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.
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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.
What is the probability the two will meet at the bar? When the first student arrives, she opens all the lockers. But then, maybe Alice doesn't need the best strategy.. Oscar Espinoza. This is easy because ;;: Two friends agree to meet up in a bar between midnight and 1 am.
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. This results in of the diners without napkins, on average. When, as in the stated puzzle, the maitre d' is not adaptive, it seemed likely to Pike and me at the time that the right strategy is to fill the even seats, then the odd ones.
However, a little more thought shows that the maitre d's best strategy is to use first the 0 mod 4 seats, then the odd s. To see this, call a diner "lonely" if when seated neither of his neighbors has yet appeared. We may assume all the lonely diners are seated first; note that there will be at most one napkinless diner between each consecutive pair of lonely diners. Suppose two consecutive lonely diners are at distance d i.
His tight napkin will be gone unless the lonely. This puzzle is an old chestnut and of the type that, at first glance, seems to offer insufficient information; why should we be able to deduce anything about Jenenel The answer is, ultimately, that Jenene is the partner of the one person not polled.
Since each person shook hands with at most eight others, the nine answers received by Mike are exactly the numbers 0 through 8. The two people say,. A and B who answered "0" and "8" must have been a couple, since otherwise their opportunity to shake one another's hand would have ruined one of those scores..
Now we examine C and D who scored" I" and "7"; since C had to shake hands with A and D had to miss B, the same argument applies and they must be a couple as well,. Similarly, the pairs scoring "2" and "6," and "3" and "5," must also be couples, This leaves both Mike and Jenene shaking hands exactly with the high scorers, for a score of "4" each. If you didn't see the argument, but guessed that the answer was 4, your intuition is on track. In fact, if there is a.
Suppose for some reason each couple had itself shaken hands, and Mike had asked everyone how many people had not shaken hands. Reasoning on the basis that the puzzle is a good one can be useful, if not entirely satisfactory, Martin Gardner, in one of his famous "Mathematical Games" columns in Scientific American, once. It seems like one would need to know the diameter of the hole, or of the original sphere, to solve the problem, but in fact, this is not the case.
The bigger the sphere, the wider the hole has to be for its length to be 6"; a calculation verifies that the volume of the remaining napkin-ring-shaped solid is the same in every case. However, you don't need to do that calculation if you trust the puzzle poser. Baxter is correct-in fact, he has understated the case; assuming no voter changes his mind, Ashford will win for sure! To see this, suppose Ashford's supporters prefer Baxter to Campbell so that Baxter would heat Campbell in the proposed two-candidate race.
Thus, in this case, Ashford will beat Baxter in the runoff. This puzzle, dreamed up by mathematician Ehud Friedgut for classroom purposes, serves as a warning: There may be more to some tiebreakers than meets the eye! This puzzle was devised by Johan Wastlund of Linkoping Universify, and looselyl inspired by historical events in Sweden.
There are two key observations: In successive turns, the number of salaried voters falls to 9, 5, 3, and 2. It is not difficult to see that the king can do no better at any stage than to reduce the number of salaried voters to just over half the previous number; m particular, he can never achieve a unique salaried voter. More generally, if the original number of citizens is n, the king can achieve a salary of n-3 dollars in k rounds, where k is the least integer greater than or equal to IDg2 n This delightful problem was posed by Andy Latto andy,latto pobox.
The idea of a third hand instead of a second clock came to me from David Gale. Let us first note that for the problem to make sense, we must assume tha. Note that we can tell what time it is when the two hands coincide, even though we can't tell which hand is which; this happens 22 times a.
This reasoning turns out to be good practice for the proof, Imagine that we add to our clock a third "fast" hand, which starts at 12 at midnight and runs exactly 12 times as fast as the minute hand,. Now we claim that whenever the hour hand and the fast hand coincide, the hour and minute hands are in an ambiguous position. Because later, when the minute hand has traveled 12 times as far, it will be where the fast hand and thus also the hour hand is now, while the hour hand is where the minute hand is now.
