Sussex as a shepherd who had taught himself mathematics and The Canterbury Puzzles, Dudeney's first book, was published in It was. The Moscow Puzzles: Mathematical Recreations This is, quite simply, the best and most popular puzzle book ever published in the Soviet. Tho knew that math could be so cool? Crammed with games, puzzles, and trivia, The Everything" Kids' Math. Puzzles Book puts the fun back into playing.

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Brilliant Maths. Puzzles for 8 - 9 year olds. Revision • Consolidation • Fun The puzzles in this book -designed to reinforce mathematical terms, concepts and. Puzzles in Thought and Logic (Math & Logic Puzzles). Read more Math and Logic Games: A Book of Puzzles and Problems. Read more. PDF | On Jan 1, , Jambulingam Subramani and others published Mathematical Puzzles- puzzles given in this book, which will certainly.

All rights reserved. No part of the material protected by this copyright notice may be reproduced or utilized in any form, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from. To Lois, for love and support; to Martin Gardner, for inspiration; and to the many misguided friends, relatives, colleagues, and reviewers who are praying that the publication of this book will bring about a diminution in my passion for puzzles. Doubt is the vestibule which all must pass before they can enter the temple of wisdom. When we are in doubt and puzzle our the truth by our own exertions, we have gained something that will stay by US and will serve 11S again. But if to avoid the trouble of the search we avail ourselves of the superior information of a friend, such knowledge will not remain with us; we have nor bought, but borrowed it. To appreciate them, and to solve them, it is necessary-but not sufficient-to be comfortable with mathematics.

It should be elementary and easy to stare. After all, to be passed orally it must be easy to remember! If the statement carries an element of surprise, so much the better,. It should boast at least one solution which is elementary and easily convincing,.

The last two points create a tension: The puzzle should have an easy solution, yet not be easy to solve, Like a good riddle, the answer should be hard to find, but easy to appreciate, Of course, in the case of the Unsolved Puzzles in Chapter 12, the difficulty is evident and the last constraint must be forgiven,.

A word on fo nnat , The puzzles are organized into chapters for convenience, classified loosely by mathematical area of statement or solution, The solutions are presented at the end of each chapter except the last ; the end of each solution is marked with a. If there is mformation about the background and source of a puzzle, it is presented here, A puzzle's statement is not repeated at the head of its solution; I want to encourage readers to tackle all the puzzles in each chapter before reading the answers,.

These puzzles are hard, Several existed as unsolved problems before someone came up with the elegant solution you will read here, The Unsolved Puzzles at the end of the book are thus a logical wind-up to the collection, perhaps only slightly harder than the others. You can take pride in any puzzles you solve, and even more in any for which you find better solutions than mine. During [rhese] periods 'of' relaxation after concentrated intellectual activiry, the intuitive mindseems to take OVer and can produce the sudden clarifying insights which give so much joy and delight.

This warm-up chapter contains a variety of puzzles not associated with a particular topic or technique. As is often the case, however, some key insight will put you on the right track. Here's one to get you started:. On a table is a TOW of fifty coins, of various denominations. Alice picks a coin from one of the ends and puts it in her pocket; then Bob chooses a coin from one of the remaining ends, and thealternation continues until Bob pockets the last coin.

Try this yourself with some coins or random numbers , perhaps just 4 or 6 of them instead of 50; it's not obvious how best to play, is it? But then, maybe Alice doesn't need the best strategy.. Here's your chance to set a precedent for yourself by trying to solve this one before reading further. Number the coins from 1 to SO and observe that no matter how Bob plays, Alice can capture all the even-numbered coins, or, if she prefers, all the odd-numbered coins.

One of these choices must at least match the other. This puzzle, passed to me by mathematician Noga Alon, was alleged to have been used by a high-tech company ill! Israel to test job candidates. In general Alice has even better strategies tha: However, if there are 51 coins instead of 50, it is usually Bob the second to play who will.

It seems paradoxical that the parity of the number of coins has such a huge effect on the outcome of this game, in which all of the action takes place at the ends. You're OIl. We'll begin with two puzzles which are a bit less mathematical, then move on to the more serious stuff. Let your imagination be your guide! It was the first day of class and Mrs. Feldman had two identicallooking pupils, Donald and Ronald Bixby, sitting together in the first row. But a check of their records showed that they had the same parents and were born !

