pairs of times from any hour (n) to midnight is the sum of 12  (n + 1) natural numbers. In the case of the puzzle n = 3; therefore 12  (3 + 1) = 8 and 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36, the required answer.
The first pair of times is 3 hr. 21+57/143 min. and 4 hr. 16+112/143 min., and the last pair is 10 hr. 59+83/143 min. and 11 hr. 54+138/143 min. I will not give all the remainder of the thirtysix pairs of times, but supply a formula by which any of the sixtysix pairs that occur from midday to midnight may be at once found:—
720b + 60a 720a + 60b min. a hr ————— min. and b hr. ——————— 143 143
For the letter a may be substituted any hour from 0, 1, 2, 3 up to 10 (where nought stands for 12 o'clock midday); and b may represent any hour, later than a, up to 11.
By the aid of this formula there is no difficulty in discovering the answer to the second question: a = 8 and b = 11 will give the pair 8 hr. 58+106/143 min. and 11 hr. 44+128/143 min., the latter being the time when the minute hand is nearest of all to the point IX—in fact, it is only 15/143 of a minute distant.
Readers may find it instructive to make a table of all the sixtysix pairs of times when the hands of a clock change places. An easy way is as follows: Make a column for the first times and a second column for the second times of the pairs. By making a = 0 and b = 1 in the above expressions we find the first case, and enter hr. 5+5/143 min. at the head of the first column, and 1 hr. 0+60/143 min. at the head of the second column. Now, by successively adding 5+5/143 min. in the first, and 1 hr. 0+60/143 min. in the second column, we get all the eleven pairs in which the first time is a certain number of minutes after nought, or midday. Then there is a "jump" in the times, but you can find the next pair by making a = 1 and b = 2, and then by successively adding these two times as before you will get all the ten pairs after 1 o'clock. Then there is another "jump," and you will be able to get by addition all the nine pairs after 2 o'clock. And so on to the end. I will leave readers to investigate for themselves the nature and cause of the "jumps." In this way we get under the successive hours, 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66 pairs of times, which result agrees with the formula in the first paragraph of this article.
Some time ago the principal of a Civil Service Training College, who conducts a "Civil Service Column" in one of the periodicals, had the query addressed to him, "How soon after XII o'clock will a clock with both hands of the same length be ambiguous?" His first answer was, "Some time past one o'clock," but he varied the answer from issue to issue. At length some of his readers convinced him that the answer is, "At 5+5/143 min. past XII;" and this he finally gave as correct, together with the reason for it that at that time the time indicated is the same whichever hand you may assume as hour hand!
62.—THE CLUB CLOCK.
The positions of the hands shown in the illustration could only indicate that the clock stopped at 44 min. 51+1143/1427 sec. after eleven o'clock. The second hand would next be "exactly midway between the other two hands" at 45 min. 52+496/1427 sec. after eleven o'clock. If we had been dealing with the points on the circle to which the three hands are directed, the answer would be 45 min. 22+106/1427 sec. after eleven; but the question applied to the hands, and the second hand would not be between the others at that time, but outside them.
63.—THE STOPWATCH.
The time indicated on the watch was 5+5/11 min. past 9, when the second hand would be at 27+3/11 sec. The next time the hands would be similar distances apart would be 54+6/11 min. past 2, when the second hand would be at 32+8/11 sec. But you need only hold the watch (or our previous illustration of it) in front of a mirror, when you will see the second time reflected in it! Of course, when reflected, you will read XI as I, X as II, and so on.
64.—THE THREE CLOCKS.
As a mere arithmetical problem this question presents no difficulty. In order that the hands shall all point to twelve o'clock at the same time, it is necessary that B shall gain at least twelve hours and that C shall lose twelve hours. As B gains a minute in a day of twentyfour hours, and C loses a minute in precisely the same time, it is evident that one will have gained 720 minutes (just twelve hours) in 720 days, and the other will have lost 720 minutes in 720 days. Clock A keeping perfect time, all three clocks must indicate twelve o'clock simultaneously at noon on the 720th day from April 1, 1898. What day of the month will that be?
I published this little puzzle in 1898 to see how many people were aware of the fact that 1900 would not be a leap year. It was surprising how many were then ignorant on the point. Every year that can be divided by four without a remainder is bissextile or leap year, with the exception that one leap year is cut off in the century. 1800 was not a leap year, nor was 1900. On the other hand, however, to make the calendar more nearly agree with the sun's course, every fourth hundred year is still considered bissextile. Consequently, 2000, 2400, 2800, 3200, etc., will all be leap years. May my readers live to see them. We therefore find that 720 days from noon of April 1, 1898, brings us to noon of March 22, 1900.
65.—THE RAILWAY STATION CLOCK.
The time must have been 43+7/11 min. past two o'clock.
66.—THE VILLAGE SIMPLETON.
The day of the week on which the conversation took place was Sunday. For when the day after tomorrow (Tuesday) is "yesterday," "today" will be Wednesday; and when the day before yesterday (Friday) was "tomorrow," "today" was Thursday. There are two days between Thursday and Sunday, and between Sunday and Wednesday.
