(.0.1937 - 25.2.2022)
(..1906 - 25.2.1964)
(..1883 - 25.2.1924)
Here is a remarkable age problem which I am sure will amuse the young folks and at the same time open up a new line of reasoning for some of the wiseacres who make a specialty of statistical calculations.
It appears that an ingenious orr eccentric teacher, as the case may be, desirous of bringing together a number of older pupils into a class he was forming, offered to give a prize each day to the side of boys or girls whose combined ages would prove to be the greatest.
Well, on the first day there was only one boy and one girl in attendance, and, as the boy's age was just twice that of the girl's, the first day's prize went to the boy.
The next day the girl brought her sister to school. It was found that their combined ages were just twice that of the boy, so the two girls divided the prize.
When school opened the next clay, however, the boy had recruited one of his brothers. It was found that the combined ages of the two boys were exactly twice as much as the ages of the two girls, so the boys carried off the honors that day and divided the prize between them.
The battle waxed warm now between the Jones and Brown families, and on the fourth day the two girls appeared accompanied by their elder sister; so it was then the combined ages of the three girls against the two boys. The girls won of course, once more bringing their ages up to just twice that of the boys. The struggle went on until the class was filled up, but our problem does not need to go further than this point. Tell me the age of that first boy, provided that the last young lady joined the class on her twenty-first birthday.
It is a simple but pretty puzzle, calling for ingenuity rather than mathematics and yielding readily to puzzle methods. (Mathematical Puzzles of Sam Loyd, Edited by Martin Gardner)
The first girl was just 638 days old, and the boy twice as much, namely 1,276 days. The next day the youngest girl will be 639 days old, and her new recruit 1,915 days, total, 2,554 days, which doubles that of the first boy, who having gained one day, will be 1,277 days old. The next day the boy, being 1,278 days old, brings his big brother, who is 3,834 days old, so their combined ages amount to 5,112 days, just twice the ages of the girls who will now be 640 and 1,916, or 2,556.
The next day, the girls gaining one day each, will represent 2,558 days, which added to 7,670 days of the last recruit, brings their sum total to 10,228 days, just twice that of the two boys, which, with the two points added for the last day, would be increased to 5,114 days.
We arrive at the 7,670 days as follows. The young lady having reached her twenty-first birthday, 21 times 365 equals 7,665 plus 4 days for four leap years, and the extra one day which is the day of her twenty-first birthday. Those who gave the boy's age as 3½ years overlooked the feature of increasing the ages of the pupils from day to day.
Here is a pretty problem which I figured out during a ride from Bixley to Quixley astride a razor-back mule. I asked Don Pedro, a native guide who walked ahead of me pulling the mule forward by its reins, if my steed had another gait. He said it had but that it was much slower, so I pursued my journey at uniform speed. To encourage Don Pedro, who was my chief propelling power, I said we would pass through Pixley, so as to get some liquid refreshments; and from that moment he could think of nothing but Pixley.
After we had been traveling for forty minutes I asked how far we had gone. Don Pedro replied: "Just half as far as it is to Pixley." After creeping along for seven miles more I asked: "How far is it to Quixley?" He replied as before: "Just half as far as it is to Pixley."
We arrived at Quixley in another hour, which induces me to ask you to determine the distance from Bixley to Quixley. (Mathematical Puzzles of Sam Loyd, Edited by Martin Gardner)
Loyd's answer makes use of the two time intervals given in the problem, but as Ronald C. Read, of Kingston, Jamaica, points out, these time intervals are not really needed in order to solve the problem. Let x be the point (between Bixley and Pixley) where the first question is asked, and y the point (between Pixley and Quixley) where the second question is asked. The distance from x to y, we are told, is 7 miles. Since the distance from x to Pixley is 2/3 the distance between Bixley and Pixley, and the distance from y to Pixley is 2/3 the distance between Pixley and Quixley, it follows that the distance between x and y, or 7 miles, is 2/3 the total distance. This makes the total distance 10½ miles. - M.G.
To show how difficult it is for the average person to leave the beaten track when thinking out some simple problem, let us take a look at the decimal system of numeration with which we are all familiar. It is safe to say that most people have given little thought to the subject. They see that any column can be built up to 9, but as soon as it gets above 9, it is carried over to the column on the left. They think it is so because it must be so, and can't help itself any more than 1 and 2 can help being 3. But this is far from the case. Primitive man originally learned to calculate upon the fingers of both hands, just as we see many people today utilizing their fingers in some every-day transaction.
Hence the introduction of the decimal system. If the human race, as has been claimed, sprang from the Angwarribo family of monkeys, who have but four fingers, and we had not taken on that extra finger, we would have continued to calculate in what is known as the octamal system.