Conversely, by the same reasoning, all ambiguouspositions occur when the hour hand and fast hand coincide. This card trick is usually credited tel mathematician William Fitch Cheney. For more information, readers are referred to an article by Michael Kleber in the Mathematical Iruelligencer, Vol. I Winter ; or Colm Mulcahy's article in the Mathematical Association of America's Math Horizons, in , which discusses variations of the trick.
Dorothy communicates to David only through the ordering of the four cards she hands him. Of course, there are only 4! The easiest way I know of to perform the trick is for Dorothy to pull a card in a suit that appears at least twice the pigeon-hole principle againl.
Dorothy pulls x, puts y first among the remaining four, and orders the three other cards so as to encode the difference x - y. For example, suppose David and Dorothy agree that the natural order of the deck is. A, "'2, If the three cards ascend e. Notice that there is some slack in the scheme; if fewer than four suits are represented among the five cards handed to Dorothy, she wi11 have at least two choices for the pulled card, It is natural to ask how much bigger a deck could be accommodated; in fact, cards is the maximum.
To see that you can do no better, imagine that the cards are numbered from 1 to n and consider the function f which assigns to any ordered 4-tuple u, v, y z with distinct entries the fifth card. Thus, the total number of 4-tup]es must be at least equal to the total number of sets of size 5; i.
To actually accomplish the trick with cards numbered 1, Obviously, Lavish spends at least as much as Frugal! But how to prove it? It seems that the best way is to show that for any Ie, the kth cheapest flight call it f taken by Lavish is at least as costly as the kth cheapest flight taken by Frugal. This seems like a stronger statement than what was requested, but it really isn't; if there was a counterexample, we could adjust the flight costs, without changing their order, in such a way that Lavish paid less than Frugal.
For convenience, imagine that Lavish ends up visiting the cities in west-to-east order. Let F be the set of Lavish's k cheapest flights, X the departure cities for these flights,and Y the arrival cities. Note that X and Y may overlap. CaB a flight "cheap" if its cost is no more than! CaU a city "good" if Frugal leaves it on a cheap flight, "bad" otherwise. If all the cities m X are good, we are done; Frugal's departures from those cities constitute k cheap flights, Otherwise, let x be the westernmost bad city in X; then when Frugal gets to z , he has already visited every city to the east of z , else Frugal could have departed x cheaply, But then every city to the east of x, when visited by Frugal, had its cheap flight to x available to leave on, so all are good, In particular, all cities in Yeast of x are good, as well as all cities in X west of x; that is k good cities in all.
Thanks to Bruce Shepherd of Ben Labs for helping me come up with the above solution, We don't know what solution was intended by the composer. This is a trick, of course, On the average, it'll take forever for you to be wiped out-the game is in your favor!
I noticed this counterintuitive fact years ago while constructing homework problems for an elementary probability course at Emory University,. The human mind was designed by evolution to deal with foraging in small bands on the African savannah, , " faulting our minds for succumbing to games of chance is like complaining that our wrists are poorly designed for getting out of handcuffs,. Probability is with us every day.
It forms the basis of the study of statistics, which in today's society plays a huge role in how decisions are made. Probability puzzles can be devastatingly counterintuitive.
Consider the following reasonable-sounding question:. In a room stand n armed and angry people. At each chime of a clock, everyone simultaneously spins around and shoots a random other person. The persons shot fall dead and the survivors spin and shoot again at the next chime; eventually, either everyone is dead or there is a single survivor. Amazingly, this probability does not tend to a limit; as n grows, the probability varies subtly, but relentlessly, according to the fractional part of the natural logarithm of n.
For a related result, see H. Our practice puzzle is honest, but closely related to the famous "Monty Hall Problem" see below which spawned a remarkable storm of confusion and controversy a decade ago. A two-headed. One of the coins is drawn at random and flipped; it comes up "heads. Obviously, the coin chosen is either the fair coin or the two-headed coin, so its other side is equally likely to be a head or a tail, right?