67.—AVERAGE SPEED.
The average speed is twelve miles an hour, not twelve and a half, as most people will hastily declare. Take any distance you like, say sixty miles. This would have taken six hours going and four hours returning. The double journey of 120 miles would thus take ten hours, and the average speed is clearly twelve miles an hour.
68.—THE TWO TRAINS.
One train was running just twice as fast as the other.
69.—THE THREE VILLAGES.
Calling the three villages by their initial letters, it is clear that the three roads form a triangle, A, B, C, with a perpendicular, measuring twelve miles, dropped from C to the base A, B. This divides our triangle into two rightangled triangles with a twelvemile side in common. It is then found that the distance from A to C is 15 miles, from C to B 20 miles, and from A to B 25 (that is 9 and 16) miles. These figures are easily proved, for the square of 12 added to the square of 9 equals the square of 15, and the square of 12 added to the square of 16 equals the square of 20.
70.—DRAWING HER PENSION.
The distance must be 63/4 miles.
71.—SIR EDWYN DE TUDOR.
The distance must have been sixty miles. If Sir Edwyn left at noon and rode 15 miles an hour, he would arrive at four o'clock—an hour too soon. If he rode 10 miles an hour, he would arrive at six o'clock—an hour too late. But if he went at 12 miles an hour, he would reach the castle of the wicked baron exactly at five o'clock—the time appointed.
72.—THE HYDROPLANE QUESTION.
The machine must have gone at the rate of seventwentyfourths of a mile per minute and the wind travelled fivetwentyfourths of a mile per minute. Thus, going, the wind would help, and the machine would do twelvetwentyfourths, or half a mile a minute, and returning only twotwentyfourths, or onetwelfth of a mile per minute, the wind being against it. The machine without any wind could therefore do the ten miles in thirtyfour and twosevenths minutes, since it could do seven miles in twentyfour minutes.
73.—DONKEY RIDING.
The complete mile was run in nine minutes. From the facts stated we cannot determine the time taken over the first and second quartermiles separately, but together they, of course, took four and a half minutes. The last two quarters were run in two and a quarter minutes each.
74.—THE BASKET OF POTATOES.
Multiply together the number of potatoes, the number less one, and twice the number less one, then divide by 3. Thus 50, 49, and 99 multiplied together make 242,550, which, divided by 3, gives us 80,850 yards as the correct answer. The boy would thus have to travel 45 miles and fifteensixteenths—a nice little recreation after a day's work.
75.—THE PASSENGER'S FARE.
Mr. Tompkins should have paid fifteen shillings as his correct share of the motorcar fare. He only shared half the distance travelled for L3, and therefore should pay half of thirty shillings, or fifteen shillings.
76.—THE BARREL OF BEER.
Here the digital roots of the six numbers are 6, 4, 1, 2, 7, 9, which together sum to 29, whose digital root is 2. As the contents of the barrels sold must be a number divisible by 3, if one buyer purchased twice as much as the other, we must find a barrel with root 2, 5, or 8 to set on one side. There is only one barrel, that containing 20 gallons, that fulfils these conditions. So the man must have kept these 20 gallons of beer for his own use and sold one man 33 gallons (the 18gallon and 15gallon barrels) and sold the other man 66 gallons (the 16, 19, and 31 gallon barrels).
77.—DIGITS AND SQUARES.
The top row must be one of the four following numbers: 192, 219, 273, 327. The first was the example given.
78.—ODD AND EVEN DIGITS.
As we have to exclude complex and improper fractions and recurring decimals, the simplest solution is this: 79 + 5+1/3 and 84 + 2/6, both equal 84+1/3. Without any use of fractions it is obviously impossible.
79.—THE LOCKERS PUZZLE.
The smallest possible total is 356 = 107 + 249, and the largest sum possible is 981 = 235 + 746, or 657+324. The middle sum may be either 720 = 134 + 586, or 702 = 134 + 568, or 407 = 138 + 269. The total in this case must be made up of three of the figures 0, 2, 4, 7, but no sum other than the three given can possibly be obtained. We have therefore no choice in the case of the first locker, an alternative in the case of the third, and any one of three arrangements in the case of the middle locker. Here is one solution:—
107 134 235 249 586 746 — — — 356 720 981
Of course, in each case figures in the first two lines may be exchanged vertically without altering the total, and as a result there are just 3,072 different ways in which the figures might be actually placed on the locker doors. I must content myself with showing one little principle involved in this puzzle. The sum of the digits in the total is always governed by the digit omitted. 9/9  7/10  5/11 3/12  1/13  8/14  6/15  4/16  2/17  0/18. Whichever digit shown here in the upper line we omit, the sum of the digits in the total will be found beneath it. Thus in the case of locker A we omitted 8, and the figures in the total sum up to 14. If, therefore, we wanted to get 356, we may know at once to a certainty that it can only be obtained (if at all) by dropping the 8.
80.—THE THREE GROUPS.
