(2004-12-02) Pick a 3-digit number where the first and last digits differ by 2 or more...Why is this always equal to 1089?
This is one of the better tricks of its kind,because the effect of reversing the digits isn't obvious to most people at first... If the 3-digit number readsabc, it's equal to 100a+10b+c and the second step gives the following result:
| (100a+10b+c) (100c+10b+a) | = 99 |a-c |
The quantity |a-c | is between 2 and 9,so the above is a 3-digit multiple of 99, namely:198, 297, 396, 495, 594, 693, 792 or 891. The middle digit is always 9,while the first and last digits of any such multiple add up to 9. Thus, adding the thing and itsreverse gives 909 plus twice 90, which is 1089, as advertised.
Henri Monjauze (2008-02-07) Pick a 2-digit number...How can the magician predict what that symbol is?
The trick will become boring (or obvious) if the same table is used repeatedly. Thus, a new table must be provided each time. Several online implementationdo this quite effectively, with nice graphics. Examples:
How fast can you discover the secret which makes this work? [ Answer ]
Art Benjamin (2010-05-29) Figure out the missing digit in a large product of two integers.
: The magician hands out a 3 or 4-digit integer chosen by a spectator in a previouspart of the show. Using a pocket calculator, another spectator multiplies thatnumber by some secret 3-digit number which he chooses freely and keeps for himself. The result is a 6 or 7-digit number. The spectator withholds one of those digitsand reveals all the others in a random order. The magician then reveals thewithheld digit!
: The number handed out by the magician is a multiple of 9 (Art Benjamintakes it from a random list of perfect squares coming from a earlier stage ofthe show; each of those has one chance in 3 of being divisible by 9 andthe earlier stage went on until a "good" number came up). The result is therefore a multiple of 9 and the sum of itsdigits is a multiple of 9. When all the digits but one are revealed, the last one is thus knownmodulo 9. This does reveal it unless it's eithera zero or a nine. In that ambiguous case, the magician will guessit to be a 9 and will almost always be right because people will rarelyskip a zero when they are told to skip any digit they like. If you'd rather not take any chances at all, then instruct people to skip a nonzero digit...
Many variants of this trick can be devised based on any obscure processwhich produces a multiple of 9. Here is one:
Ask a spectator to pick any 4-digit number and to consider the numberobtained by reading it backwards. Let the spectator secretly subtract the lesser numberfrom the larger one, add 54 and multiply the result by a 3-digit numberfreely chosen by the spectator...
Ask how many digits there are in the final result and ask the spectatorto keep one nonzero digit secret and to reveal the otherdigits in scrambled order. (Count ostensibly on your fingers how many digits you are given to make sure you'reonly missing one.)
You may then call the remaining digit with perfect accuracy.
(2013-09-27) Guess with perfect accuracy which one of three cards was chosen.
The following effect can be repeated as many times as neededto convince the spectators that you can read their minds withperfect accuracy as they pick one of three choices.
Effect :
Put three cards face-up on the table. Ask a spectator to choose one mentally and remember its position. Flip the three cards over.
Turn around and instruct the spectator to show the other spectatorswhich card he has chosen, then have him switch the two other cards behind your back.
Now face the table again and instruct the spectators to switchcards as many times as they wish in front of your eyes.
Reveal the card originally chosen by the spectator.
Secret :
Before you flip the cards and turn around, remember which card is in themiddle. When you face the table again, focus of the card whichis now in the middle and keep track of its position as spectators movecards around.
Flip that card.
If this is the card you had memorized, simply announce that thisis the card the spectators have chosen (this is so because the spectatorclearly hasn't switched the middle card behind your back in this case). Otherwise, the chosen card can be neither the one you had memorized northe one you're now seeing. It's the third card.
In a 1992 TV show, David Copperfield turned simple-minded mathematicalproperties into something wonderful,for an audience who was (skillfully) led to expect magical things to happen.
Copperfield first asks you to take N steps forward and N steps back. It doesn't matter what N is, does it? Later, he says to go halfway around a circle in whicheverdirection you choose (another type of irrelevant choice).
Regardless of the details of that show, it should be clearthat a magician can only make predictions about outcomes which donot depend on the choices of his many spectators. However, surprisingly many people wantto believe in some irrational explanation. This is what really scares me.