How could this be? A downstairs panel contains three on-off switches, one of which controls the lamp ill. Your mission is to do something with the switches, then determine after one trip to the attic which switch is connected to the attic lamp. There's a gasoline crisis, and the fuel stations located ! Prove that if you start at the right station with an empty tank, you can make it all the way around. You are presented with two fuses lengths of string , each of which will bum for exactly 1 minute, but not uniformly along its length.

Can you use them to measure 4S seconds? A large rectangle in the plane is partitioned into smaller rectangles, each of which has either integer height or integer width or both. Prove that the large rectangle also has this property. There is a set of weights on the scales, and on each weight is the name of at least one pupil. On entering the classroom, each pupil moves all the weights carrying his or her name to the opposite side of the scale, Prove that there is some set of pupils that you, the teacher, can let in which will tip the scales to the left,.

Fifty accurate watches lie on a table. Prove that there exists a moment in time when the sum of the distances from the center of the table to the ends of the minute hands is more than the sum of the distances from the center of the table to the centers of the watches.

Alice begins by marking a corner square of an 17, x 17, chessboard; Bob marks an orthogonally adjacent square. Thereafter, Alice and Bob continue alternating, each marking a square adjacent to the last one marked, until no unmarked adjacent square is available at which time the player whose turn it is to play loses. For which n does Alice have a winning strategy?

For which n does she win if the first square marked is instead a neighbor of a corner square? However, one student noticed that if the problem had instead specified that.

An odd number of soldiers are stationed in a field, in such a way that all the pairwise distances are distinct. Each soldier is told to keep an eye on the nearest other soldier.

Let S be the union of k: Suppose S has the property that for every real number d in [0,1], there are two points in S at distance d. Alice and Bob alternately choose numbers from among I, 2, The first to obtain 3 numbers which sum to 15 wins. Does Alice the first to play have a winning strategy? This puzzle swept the world like a flu epidemic about a decade ago; I don't know its original source. It really is impossible to tell which switch is connected to the attic lamp if all you get is one bit of information from your trip to the attic.

However, you can get more information with your hand! Turn on Switches I and 2, wait a few minutes, then turn off Switch 2 before ascending to the attic.

If the bulb is off, but warm, you conclude that Switch 2 is the winner. If you can't reach the bulb, but have enormous patience, you can achieve the same effect by turning Switch 2 on and then waiting a couple of months before turning on Switch 1 and visiting the attic. If the bulb is burnt out, Switch 2 is the culprit. This puzzle has been around for a. The trick is to imagine that you begin at Station 1 say with plenty of fuel, then proceed around the route, emptying each station as you go.

When you rerum to Station 1, you will have the same amount of fuel in your tank as when you started. As you do this, keep track of how much fuel you have left as you pull into each station; suppose that this quantity is minimized at Station k. Then, if you start at Station k with an empty tank, you will not run out of fuel between stations.

Simultaneously light both ends of one fuse and one end of the other; when the first fuse burns out after half a minute , light the other end of the second. When it finishes, 45 seconds have passed.

This and other fuse puzzles seem to have spread like wildfire a. Recreational mathematics expert Dick Hess has put together a miniature volume called Shoelace Clock Puzzles devoted to them; he first heard the one above from Carl Morris of Harvard University. Hess considers multiple fuses shoelaces, for him of various lengths, but lights them only at ends. If you allow rnidfuse ignition and arbitrary dexterity, you can do even more. Bit of a mad scramble at the end, though. You'll need infinitely many matches to get perfect precision.

Coloring the grid squares alternately black and white, as on a chessboard, we see that each small rectangle is exactly half white and half black. The same, consequently, is true for the big rectangle. Hence, the width would have to be integral, Q.

Your author is responsible for the following solution, not found in Wagon's article. Letting E be less tha: Place the lower left point of the big rectangle at the origin. Either there is a green path from the left side of the big rectangle to the right side, or a red path from bottom to top; suppose the former.

Every time the green path crosses a vertical border of the partition, it is at an integral coordinate; thus, the big rectangle has integral width.

Similarly a red path from bottom to top forces integral height. Consider all subsets of students, including the empty set and the full set.

Each weight will be on the left half the time, so the total weight on the left for all these subsets is the same as the total weight on the right. Since the empty set results in a tip to the right, some other set must tip to the left. Considering just one watch, we claim that during the passing of an hour, the average distance from the center 0 of the table to the tip M of the minute hand exceeds the distance from C to the center W of the watch, This is so because if we draw a.

