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- Mathematics In Europe - Gergonnes trick

or not various extensions in many recreational mathematics books e g in Gardner s Mathematics Magic and Mistery Here we present the basic version of the trick using 27 cards The 27 cards are dealt face open in three columns with 9 cards in each The performer asks a spectator to choose a card but to name only its column Say the spectator has chosen the card the top card in the rightmost column in the picture above he names the rightmost column as the chosen one The performer collects the cards and then deals them out again In the described case the cards will be rearranged as shown in the next picture The spectator is asked to name the column with his card again in our example case this would now be the leftmost column The procedure performer collects the cards deals them out again and asks the spectator for the column with his card is repeated In our case the third layout of cards would look like this Now the spectator would again name the leftmost column Once more the performer collects and deals the cards but now he does not ask questions Instead he uses his mathemagical powers concentrates and reveals that the chosen card is now in the center of the layout In fact the previous procedure will always bring the chosen card to the central position provided the performer takes care of the following rules always pick up the cards by columns put the indicated column between the other two and deal the cards out by rows Why does it work In this basic version the secret is a basic principle of elimination indicating the column reduces the number of possible cards to 9 by putting this column in the center and dealing out by

Original URL path: http://mathematics-in-europe.eu/tr/78-enjoy-maths/recreational-mathematics/689-gergonnes-trick (2013-11-18)

Open archived version from archive - Mathematics In Europe - Two rows of pebbles

the pebbles in two rows so that the bottom row contains one pebble more than the top one For example if the spectator chose to work with 17 pebbles he would arrange them like this Now the mathemagician asks the spectator to name a positive integer number smaller than the number of pebbles in the top row In our example the spectator would choose a number smaller than 8 say 6 The mathemagician instructs him to remove that many pebbles from the top row In our example this leaves the spectator with Then the mathemagicians asks the spectator to remove from the bottom row as many pebbles as there are left in the top row In our example the spectator removes 2 pebbles from the bottom row and obtains situation Finally the mathemagician instructs the spectator to remove all the pebbles that still remain in the first row After that the mathemagician guesses correctly of course the remaining number of pebbles In the described example the last step would be removing the three pebbles from the top row and the mathemagician would announce that there are 7 pebbles left The background of the trick is simple basic algebra We encourage the reader to try to work it out by him or herself For those who prefer to have the full description of the trick or want to check if their solution is correct the following paragraph gives the full explanation First note that any odd number of pebbles can be divided in two rows differing by 1 in the number of pebbles So if we denote by n the number of pebbles in the top row at the beginning of the trick the bottom row contains n 1 pebbles in the described example n 8 Denote by m the number

Original URL path: http://mathematics-in-europe.eu/tr/78-enjoy-maths/recreational-mathematics/618-two-rows-of-pebbles (2013-11-18)

Open archived version from archive - Mathematics In Europe - The four ducks trick

two principal questions to think about when trying to explain and learn the trick How does the mathemagician know how to instruct the participator in the removal of the ducks And what about using other numbers of switches besides 5 In fact consideration of the second question leads to the answer to the first one If you would like to work it out yourself try what happens if 1 2 3 4 switches are required Think about possible final configurations of the ducks in each of the cases As a hint we give you the following number the starting positions of the ducks with 1 2 3 and 4 Now if you do not want the trick disclosed to you before you have tried to find out about it on your own please do not read further All other readers will find the explanation in the following paragraphs If the starting positions are numbered 1 to 4 say 1 for the starting position of the yellow duck in the first of the pictures above then if the chosen duck started from an even position after 5 switches it will end up in an odd position and vice versa In our example the chosen pink duck started from position 3 so after 5 switches it must end up in position 2 or 4 This enables the performer to eliminate one of the two ducks at the ends If the starting position was odd like in our example the performer can be sure that the duck is now not on position 1 and can give the instruction to remove the duck at the corresponding end After this the chosen duck is sure not to be in the middle of the three ducks left on a table and consequently another switch with an in fact the adjacent duck brings it to the middle position Now it is easy to instruct the participator to remove the flank ducks and end up with the chosen duck Finally note that the performance would be exactly the same if instead of 5 any other odd number of switches was used If one uses an even number of switches then the chosen duck will end in a position of same parity as its starting position even if starting from an even one and odd if starting from an odd one and it is easy to adapt the performance to this case There is more maths behind the trick than meets the eye Although the trick can be explained on a pure logical basis like above the explanation can be connected to a relatively advanced mathematical notion of odd and even permutations A permutation is a rearrangement of a number of things in our example every rearrangement of the four ducks is one permutation of the ducks It relates to the relative positions of the things and not to the things itself any rearrangement of the numbers 1 2 3 and 4 can be related to a rearrangement of