Besides the visual effects and the drama,the challenge in designing such a collective effect is also to deviseinstructions that everyone can follow...
(2004-04-03) Mental magic for classroom use... [Single-use collective mentalism]
The teacher tells the class that a crowd can be driven to think about thesame thing; very few people will escape the mental picture shared by all others...
Each student in the class is asked to think about a small number and is theninstructed to perform the following operationssilently.
Double the number.
Add 8 to the result.
Divide the result by 2.
Subtract the original number...
Convert this into a letter of the alphabet. (1=A, 2=B, 3=C, 4=D, etc.)
Think of the name of a country which starts with this letter.
Think of an animal whose name starts with the country'ssecond letter.
Think of thecolor of that animal...
The teacher then announces to a puzzled classroom that their collective thinkingmust have gone wrong, since "there are no grey elephants in Denmark"...
Thetrick works in most parts of the World,but I wonder how many students from the Caribbeans would thinkof an "ostrich in Dominica" instead.
Michael Jørgensen(2004-03-24) ofFitch Cheney How to reveal one of 5 random cards by showing theother 4 in order.
The 4! = 24 ways of showing 4 given cards in order would not be enough to differentiateamong the remaining 48 cards of the pack. However, since we may choose what card is offered for guessing,we have an additional choice among 5. The resulting 120 possible courses of action aremore than enough to convey the relevant information. Here's one practical way to do so:
Consider two cards of the same suit (among 5 cards, at least one such pair exists). Let's call them the base card and the hidden card,in whichever order makes it possible to go from the base card to the hidden card card by counting at most6 steps clockwise on a circle of the 13 possible values. (King is followed by Ace, Ace is followed by 2, 3, 4, etc.)
We offer thehidden card up for "guessing". By revealing thebase card first, we are telling the suit of thehidden card and we also set the point where a count of up to 6 "clockwise"steps is to begin to determine thehidden card.
The order in which the remaining 3 cards are presented can be used to revealthis count, as there are 6 possible permutations of 3 given cards. Using some agreed-upon ordering of the cards in a deck,we hold a high card (H), a medium card (M) and a low card (L). Some arbitrary code is used, like:
This trick is credited to Dr. William Fitch Cheney, Jr.(Fitch the Magician, 1894-1974) who earned the first math Ph.D.ever awarded by MIT (1927).
The puzzle is presented in the 1960 book of Wallace Lee entitled Math Miracles (chapter 14, as quoted byMartin Gardner) and was popularized by the magician Art Benjamin in 1986. It was used in a 1994 job interview and subsequently appearedon therec.puzzles newsgroup, where Bob Vesterman posted the particularsolutionpresented above (1994-04-25).
In 1995, Robert Orenstein implemented Vesterman's encoding for online play atwww.anamorph.com/docs/ct/cards.html (a dead link resuscitated from the 2007archives,courtesy ofdeadURL.com, on 2010-04-25). For many years, that page was apologizing for having "temporarily" shut down its(terse) interactive features, since 2002-08-15. Fortunately, that part wasrevivedin the same terse form, by Tom Ace, an admirer of the trick who happens to be a software engineer.
Eric Farmer(2004-03-25) Reveal n random cards (from a deck of d) by showing only k of them...
The previous article deals with k=4, n=5, d=52. The case k=3, n=8, d=13 is called Devil's Poker : The Devil chooses 5 cardsof a single suit and you present 3 of the remaining 8 cards one by oneto an Angel who must guess the Devil's hand,using a prior convention between you and the Angel.
We have k! C(n,k) = n!/(n-k)! possible actionsto reveal one of C(d-k,n-k) compatible possibilities. This task isonly possible if the former exceeds the latter,which means that n!(d-n)! must be greater than or equalto (d-k)! .
That means k = n-1 in the case considered by Michael Kleber in the aforementionedMathematical Intelligencer article (PDF). In that case, the above inequality boils down to:
d < n! + n
Kleber goes on to show that this necessary conditionis sufficient to establish a working strategy.
(2006-05-01) Kruskal's card trick.