You can think of it this way: If the coin were fair, it might have come up "tails," whereas the two-headed coin had no choice; hence, there is a presumption in favor of the twoheaded coin. This notion is known to bridge players and a century ago, to whist players as the "principle of restrictedchoice,". To make it plainer, suppose the coin is tossed ten times and comes up "heads" every time.
It could still be the fair coin, but we'd guess it was the two-headed coin. That presumption exists even after a single flip. One way to calculate the odds in a straightforward manner is simply to think of the six sides of the coins as labeled: When a: Of the three heads, HI and H2 have a: Who knows.
I used to perform this experiment myself when teaching elementary probability at Stanford and Emory Universities. The Monty Hall Problem is based on the TV show, "Let's Make a Deal," on which some contestants were asked to choose one of three doors in search of a valuable prize. Host Monty Hall, who knew where the prize was, would open a second door instead: The contestants were then given the option of sticking with their original choice or switching to the third door.
I watched this show occasionally as a kid, and I remember audiences shouting to the contestant to "STAY! Of course, she should switch. If the game is played times, the right door will be chosen initially about of those times; the other games will be won by the contestant who switches! If all this is obvious to you, do not despair.
The remaining problems may yet test your confidence in your own probabilistic intuition. Peter Winkler - Mathematical Puzzles Uploaded by prabhatravi. Flag for inappropriate content. Related titles. The Moscow Puzzles - Mathematical Puzzles. Jump to Page. Search inside document. Mathematical recreations. L Tide. QA9SW Calton These puzzles are not for everybody. You wiIJ not need a professional acquaintance with mathematics. Amateur mathematicians. Scientists of all kinds. As of this writing, I have been a professional mathematician for 28 years 14 in academia, 14 in industry and have collected mathematical puzzles since my own high school days in the '60s, What you see here are only my favorite hundred or so puzzles, To make it into this book, a puzzle should satisfy most of the following criteria, flmusement: Universality; It should suggest some general mathematical truth, Complex logic puzzles, algebraic puzzles of the type "In two years, Alice will be twice as old as Bob was when",," puzzles relying on properties of particular large numbers, and many other kinds of cleverly devised problems are ruled out elcganclZ: If the statement carries an element of surprise, so much the better, 'Eliffieulty; It should not be obvious how to solve: 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.
Good luck! Here's one to get you started: Prove that Alice call. However, if there are 51 coins instead of 50, it is usually Bob the second to play who will 2 Malhemalieal Puzzles have the advantage, despite collecting fewer coins than Alice. Gasoline Crisis There's a gasoline crisis, and the fuel stations located ! If what is x?
However, one student noticed that if the problem had instead specified that he would have obtained the same answer: Just what is -J2 , anyway? Can you prove it? Soldi lrs in the Field An odd number of soldiers are stationed in a field, in such a way that all the pairwise distances are distinct.
Prove that at least one soldier is not being watched. Bogs A classic brain-teaser.
They were triplets, of course. The third Arnold? Q This and other fuse puzzles seem to have spread like wildfire a.
Hence, the width would have to be integral, Q Your author is responsible for the following solution, not found in Wagon's article. The "averaging" technique used here comes up often: 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, Source: 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.
Q Source: In fact, the limit does exist; the sequence is increasing and bounded above. This is easy because ;;: Here is a practice problem: This continues until students have passed through. Learn more about Kindle MatchBook. Books In This Series 3 Books. Page 1 of 1 Start Over Page 1 of 1. Previous page. Presh Talwalkar. Next page. Complete Series. Kindle Cloud Reader Read instantly in your browser.
Customers who bought this item also bought. Page 1 of 1 Start over Page 1 of 1. Math Puzzles Volume 2: Math Puzzles Volume 3: The Joy of Game Theory: An Introduction to Strategic Thinking. The Best Mental Math Tricks. The Irrationality Illusion: 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. Amazon Giveaway allows you to run promotional giveaways in order to create buzz, reward your audience, and attract new followers and customers. Learn more about Amazon Giveaway. This item: Math Puzzles Volume 1: Set up a giveaway. What other items do customers buy after viewing this item? The Original Area Mazes: Kindle Edition.