Original URL path: http://mathematics-in-europe.eu/tr/78-enjoy-maths/recreational-mathematics/617-the-four-ducks-trick (2013-11-18)

Open archived version from archive - Mathematics In Europe - Inductive Magic

if you want you can use the paper and pen Until the end you shouldn t give me any information on your actions and calculations Say we have 9 marbles on the table First divide the marbles in two heaps Then multiply the numbers of marbles in each heap and remember or write down this number Say the participator decides to divide the 12 marbles in one heap with 3 and the other with 6 marbles Heap 1 Heap 2 Score 3 6 18 Now subdivide one of the two heaps in two multiply the numbers in the two new heaps and add to your last number Say the participator decides to divide the first heap with 6 marbles into heaps of 2 and 4 marbles each so he has to add 2 4 8 to his last number 18 Heap 1 Heap 2 Heap 3 Score 18 8 26 Now continue with the process of dividing one of the heaps on the table in two adding the product of the two numbers in these two new heaps to your last total until all your heaps contain only one marble Each of the imaginary actions of the participator including the first two are represented by a row in the table below In each row the heap that is subdivided to obtain the next row is coloured in red Heap 1 Heap 2 Heap 3 Heap 4 Heap 5 Heap 6 Heap 7 Heap 8 Heap 9 Score 18 26 27 30 32 33 35 36 When the participator says he has finished subdividing and calculating although his actions have been quite arbitrary the mathematigian surprises the public by knowing the final result of the calculations How is this possible Mathematical induction explains it all Surprisingly or for a mathematician not so the only thing the final result depends upon is the starting number of marbles so this is all the mathemagician has to know When starting with 9 marbles one always ends up with the final score being 36 If you do not believe it try it And then try to prove it this is also a note for the mathematics teachers who are reading this article the trick as are many of the mathemagical class is very suitable both for performance in the classroom and as a starting point for a problem based or even discovery style lesson Although it is much more interesting to discover why this trick works and how to calculate the correct final score for other starting numbers of marbles besides the possibility of trusting the author that the result depends only on that number and performing the trick once yourself to discover the final score experimentally we present the secret for the lazy ones among the readers And those who want to do it themselves can just finish reading here The maths behind the trick is as follows Denote the starting number of marbles by For 1 we cannot do any subdivison so the

Original URL path: http://mathematics-in-europe.eu/tr/78-enjoy-maths/recreational-mathematics/610-inductive-magic (2013-11-18)

Open archived version from archive - Mathematics In Europe - Magical mathematics: The Integers

100 euro notes by predicting the remainder Indeed it happens frequently that a mathematical fact somehow finds its way into a magician s hat One simply has to find mathematical results that contradict everyday experience and that also have their basis hidden in the depths of some theory Here is a piece of advice Magic is like perfume the packaging is at least as important as the contents No one should be suggesting that the chosen three digit number is to be multiplied by 1001 such a multiplication is equivalent to writing the number twice in succession but then the whole trick would fall flat Those looking for a variant from dividing by 7 can substitute 11 or 13 since 1001 has these numbers as factors as well It will just make the calculation of the remainder a bit more difficult Advanced Variants 1001 100001 Is there a reason that we have to write down precisely a three digit number Could we achieve a similar result with two or four digits Let us consider a two digit number n written in the form xy If we write the number twice in succession then we obtain the four digit number xyxy which is equivalent to multiplying the original number by 101 But 101 is a prime number and so the divisors of xyxy are the divisors of xy together with 101 Since in performing this magic trick we know nothing about the number xy we can say only that there will be zero remainder on division by 101 But asking for division by 101 gives the trick away or at least strongly suggests what is at work and furthermore dividing by 101 may be too difficult for your friends and acquaintances We conclude then that starting with a two digit number

Original URL path: http://mathematics-in-europe.eu/tr/78-enjoy-maths/recreational-mathematics/609-magical-mathematics-the-integers (2013-11-18)