This trick is attributed to the physicistMartin David Kruskal (1925-2006). It illustrates a statistical feature which is amazing enough whenone first encounters it. Here's one way to present the effect:
If we use a regular deck of cards, we either remove the face cards or attributeto them the same value (1) as aces. Beforehand, a player chooses secretly a special number N from 1 to 10. As the cards from the deck are revealed one by one, the player counts cards and considersthe N-th card revealed to be his new special number and keeps countingN cards from that one, and so forth... All told, only a few cards are thussingled out as special. The majority are not...
Yet, toward the end of the deck the dealer (themagician) canconfidently point out that one particular card is "special"...
The same trick can be demonstrated by a clever dealer who just looks at the cardsbefore dealing them and announces that a specific card (which may then be flippedover and replaced in the deck) will turn out to be special. You mayplay this version onlinewith a computer which (honestly) shuffles the deck. Allow yourself to be baffled a few times before reading on...
Well, the explanation is simply statistical. For the sake of simplicity, let's consider therelated case of an infinite sequence of cards, each bearing a positive integer n with probability pn .
Two subsequences extracted with the above rules from an infinite sequence of digits(0 to 9) will eventually coincide,because if they coincide once they coincide forever (think about it).
(2008-01-25) In the U.S. Declaration of Independence, all paths lead to God.
In the May 1999 issueof Games Magazine, Martin Gardner published the followingpuzzle, among a small collection of some magic tricks with numbers. It involves the first sentences of the US Declaration of Independence :
When in the Course of human Events, it becomes necessary for one People to dissolve the Political Bands which have connected
Nature'sGod entitle them, adescent Respect to the Opinions ofMankind requiresthat they should declare thecauses which impelthem to the Separation.
You are instructed to pick any word in the first (red) section of the text. Then, skip as many words as there are letters in your chosen word. For example, if you picked the fourth word ("Course") you have to skip 6 words("of human Events, it becomes necessary") to end up on the word "for"... Iterate the sameprocess, by skipping as many words as there are letters in the successivewords you land on.
What's the first word you encounter in the last (green) section? Answer: God. Always. (The sequence would continue with the words:descent,that,causes.)
The "magic" is based on theKruskal principle discussed above... You will ultimately land on God by starting with most words in the middle (yellow) section. The words thatdo work have been underlined for you. You may check that this underlining is correctby working it out (backwards) for yourself, starting with the last yellow words ("and", "of") which do land on God in one step. As any word which leads to an underlined word gets underlined itself,almost all words in the yellow section end upbeing underlined. This includes the first 17 words of that yellow section. Since all words of the red section have less than 17 letters, thatsolid chunk of underlined words can't be jumped over and, therefore,all paths starting in the red section will ultimately lead to the word "God"in the green section. (Actually, any word up to the word "Station" is a valid beginning of a sequencewhich ends up on the word "God".)
(2009-01-08) (Si Stebbins, 1898) A predictable deck of cards that looks disordered.
The ordering illustrated above and presented at rightis also revealed at the end of avideoposted by FurrukhJamal presenting two related magic tricks.
Such a deck can be cut many times, but not shuffled (seasoned illusionists could use false shuffling ).
The value of the Nth card from the top (face down) is:
For example, if the bottom card is the jack of diamonds (B=11, S=0) thenthe tenth card (N=10) is a deuce (since 11+3.10 is 41, which is equal to 2 modulo 13). It's the deuce of hearts because 0+10 is equal to 2 modulo 4.
One trick is to have a spectator cut the deck. Yousecretly look at the bottom card and call the card 3 units higher in the next suit (from the "CHaSeD" sequence Clubs, Hearts, Spades, Diamonds) before revealing the top card.
Find a Specific Card by Counting :
Conversely, the position N of the card x of suit y can be obtained from theChinese Remainder Theorem (a result N=0 would denote the bottom card). Since 3N is x-B modulo 13, N is -4(x-B) modulo 13 (: -4x3 is -12 or +1 modulo 13). With that value of N modulo 13 and the value of N modulo 4 (namely y-S) we may apply ourexplicit formulato solve the Chinese Remainder Problem and obtain N modulo 52 = 4x13,namely:
For example, if the bottom card is the jack of diamonds (B=11, S=0) then the queen of hearts (x=12, y=2) is at the following position (modulo 52):
N = 13 (2-0) 4 (12-11) = 22
The king of spades is at N = 13 (3-0) 4 (13-11) = 31
The queen of diamonds is at N = 13 (0-0) 4 (12-11) = -4 = 48
The ace of clubs is at N = 13 (1-0) 4 (1-11) = 53 = 1 (Isn't it?)