Of course, if we sum over all of the watches we reach the same conclusion, and it follows that sometime during the hOUT the desired inequality is achieved, Q. The requirement that the watches be accurate is to ensure that each minute hand moves at constant speed, It doesn't matter if those speeds differ, unless our patience is limited to one hour,. One additional note: If you set and place the watches carefully, you carl ensure that the sum of the distances from the center of the table to the ends of the minute hands IS always strictly greater than the sum of the distances from the center of the table to the center of the watches,.

If 11, is even, Bob has ai simple winning strategy no matter where Alice starts. He merely imagines a covering of the board by dorninoes, each domino covering two adjacent squares of the board, Bob then plays in the other half of each domino started by Alice, Note that this works for Bob even when Alice is allowed to mark any square she wants at each movel. When 11, is odd, and Alice begins in a corner, she wins by imagining a domino tiling that covers every square except the comer in which she starts.

However, Alice loses in the odd n case when she must start in the square adjacent to the comer. Suppose the comer squares are black in a checkerboard coloring, so that her starting square is white. There is a domino tiling of the whole board minus one black square; Bob wins by completing these dominoes. Alice can never mark the one uncovered square because all the squares she marks are white.

To show the former, we name the sequence 31, This is easy because. To get the bound, observe that if we replace the top v2 in any s, by the larger value 2, the whole expression collapses to 2. Now that we know the hmit exists, let us call it v;. This problem, from the 6th All Soviet Union Mathematical Competition in Voronezh, , is most easily solved by considering the two soldiers at shortest distance from each other.

Each of these watches the other; if anyone else is watching one of them, then we have a soldier being watched twice and therefore another not being watched at all.

Otherwise, these two can be removed without affecting the others. Since the number of soldiers is odd, this procedure would eventually reduce to one soldier not watching anyone, a contradiction.

Suppose the lengths of the intervals in S are S 1, Summing over all pairs of intervals, each Si appears k-l times, so the total measure of the distances obtainable by choosing points from two different intervals is at most k -1 5. Distances obtainable by taking two points from the same interval run from 0 only up to the maximum of the lengths Si, so altogether the measure of the distances is at most ks; from ks The argument is tight only if the maxiimum Si is equal to 5, i.

The quick way to solve the puzzle is to imagine that Alice and Bob are playing on the following magic square:. Since it is exactly the rows, columns, and main diagonals which sum to 15, they are playing Tic- Tac- Toe! Everyone knows that best play in Tic-Tao-Toe leads to a draw, so the answer to OUT question is no, Alice does not have a winning strategy,.

Pericoloso Sporgersi, but, rather suspiciously, the phrase is found also on Italian railroad trains, warning passengers not to lean out the wmdow. Numbers are an endless source of fascination, and for some, a lifelong disease. Some people can be captured even by the properties of particular numbers; many intriguing puzzles have been concocted to take advantage of special properties, often by requiring deductions from what a. The spirit of this collection, however, suggests striving for greater universality.

Our number-theoretic problems are about numbers in general, not particular ones. In most cases, you will need little more to solve them than the fact that every positive integer is uniquely expressible as the product of powers of primes. Lockers numbered 1 to lOO stand in a row at the school gym. When the first student arrives, she opens all the lockers. The second student then goes through and recloses all the even-numbered lockers; the third student changes the state of every locker whose number is a multiple of 3.

The state of locker n is changed when the kth student passes through, for every divisor k of n. The exception is when n is a perfect square, in which case there is no other divisor to cancel the effect of the ynth student; therefore, the lockers which. We'll start with a couple of observations about base 10 representation of integers, and finish with a surprisingly subtle dinner table conundrum.

Let n be a natural number, Prove that a n has a nonzero multiple whose representation base 10 contains only zeroes and ones: Given 25 different positive numbers, prove that you can choose two of them such that none of the other numbers equals either their sum Or their difference,.

A set S contains 0 and 1, and the mean of every finite nonernpty subset of S, Prove that Scontanns all the rational numbers in the unit interval. At the stockholders' meeting, the board presents the month-bymonth profits or losses since the last meeting. O, "that we made a profit over every consecutive eight-month period. Each number from 1 to is written out in formal English e. What's the first odd number in the list?

For Part a , we employ the famous and useful "pigeon-hole principle": If there are more pigeons than holes, then some hole must contain at least two pigeons. Subtract one from the other!

As pointed out to me by David Gale, as long as n is not a multiple of 2 or 5, you can even find a.