Open archived version from archive - Mathematics In Europe - Order amidst Chaos

of oppositely colored cards as if by magic even though you are actually just drawing the cards from top to bottom See Figure 3 Figure 3 and now present the cards a pair at a time The mathematics behind this is interesting The fact that the cards are arranged pairwise with different colors after the three random events can be proved with combinatorial methods In this connection mathematicians speak of an invariant The magician Gilbreath who invented the trick at the beginning of the previous century seems to have discovered it by trial and error A Variant of the Trick For those who would like to add this trick to their repertoire here is a variant Recall that the original goes according to the following outline Prepare the pack of cards even number of cards of alternating colors Cut the deck and then shuffle the cards Cut the deck at a point where there are two cards of the same color and reassemble the deck Then each pair cards 1 and 2 cards 3 and 4 etc contains two cards of opposite colors And now the variant The deck of cards is prepared just as in the original version and it is again cut Warning This time you must somehow inform yourself as to whether the cards on the bottoms of the two halves are of the same color or different colors This could be done for example as the cards are handed to the shuffler The next step is again as previously the two halves are shuffled together And that is it you do not need to cut the pack again The advantage over the first variant is that you don t need to have someone look at the cut deck to see where two cards of the same

Original URL path: http://mathematics-in-europe.eu/tr/78-enjoy-maths/recreational-mathematics/612-order-amidst-chaos (2013-11-18)

Open archived version from archive - Mathematics In Europe - Magical Invariants

The key here is to make the distance between cards of the same suit king of spades and queen of spades king of clubs and queen of clubs and so on exactly four If you flash this arranged hand of cards in front of your audience it appears to be more or less randomly arranged No one will suspect your skulduggery and if you cut the pack a few times Figure everyone will believe that the cards have been thoroughly mixed Figure 1 The prepared cards But you are operating with the knowledge that the distance between cards is an invariant four cards below the first card is its partner It is therefore easy to produce a pair from under a cloth or under the table of course making it seem that you are struggling mightily You can repeat this process and produce a second pair though this time the distance between them is three and then a third pair separated by two and finally the last pair This trick relies on the existence of some kind of order amidst seeming chaos In mathematics the search for the unchanging has become a sort of leitmotiv in research Once a set of permissible transformations has been described a systematic search begins for quantities that remain unchanged under those transformations This idea has been of particular importance as a unifying principle in many branches of geometry It was proposed in 1872 by the mathematician Felix Klein and has had great influence over research ever since Figure 1 2 The cards after being cut The Back Story The Distance Modulo the Number of Cards Is Invariant Using the notion of modular arithmetic introduced in another chapter of this book we can formulate the principle that underlies the trick somewhat more mathematically If n cards are stacked and two of these cards are in positions a and b counted from the top then b a modulo n is an invariant after cutting the pack any number of times the difference between the positional values has not changed modulo n To see this one has to use modular calculations for negative numbers as well as positive but that is really no problem After all as everyone knows the day of the week seven days ago is the same as the day today and the day 13 days ago was Tuesday if today is Monday Mathematically 13 modulo 7 is equal to 1 One must be aware of this subtlety in order to interpret the invariant relationship presented above correctly An example In the example that follows we are going to use the fact that 7 modulo 10 is equal to 3 In a pack of ten cards the ace of hearts and jack of clubs are in positions 2 and 5 The difference is 3 The pack is cut at position 2 Now the ace of hearts is at position 10 the bottom of the pack and the jack has moved up to position 3

Original URL path: http://mathematics-in-europe.eu/tr/78-enjoy-maths/recreational-mathematics/613-magical-invariants (2013-11-18)

Open archived version from archive - Mathematics In Europe - A Beer Trick

the obtained number is obviously much above your daily average we ll take it as an approximation of your weekly consumption In fact I m more interested in finding estimates of how many of their favourite drinks people drink yearly The year has 52 weeks but as I m only interested in estimates it is enough if you multiply your last number with 50 50 13 650 Did it Do you wan t to know the result No not yet Did you celebrate your birthday this year already If no add 1760 to your last number otherwise add 1761 No 650 1760 2410 Now for your last exercise in arithmetics subtract the year of your birth from the previous result The full four digit year please Tell me the result when you re done I m born in 1971 so 2410 1971 439 My final result is 439 You are 39 the last two digits years old and drink 4 the number obtained removing the last two digits of the result of your favourite drinks daily How the mathematician becomes a mind reader The explanation of the trick is simple algebra of the level everybody learns in school If you the reader are a mathematics teacher you can easily make your students discover the following explanation or parts of it e g how the number to be added before subtracting the year of the birth depends on the calendar year in which the trick is performed themselves The explanation is as follows If m is the daily number of drinks the first part of the trick calculates first 2 m then 2 m 5 and then 2 m 5 50 100 m 250 By adding a number k and subtracting the birth year n we obtain the final score 100

Original URL path: http://mathematics-in-europe.eu/tr/78-enjoy-maths/recreational-mathematics/614-a-beer-trick (2013-11-18)

Open archived version from archive