Preparation : , , 7
Build the whole deck (face up) from top cards in the order: ª
(2012-04-28) A nice way to reveal the Nth card from astacked deck.
(2009-01-11) Tell the age of people (between 0 and 63) from the cards they pick.
Some traditional magic age cards forgo the numbers 61, 62 and 63 (so that only 29 or 30 numbers per card are required, which are printedin a 5 by 6 pattern, with or without a star in the 30th position). Full-range cards (with 32 numbers printed on each card) are more satisfying. Here they are:
A spectator thinks of a number (up to 63) and tells you on what cards it is.
You call the exact number!
The weight of each card is the smallest number printed on it. Any number is equal to the sum of the weights of the cards it appears on. For example:
52 = 32 + 16 + 4
This is just a straight consequence of binarynumeration. Each card actually shows all the numberswhich have a "1" in their respective binary representations at a given position. The binary representation of 52 being 110100, it appears on 3 cards and is equalto the sum of the 3 relevant powers of 2. Voilà.
(2009-01-14) Tell the age of people (between 0 and 80) from the colors they pick.
This is my own improvement (2009-01-14) over traditional "age cards".
The introduction of black and red colors allows a larger range of numbers (80 instead of 63) using fewer cards (just 4 cards instead of 6).
Those cards are based on ternary numeration: In base 3, all numbers less than 81 are represented by 4 digits or less. Each card shows the 54 numberswhich have a nonzero digit at a specific ternary position. If the digit is 1, the number is listed in . If the digit is 2, the number is listed in black.
(2009-04-05) Ask 3 questions to find one card among 27 (or fewer).
This is a classic no-brainer. Deal any odd number of cards up to 27in three equal piles (this means you're dealing 15, 21 or 27 cards,according to taste). Ask what pile the chosen card belongs toand collate the cards so the chosen pile is in the middle. Deal and collate again in the same way. Deal one last time. The chosen card will be in the middleof the selected row. Reveal it in whatever dramatic way you like...
For a very fast effect, use just 9 cards and deal only twice (although the underlying math for this 2-step trick becomes rather obvious).
(2013-07-28) What were the original positions of the 3 remaining cards?
For once, let's explain the effect before presenting a video performance.
One method of eliminating half the cards in a face-down deck isto flip-over every other card, starting with the topmost one, to form twopiles and get rid of the face-up pile...
The second step leaves 13 cards face-down, in the following order:
2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50.
The third step eliminates every other card and reverses the order again:
46, 38, 30, 22, 14, 6.
The final elimination leaves three cards face-down: 6, 22 and 38. In other words, the first of the "Final Three" was initially at position 6 (i.e., with 5 face-down cards on top of it) and the other twoappeared thereafter at regular intervals of 15 intervening cards...
In the video presentation below, the number 15 is prominent and the number 5is the result of subtracting 4 from 9 = 52-(10+1+15+1+15+1)... There are also two "false cuts" to give an impression of randomness.
(2009-01-13) You have two choices. Choose either 2 or 3...
Multiply your chosen number by any odd numberand multiply the number you did not choose by any even number. Add those two products together.
(2009-03-26) (cf.A024222) 8 perfect faro shuffles leave a deck of 52 cards unchanged.
Even number of cards :
In a perfect faro shuffle of an even number of cards,the deck is split into equal halves which are then interweaved.
There are two ways to do the interweaving. In an out shuffle, both the top card and the bottom card are unchanged In a so-called in shuffle neither is (the top cardbecomes second and the bottom card becomes next-to-last).
Out-shuffling 2n+2 cards is equivalent to in-shufflingthe inner 2n cards.
Shuffling decks with an odd number of cards :
When the deck consists of an odd number of cards,the deck is split into a pack of n+1 cards and a pack of n cards.