There are nine solutions to this puzzle, as follows, and no more:—
12 x 483 = 5,796 27 x 198 = 5,346 42 x 138 = 5,796 39 x 186 = 7,254 18 x 297 = 5,346 48 x 159 = 7,632 28 x 157 = 4,396 4 x 1,738 = 6,952 4 x 1,963 = 7,852
The seventh answer is the one that is most likely to be overlooked by solvers of the puzzle.
81.—THE NINE COUNTERS.
In this case a certain amount of mere "trial" is unavoidable. But there are two kinds of "trials"—those that are purely haphazard, and those that are methodical. The true puzzle lover is never satisfied with mere haphazard trials. The reader will find that by just reversing the figures in 23 and 46 (making the multipliers 32 and 64) both products will be 5,056. This is an improvement, but it is not the correct answer. We can get as large a product as 5,568 if we multiply 174 by 32 and 96 by 58, but this solution is not to be found without the exercise of some judgment and patience.
82.—THE TEN COUNTERS.
As I pointed out, it is quite easy so to arrange the counters that they shall form a pair of simple multiplication sums, each of which will give the same product—in fact, this can be done by anybody in five minutes with a little patience. But it is quite another matter to find that pair which gives the largest product and that which gives the smallest product.
Now, in order to get the smallest product, it is necessary to select as multipliers the two smallest possible numbers. If, therefore, we place 1 and 2 as multipliers, all we have to do is to arrange the remaining eight counters in such a way that they shall form two numbers, one of which is just double the other; and in doing this we must, of course, try to make the smaller number as low as possible. Of course the lowest number we could get would be 3,045; but this will not work, neither will 3,405, 3,45O, etc., and it may be ascertained that 3,485 is the lowest possible. One of the required answers is 3,485 x 2 = 6,970, and 6,970 x 1 = 6,970.
The other part of the puzzle (finding the pair with the highest product) is, however, the real knotty point, for it is not at all easy to discover whether we should let the multiplier consist of one or of two figures, though it is clear that we must keep, so far as we can, the largest figures to the left in both multiplier and multiplicand. It will be seen that by the following arrangement so high a number as 58,560 may be obtained. Thus, 915 x 64 = 58,560, and 732 x 80 = 58,560.
83.—DIGITAL MULTIPLICATION.
The solution that gives the smallest possible sum of digits in the common product is 23 x 174 = 58 x 69 = 4,002, and the solution that gives the largest possible sum of digits, 9x654 =18x327=5,886. In the first case the digits sum to 6 and in the second case to 27. There is no way of obtaining the solution but by actual trial.
84.—THE PIERROT'S PUZZLE.
There are just six different solutions to this puzzle, as follows:—
8 multiplied by 473 equals 3784 9 " 351 " 3159 15 " 93 " 1395 21 " 87 " 1287 27 " 81 " 2187 35 " 41 " 1435
It will be seen that in every case the two multipliers contain exactly the same figures as the product.
85.—THE CAB NUMBERS.
The highest product is, I think, obtained by multiplying 8,745,231 by 96—namely, 839,542,176.
Dealing here with the problem generally, I have shown in the last puzzle that with three digits there are only two possible solutions, and with four digits only six different solutions.
These cases have all been given. With five digits there are just twentytwo solutions, as follows:—
3 x 4128 = 12384 3 x 4281 = 12843 3 x 7125 = 21375 3 x 7251 = 21753 2541 x 6 = 15246 651 x 24 = 15624 678 x 42 = 28476 246 x 51 = 12546 57 x 834 = 47538 75 x 231 = 17325 624 x 78 = 48672 435 x 87 = 37845 ——— 9 x 7461 = 67149 72 x 936 = 67392 ——— 2 x 8714 = 17428 2 x 8741 = 17482 65 x 281 = 18265 65 x 983 = 63985 ——— 4973 x 8 = 39784 6521 x 8 = 52168 14 x 926 = 12964 86 x 251 = 21586
Now, if we took every possible combination and tested it by multiplication, we should need to make no fewer than 30,240 trials, or, if we at once rejected the number 1 as a multiplier, 28,560 trials—a task that I think most people would be inclined to shirk. But let us consider whether there be no shorter way of getting at the results required. I have already explained that if you add together the digits of any number and then, as often as necessary, add the digits of the result, you must ultimately get a number composed of one figure. This last number I call the "digital root." It is necessary in every solution of our problem that the root of the sum of the digital roots of our multipliers shall be the same as the root of their product. There are only four ways in which this can happen: when the digital roots of the multipliers are 3 and 6, or 9 and 9, or 2 and 2, or 5 and 8. I have divided the twentytwo answers above into these four classes. It is thus evident that the digital root of any product in the first two classes must be 9, and in the second two classes 4.
Owing to the fact that no number of five figures can have a digital sum less than 15 or more than 35, we find that the figures of our product must sum to either 18 or 27 to produce the root 9, and to either 22 or 31 to produce the root 4. There are 3 ways of selecting five different figures that add up to 18, there are 11 ways of selecting five figures that add up to 27, there are 9 ways of selecting five figures that add up to 22, and 5 ways of selecting five figures that add up to 31. There are, therefore, 28 different groups, and no more, from any one of which a product may be formed. 