The reason is that the above argument produces a multiple of n of the form III For Part b , it's perhaps easiest to show by induction on k that there is a k. A similar problem, from the l st All Soviet Union Mathematical Competi tion, Thilisi, , asks for a proof that there exists a number divisible by 5l JO not containing any 0 in its decimal representation, One approach is to assume otherwise and let k be the biggest power of 5 available, Let n have the most factors of 5 say, j:: A substitute high school math teacher at Fair Lawn Senior High School, N ew Jersey, told me that some WWII prisoner of war entertained himself by trying various sequences of four numbers to see how long he could get them to survive under the above operations.

Since this covers all cases, we see that when working with ordinary integers, at most four steps are required to make all the numbers even; at that point, we may as well divide out by the largest common power of two before proceeding.

A little analysis via polynomials over the integers modulo 2 shows that the salient issue is whether n is a power of 2. If the restriction to integer entries is relaxed, there is, incredibly, a unique up to the obvious alterations sequence of fOUT positive reals which fails to terminate, as shown recently by Antonio Behn, Chris Kribs-Zaleta, and Vadim Ponornarenko.

This puzzle is adapted from one which appeared on the International Mathematical Olympiad, submitted by a Vietnamese. What's needed, of course, is a maximum-length sequence of numbers such that every substring of length 8 adds up to more than 0, but every substring of length 5 adds up to less than 0, The string must certainly be finite, in fact less than 40 in length, else you could express the sum of the first 40 entries both as the positive sum of 5 substrings of length 8 and the negative sum of 8 substrings of length 5,.

Then you can have a string of length x itself, with entries that alternate between, say, x-I and -x. But there are two z-substrings and together they imply that the middle two numbers are both positive, a contradiction.

Observe that any consecutive b of them can be expressed as an x" substring of the full string, with a y"substrings removed; therefore, it has positi ve sum. It follows that feb, y - b 2'. It takes only two distinct values, and it is periodic with periods both x and y.. Call the two values u and v, and imagine at first that we assign them arbitrarily as the first y entries of OUt string. Then these assignments are repeated until the end of the string, making the string perforce periodic in y, To be periodic in X as well, we only need to ensure that the last y- 2 entries match up with the first y-2, which entails satisfying y-2 equalities among the original y choices we made, Since there are not enough equalities to force all the choices to' be the same, we can ensure that there is at least one u and one v,.

The industrious reader will not find it difficult to generalize the above arguments to the case where x and y have a greatest common divisor gcd x,y other than 1. This is just a matter of carefully and systematically considering the successive words involved in the description of a number, The earliest actual digit is of course "eight," but the earliest available.

The idea for this silly puzzle came when HerbWiilf University of Pennsylvania asked me for the first prime in the dictionary, This question has been attributed to the computer guru Donald Knuth Stanford University , and reasoning as above, followed by some checking on a computer, will lead you to 8,,, If a puzzle begins with "How many ways are there to Combinatorial reasoning is useful in the following quite eclectic list of puzzles and in many other puzzles in this book.

Our practice problem does fit the classical mold, however,and makes use of the most fundamental of combinatorial techniques: How many ways are there to write the numbers 0 through 9 in a row, such that each number other than the left-most is within one of some number to the left of it? On the face of it, this problem does not seem amenable to multiplying numbers of options because the number of options dependson previous choices.

For example, there are ten choices for the left-most digit, but if we start by writing "3" on the left, there are two choices for the next digit; if we start with "0" Or "9," there is only one choice. If you know how to sum binomial coefficients, you can nonetheless analyze the problem in this manner, but there's a better way. Observe that the sequence must terminate with a "0" or "9," and as we move from right to left, we always have a choice between writing the highest unused digit or the lowest-until we hit the left end, of course, where these two choices coincide.

Thus, there are two choices at each of nine opportunities. The rest of the solutions are up to you, One hint Keep your eyes open for more applications of the pigeon-hole principle! Prove that every set of ten distinct numbers between 1 and contains two disjoint nonempty subsets with the same sum,.

At a mathematics conference banquet, 48 male mathematicians, none of them knowledgeableabout table etiquette, find themselves assigned to a big circular table. On the table, between each pair of settings, is a coffee cup containing a cloth napkin. As each person is seated by the maitre d' , he takes a napkin from his left or right; if both napkins are present, he chooses randomly but the maitre d' doesn't get to see which one he chose.

In what order should the seats be filled to maximize the expected number of mathematicians who don't get napkins? Mike and Jenene go to a dinner party with four other couples; each person there shakes hands with everyone he or she doesn't know.

Later, Mike does a survey and discovers that everyone of the nine other attendees shook hands with a different number of people. Ashford, Baxter, and Campbell run for secretary of their union, and finish in a three-way tie.