If the larger pack is on the bottom, then the bottom card always remains unchanged andwe are simply faced with a faro shuffling of only the 2n top cards. So, only the case where the top pack is larger need be considered. For an out shuffle that case is equivalent to an out shuffle of 2n+2 cards (in an outshuffle of an even number of cards, thebottom card stays in place). All told, the only case where the faro shuffling ofan odd number of cards does not reduce trivially to the shuffling of an evennumber of cards is the following one:
In-shuffling of 2n+1 cards,cutting n+1 cards from the top.
In such a shuffle, there's a pair of adjacent cards fromthe middle of the pack which remain adjacent at the bottom of the pack after theshuffle. It's much less regular than the other type of faro shuffling. Yet, some patterns appear:
The number s of such shuffles needed to return a deck of n cards to its original state is a complicated function of n. Remarkably, if n is 3 units below a power of 2, then s is a simple quadratic function of the exponent (usually, the ratio s/n is then much smaller than for any lesser values of n).
Some examplesof s in-shuffles leaving n cards unchanged :
n
5
13
29
61
125
253
509
1021
2045
4093
...
2k -3
s
6
12
20
30
42
56
72
90
110
132
...
(k-1) k
n
7
15
31
63
127
255
511
1023
2047
4095
...
2k -1
s
10
56
90
132
182
240
306
380
462
552
...
2 k (2k-1)
(2016-10-19) In at most 7 shuffles, put the top card at any given position in the deck.
If n is the number of cards above a given card in the top half ofthe deck, then the number of cards above that same card becomes:
2 n after an out-shuffle.
2 n + 1 after an in-shuffle.
Let's encode a sequence of perfect shuffles with a binary number Q wherea "0" bit corresponds to an out-shuffle and a "1" to an in-shuffle. For example, an in-shuffle followed by two out-shuffles corresponds to "100" (which is the binary expansion of the integer Q = 4).
If Q is less than the number of cards in the deck, the above remark can be used to show (byinduction on the number of shuffles) that a sequence of shuffles of code Q moves the topmost card Q places down (i.e., after shuffling, there are Q cards above the card which was originally on top).
(2012-11-02) Equal numbers of heads. Always !
This classical trick can be done with ordinary coins (each side is either heads or tails ). However, it's simpler and more spectacular with coins whose sides are easy to tell apart from a distance. Othello/Reversi pieces (discs) are ideal for this: They have a white side and a black one...
The Effect :
Put all the discs on the table, flip some of them over, shuffle them. Ask your spectators to do the same. Explain the difference betweenshuffling the coins (sliding only) and flipping them over. Now, turn around and tell the spectators to shuffle the coins behindyour back ("no flipping") then announce that you willseparate the whole mess into two heaps containing the same number of white discs "using your sense of touch alone".
You do just that, very quickly (using both hands to go faster). Then put your hands up in the air and turn around (in that order) to check with the spectators that you've accomplished the improbable.
Do it several times and the improbable will look like the impossible.
The Secret :
Before turning your back, you count the number W of white discs.
What you do behind your back is simplypick W discs randomly and flipthem over as you put them flat on the table to form a separate heap.
Why it works :
Consider any heap of W discs taken from a set that originally contained W white discs and any number of black ones.
If x is the number of white discs in that heap,then there are W-x black discs in it, which isprecisely the number of white discs that you left in the rest of the set.
If you flip over all the discs in your heap, there are now as many white discsin it (namely, W-x) as in the rest of the set.
(2015-02-21) Some card patterns are preserved when a deck is shuffled by a spectator.
The general principle is that a certain type of shuffling which ends with an honest riffle-shuffle (performed by a spectator) will always preserve one particular aspect of acyclic deck with respect toa separation of the deck into n equal classes (e.g., 2 colors, 4 suits, 13 values): Even after such a shuffle, every "slice" of n cards still contains one card of every class !
The Gilbreath Shuffle : Starting from a cyclic deck (cut any number of times) the first step of a proper Gilbreath shuffle consistsin separating the deck into two piles by dealing out one of them, thereby reversingthe order of the cards in it (you may pretend that you're counting the cards to "make sure"the piles are exactly even, although that's actually irrelevant). The second step is to let a spectator riffle-shuffle those two piles together once.