From a mathematical standpoint, it can be shown that the decimal system is not as perfect as some of the others, and that for some purposes the septamal, which only runs up to 7, is better. In that notation 66 would mean six 7's and six 1's, so the addition of 1 more would change it to 100, which would be equal to only 49 in our decimal notation.
You see, 1 added to the 6 in the unit column would change it to 7, so we would have to place a 0 and carry 1 on to the other 6 which in turn becomes a 7, so we place another 0 and carry the 1 to the third column, making it 100, which stands for 49. In this same way, 222 represents 114 - two units, two 7's and two 49's.
Assuming the octamal system to be the popular notation in the Angwarribo days of our four-fingered ancestors, when they counted up to eight and knew nothing about 9's or 10's, how would you write down the year 1906 so as to show the number of years which have elapsed since the Christian era began? It is a pretty problem which will clear the cobwebs from your brain, and introduce you to some elementary principles involved in converting from one number system to another. (Mathematical Puzzles of Sam Loyd, Edited by Martin Gardner)
In the octamal system 1906 is written 3562, which represents two units, six 8's, five 64's and three 512's. The simple procedure for arriving at this number is first to divide 1906 by 512 to obtain 3. The remainder, 370, is then divided by 64 to obtain 5. The remainder, 50, is divided by 8 to get 6, and the final remainder of 2 is of course the last digit of the answer. Had we wished to convert 1906 to the septamal system we would have followed a similar procedure, dividing by the successive multiples of 7. - M.G.
1. Once when entertaining at a children's birthday party, I went into the kitchen and found three peaches, three plums, and three paper bags. In one hag I put two peaches, in one I put two plums, and in the remaining bag (the "mixed bag "). I put a peach and a plum. I brought the three bags into the living room, where the company was assembled, picked three children named Arthur, Lillian, and Robert, and gave them each a bag. I then explained to the company that one of the three held two peaches, one held two plums, and one held the mixed bag - one peach and one plum - but didn't tell them who held what. I then said tothe three children: "I want each of you to peek into your bag and tell the company what you have, but I want each of you to lie!" Arthur said, "I have two peaches"; Lillian said, "I have two plums"; and Robert said, "I have one peach and one plum."
"Very good," I said. "I see that each of you has lied, as I requested. From now on, I want each of you to tell the truth." I then gave pencils and paper to the company and told them: "You are to figure out a strategy whereby you successively ask each of the three to pull out one fruit that he or she is holding and show it to you, until you can deduce which bag is really the mixed bag.
Remember, the answers you get will now be truthful. The first one to come up with a strategy involving the minimum possible number of necessary questions will be allowed to try it and, if successful, will win a prize." After a time, Violet, a very clever girl, came up with a minimal strategy, and it worked! What is the strategy, and what is the minimum number of necessary questions?
2. After this I gave the group the problem of two men, four miles apart, starting at the same instant, walking toward each other, one at the rate of three miles an hour and the other at the rate of two miles an hour. How far apart will they be two minutes before they meet?
3. Next I gave the problem about a certain trust fund that provided a weekly allowance to a certain club. The allowance was divided equally among the members. One day a new member entered the club. The weekly allowance did not change, but thanks to the new member, each got $5 a week less than formerly (the allowance was still distributed equally among the members). This situation lasted for one year. Then two members dropped out, and as a result each remaining member got $12 a week more than in the year before.
How many members were originally in the club, and how much was the weekly allowance to the club? (DISCOVER Magazine)
1. Only one question is necessary, and that is sufficient to determine the contents of each of the three bags! You should ask Robert (who falsely claimed to have the mixed bag) to show you one of the fruits he has in his bag. Suppose he pulls out and shows you a plum. Then you know that the other fruit in Robert's bag is also a plum (since the bag is not mixed).
Arthur's bag doesn't really have two peaches, as Arthur falsely claimed, nor can it have two plums (since there are only three plums altogether, and Robert has two of them); hence Arthur must have the mixed bag (and hence also Lillian has two peaches). On the other hand, if Robert showed you a peach instead of a plum, then by symmetric reasoning, Robert would have two peaches. Lillian would have the mixed bag, and Arthur would have two plums.
2. First of all, the distance between the two men is decreasing at the rate of 5 miles an hour, which is 1/12 mile per minute, which is 1/6 mile per two minutes. So two minutes before they meet, they will be 1/6 mile apart. (Actually, the distance between them when they first start has no bearing on the answer, except that they should start at least 1/6 mile apart.)
3. a) Let y be the number of people initially in the club, and let x be the number of dollars that each person receives per week. Then xy is the weekly allowance to the club.
b) After the new member joined, there were y + 1 people, each getting x- 5 dollars per week, so (y+1)(x-5)=xy. Thus xy-5y+x-5 =xy, so x-5y-5=0 (a linear equation).
c) Then two members left, so there were y-1 members, each getting (x-5)+12 dollars per week, which is x+7 dollars per week. Thus (y-1)(x+7)=xy, which yields xy+7y-x-7= xy, which in turn yields 7y-x-7=0. Solving this equation simultaneously with the equation x-5y-5=0 (from part b above), we get y=6 and x=35. Thus there were originally six members in the club, each getting $35 a week, totaling a $210-a-week allowance to the club.