To break it, they solicit the voters' sec, and choices, but again there is a three-way tie. Ashford now steps forward and notes that, since the number of voters is odd, they can make two-way decisions; he proposes that the voters choose between Baxter and Campbell, and then the winner could face Ash, ford in a runoff.

Baxter complains that this is unfair because it gives Ashford a better chance to win than either of the other two candidates. Is Baxter tight? After the revolution, each of the 66 citizens of a certain country, including the king, has a salary of one dollar.

The king can no longer vote, but he does retain the power to suggest changes-namely, redistribution of salaries. Each person's salary must be a whole number of dollars, and the salaries must sum to Each suggestion is voted on, and carried ifthere are more votes for than against. Each voter can be counted on to vote "yes" if his salary is to be increased, "no" if decreased, and otherwise not to bother voting. The king is both selfish and clever. What is the maximum salary he can obtain for himself; and how long does it take him to get it?

The hour and minute hands of a clock are indistinguishable. How many moments are there fin a day when it is not possible to tell from this clock what time it is? David and Dorothy have devised a clever card trick. While David looks away, a stranger selects five cards from a bridge deck and hands them to Dorothy; she looks them over, pulls one out,and hands the remainder to David.

David now correctly guesses the identity of the pulled card. How do they do it? What's the biggest deck of cards they could use and still perform the trick reliably? Between every pair of major cities in Russia, there's a fixed air travel cost for going from either city to the other. Traveling salesman Alexei Frugal begins in Moscow and tours the cities, always choosing the cheapest flight to a city not yet visited he does not need to return to Moscow.

Salesman Boris Lavish also needs to visit every city, but he starts in Kaliningrad, and his policy is to choose the most expensive flight to an unvisited city at each step. When six dice are rolled, the number of different numbers which can appear can range from 1 to 6. Suppose that once every minute, the croupier rolls six dice and you bet: The trick to this puzzle, based on a. This problem can be traced to a particular event. Princeton mathematician John H.

Conway asked how many diners would be without napkins if they were seated in random order see Chapter 11 , and Pike said: If the maitre d' sees which nap kin is grabbed each time he seats a diner computer theorists would call him an "adaptive adversary" , it is not hard to see that his best strategy is as follows. If the first diner takes say his right napkin, the next is seated two spaces to his right so that the diner in between may be trapped.

If the second diner also takes his right napkin, the maitre d' tries again by skipping another chair to the right. If the second diner takes his left napkin leaving the space between him and the first diner napkinless , the third diner is seated directly to the second diner's right. Further diners are seated according to the same rule until the circle is closed, then the remaining diners some of whom are doomed to be napkinless are seated.

Editorial Reviews About the Author Presh Talwalkar runs the YouTube channel and blog Mind Your Decisions, which features videos and articles in math and game theory that have been viewed millions of times. He studied Economics and Mathematics at Stanford University. Product details File Size: Unlimited Publication Date: December 24, Sold by: English ASIN: Enabled X-Ray: Not Enabled.

Share your thoughts with other customers. Write a customer review. Top Reviews Most recent Top Reviews. There was a problem filtering reviews right now. Please try again later. Paperback Verified Purchase. I really don't care for Sudoku's. I believe Sudoku's are s a good and healthy thing to do for one's self if they need a bit of problem solving engineering in their lives , see the fun and benefit, but really sitting down and being mentally active by problem solving in a sedentary position for so long makes me not too excited.

They might 'care' but really some people just don't like video games or the ones that I play , and they just politely nod. This book has neither of these two issues. It is fun, active, easy to bring up some 'problems' when you are in the break room with your coworkers, talking with students in your classes, or just taking mins to solve a problem with your loved ones. It's fun. Kindle Edition Verified Purchase.

This little book is a treasure of old and new problems, presented in a way that allows me to peruse it on the Tube, and spend some time at home squeezing my brain cells and working out the math rigorously. My favourite is the game theory section, although I will get back to the probability part which I found harder than the rest, but then probability has never been my cuppa. On a recent family vacation to China, I worked through these puzzles with my wife, parents and in-laws. The level of difficulty was just right.

The problems were challenging and it was a lot of fun trying to see who could get the right answer first. To get a sense of what the puzzles are like you should check out Presh's blog at Mind Your Decisions.

Fun and educational. If you like to spend a few days solving tricky math puzzles, this is the book for you! Iam a math teacher and I am always looking for some new. See all 14 reviews.

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