Figure The Ancient Figures:
1. Four Romans-Virgil, Catullus, Lucretius, and Julius Caesar - were seated around a table, and each was drinking a different beverage. Use the following clues to figure out who drank which beverage.
a. The man who drank juice sat across from Catullus.
b. The man who sat to the right of Lucretius drank water.
c. Virgil never drank milk.
d. Julius Caesar sat across from Virgil.
e. The man who drank wine sat to the left of the man who drank milk.
2. Five Greek soldiers were marching in single file, and each was carrying a different weapon.
Use the following clues to figure out who carried which weapon and in what order they were marching.
a. Themistocles marched just behind Alexander.
b. Alcibiades was just behind Thucydides and ahead of Philip.
c. The man with the sword was just in front of the man with the spear, but he was not first in line.
d. The last man in line carried neither the bow and arrow nor the slingshot.
e. Alcibiades carried neither the sword nor the sarissa.
f. The man who marched first carried neither the sarissa nor the slingshot.
g. The man carrying the spear came before Alcibiades.
Fill each yellow square with an operation (+,-,x,/) and each green square with a different number (from 1 to 9) so that each row and column makes a true equation.
Every number 1 to 9 will appear just once. Use parentheses when needed.
Figure The Ancient Figures:
1. The Romans and their beverages:
- Virgil drank water.
- Lucretius drank juice.
- Catullus drank wine.
- Julius Caesar drank milk.
2. The soldiers and their weapons:
- Alexander, bow and arrow.
- Themistocles, sword.
- Thucydides, spear.
- Alcibiades, slingshot.
- Philip, sarissa.
Note: The "/" in the middle column could also be an "x".
Pick up two adjacent glasses at a time and in four moves change the positions so that each alternate glass will be empty.
For readers interested in parlor tricks, here is an amusing puzzle which can be used advantageously to amuse the guests after a banquet or at an evening party. In the former case eight wine glasses - four empty and four partially filled - illustrate the trick to perfection.
In this, as in all exhibitions of a similar character, everything depends upon the skill and clever acting of the performer. He must have his little book down to perfection, so as to be able to do the trick forwards or backwards without the slightest hesitation, while by the aid of a ceaseless flow of conversation he impresses upon his hearers the fact of its being the most simple little trick that ever happened, which anyone can do unless he be a natural born muttonhead or hopelessly befuddled.
It really looks so simple that almost anyone will be lured into accepting an invitation to step up and test his sobriety by showing how readily he can perform the feat and then the fun begins - for it will rattle ninety-nine out of a hundred.
The problem is stated below the sketch. The glasses in the picture are numbered to make it easy to describe the correct procedure. (Edited by Martin Gardner)
That odd little sleight-of-hand performance with the four empty and four full glasses can readily be remembered by the following rule: One long move, two short ones, then one long one. First move 2 and 3 to the extreme end; then fill the gap with 5 and 6. Fill gap with 8 and 2; then finish with 1 and 5.
STUDENTS of geometry will find here an interesting elementary problem which can best be solved by experimental puzzle methods, although there is a scientific rule for getting the correct answer which bears a close resemblance to the famous forty-seventh proposition of Euclid. The joiner has a piece of board four feet long by two feet wide, with a corner clipped off. The puzzle is to divide the board into the fewest number of pieces, so that without any waste they will fit together and make a perfect square top for the table shown in the picture.
In this particular case the missing piece has been cut off at an angle of fifteen degrees, but when you solve the puzzle you will find that the cutting procedure will work just as well when this angle is greater or less than shown here.
The best answer requires only two straight cuts and the turning over of one piece, a practical piece of carpentering which some of the followers of Euclid did not think of.
It makes no difference if the angle from D to B be more or less acute. Simply draw a line from the center of the left side to the midle of line BD. Then draw a perpendicular line from corner G to line EC. Turn over piece A and the three pieces will form the square as shown.
Close the sedan chair by cutting it into the smallest number of pieces.
"SPEAKING about modes of conveyance in China," says a writer who has spent most of his life in the Flowery Kingdom, "one soon gets used to being carried around in a sedan chair, which is far more comfortable and expeditious than a hack. These chairs are made of rattan wicker and remind you very much of those little Chinese puzzle boxes made of colored straws and so cleverly put together that you cannot discover where they are joined."
All this is suggestive of a clever puzzle, for those sedan chairs will close up to make a covered box when it rains, yet the closest examination will not detect where the pieces are joined. To illustrate the puzzle, you are asked to cut the chair into the fewest possible pieces which will fit together and form a perfect square.
This is the first of many "dissection" problems included in this collection. It may interest the reader to know that there is a proof by David Hilbert that any polygon can be sliced into a finite number of pieces which can be rearranged to form any other polygon of equal area. Such dissections are of little interest, however, unless the number of required pieces is small enough to make the dissection elegant and surprising.
Almost all simple and regular polygons (except the pentagram or five-pointed star which offers formidable difficulties) have been exploited in dissection puzzles of great ingenuity. For a recent and excellent discussion of dissection theory see a series of articles by the mathematics staff of the college, University of Chicago, in The Mathematics Teacher, May, October, and December, 1956; February and May, 1957.-M.G.
Divide the field into four identical parts, each containing a tree.
THE TOWN of Four Oaks derives its name from the fact that one of the early settlers, who owned a large tract of land, left it to four sons with the stipulation that they "divide it into equal portions, as indicated by the positions of four ancient oaks which had always served as landmarks."
The sons were unable to divide the land amicably, since the four trees really furnished no clue to guide them, so they went to law over it and squandered the entire estate in what was known as the "battle of the four oaks." The person who told me this story thought it might form the groundwork for a good puzzle, which it has done, so far as suggesting a theme is concerned.
The picture represents a square field with four ancient oak trees, equal distance apart, in a row from the center to one side of the field. The property was left to four sons who were instructed to divide the field into four pieces, each of the same shape and size, and so that each piece of land would contain one of the trees. The puzzle is an impromptu one, gotten up on the spur of the moment, so it is really not very difficult. Nevertheless it is safe to say that everyone will not hit upon the best possible answer.
Rearrange the eight pieces to form a perfect checkerboard.
IN the history of France is told an amusing story of how the Dauphin saved himself from an impending checkmate, while playing chess with the Duke of Burgundy, by smashing the chess board into eight pieces over the Duke's head. It is a story often quoted by chess writers to prove that it is not always politic to play to win, and has given rise to a strong line of attack in the game known as the King's gambit.
The smashing of the chess board into eight pieces was the feature which always struck my youthful fancy because it might possibly contain the elements of an important problem. The restriction to eight pieces does not give scope for great difficulty or variety, but not feeling at liberty to depart from historical accuracy, I shall give our puzzlists a simple little problem suitable for summer weather. Show how to put the eight pieces together to form a perfect 8 x 8 checkerboard.
The puzzle is a simple one, given to teach a valuable rule which should be followed in the construction of puzzles of this kind. By giving no two pieces the same shape, other ways of doing the puzzle are prevented, and the feat is much more difficult of accomplishment.
How old is the mother?
AGE PUZZLES are always interesting, and possess a certain fascination for young folks who are at all mathematically inclined. As a rule, they are extremely simple, but in this problem the data is so meager, and the proposition so different from what is expected, that the query actually appears startling. One of the trio in the picture was having a birthday anniversary. This aroused Master Tommy's curiosity regarding their respective ages, and in response to his queries his father said:
"Now, Tommy, our three ages combined amount to just seventy years. As I am just six times as old as you are now, it may be said that when I am but twice as old as you, our three combined ages will be twice what they are at present. Now let me see if you can tell me how old is mother?"
Tommy, being bright at figures, readily solved the problem, but then he had the advantage of knowing his own age and could guess pretty closely the ages of the others. Our puzzlists, however, have merely the data regarding the comparative ages of father and son, followed by the startling question, "How old is mother?"
The mother's age is 29 years and 2 months. Tommy's age is 5 years and 10 months, and the father is 35 years old.
Cut the mosaic into parts which will form two squares.
It is not generally known that the celebrated piece of Venetian mosaic by Domenichino, known as the Guiđo collection of Roman heads, was originally divided into two square groups, discovered at different periods. They were brought together and restored to what is supposed to be their correct form, in 1671. Apparently by accident it was discovered that each of the two squares consisted of pieces which would fit together into one 5 x 5 piece as shown.
It is a pretty puzzle, and as many puzzles, like mathematical propositions, can be worked backward to advantage at times, we will reverse the problem and ask you to divide the large square into the fewest number of pieces which can be refitted into two squares.
This puzzle differs from the Pythagorean principle of cutting lines on the bias. We know that two squares can be divided by diagonal lines to produce one larger square, and vice versa, but in this puzzle we must cut on the lines only, so as not to destroy the heads. It may also be mentioned, incidentally, that students who have mastered the Pythagorean problem will not find much difficulty in discovering how many heads there must be in the smaller squares.
Problems of this kind, which call for the "best" answer in the "fewest number of pieces," offer great scope for cleverness. In this problem the best solution does not destroy any heads or turn any of them upside down. (Mathematical Puzzles of SAM LOYD, Martin Gardner)
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