Why it works : After a Gilbreath shuffle, the situation is exactly identical to what would be obtained if wehad built the deck by successively picking a card either from the top or the bottom of theoriginal deck (at random). Since slices of n cards are ordered alike at the top or the bottom of our cyclicdeck, we always obtain a full slice of different cards after n picks,irrespective of the random choices made. (The same basic situation repeats aftera whole number of slices have been so obtained.)
In the case n = 2 (for black/red colors only,as presented by Gilbreath in 1958) it's an overkill to deal out one pile completely... It suffices to make sure that the bottom cards in the two pileshave different colors (while preserving alternating colors). This easy version is most often used by magicians (see video clips in the footnotes below).
Here's my way to apply the full Gilbreath principle with a 52-card deck. Slices of sizes 13 and 4 are used simultaneously in this example:
Effect (2015-03-09) :
Ask a spectator to take 12 cards off the top of theshuffled deck and to secretly remove one of them. The other 11 cards are then exposed. You stress that you clearly can't tell their card for sure without seeing all the other 51 cards. Yet, you guarantee to call it with perfect accuracy,by just peeking at the four top remaining cards... Then, you do just that!
Secret :
As the deck is initiallycyclically stacked,the second Gilbreath principle applies: The correct suit is the only suit which appears just 2 times (not 3) among the first 11 cards shown. The value of the hidden card is the only value missing from the first 12 cards revealed (the 11 cards originally shown and the first card turned over from the top of the remaining deck).
Yes, only the top card is needed; the extra ones just help conceal the secret!
Also, the first 11 cards allow you to determine the correct suit and leave a choice of only two values. The correct value is whichever of those two differs from the value of thetop card (to be revealed next). Thus, you can be ready to call the spectator's card very fast when the top cards are turned over (all at once). This speed-flourish enhances the effect's impact.
(2015-02-23) A self-working effect, originally designed by Paul A. Lelikis c. 1970.
This trick works by relating the number of cards in several piles to the values of their top cards. The video byBrian Brushwood (link in the footnote) demonstrates the effect and shows howto perform it (same thing really, for this particular effect) but doesn't provide an explanation... Here's one:
Well, in the last stage of the trick, each of the three face-down piles on the table contains 14-n cards, where n is the value (between 1 and 13) of the top card. If you deal out the number of cards indicated by that top card for two piles,each of them will have consumed a total of 14 cards. Therefore, from the original 52 cards,24 cards are left (52-28) which are either in your hand or in the remaining pileof 14-n cards, where n is the value of the face-down top card, which youhave to determine...
So, if you have removed 10 cards from your hand at the outset,there's a total of 14 cards left. As 14-n of those are in the "unknown" pile,you are holding exactly n cards. Counting them reveals the value of the pile's topcard.
Many variants of the same mathematical trick start with the construction of several pilescontaining 14-n cards ( n being the value of each pile's own top card). The key point is that, when one such pile is finally chosen,the number of cards outside of it is simply 38+n. Thus, the real challenge in designing such an effect is to come up with some interesting methodto get rid of 38 cards before the final count (in a sufficiently obscure way).
(14-n) + (38+n) = 52
(2015-04-15) Stu Ungar (1953-1998) Effect: Tell the last card of a deck when all the others have been shown.
The secret is to assign a unique number to each card an know what the total for the whole deck is. Mentally keep the running total of all the cards which have been shown. When only one card remains, its number is obtained by subtracting that total fromthe known total for the whole deck.
If we use numbers from 1 to 52, the total for the whole deck is:
(52+1) 52 / 2 = 1378
The formula behind that calculation was discovered byGauss when he was seven years old.
In practice, it's easier to assign cards in each suit values from 0-12,20-32, 40-52 and 60-72. The grand total for that scheme is 1872.
Better yet, use modular arithmetic modulo 4 or 13 to add separately the suits (the total for the whole deck is 2 modulo 4) and the values (the total for the whole deck is 0 modulo 13).
(2019-06-25) Effect: A sum of differences which is always equal to 25.
An effect in search of a good way to present it.
(2021-07-19) Effect: Tell a sum of rolls, in spite of a secret choice.
Presentation : Behind the magician's back, a spectator rolls threedice and adds the pips, then choses one die and adds the number on its bottom face. Finally, the chosen die is rolled again and the result is added to the total.
Effect : The magician turns around and names the total secretly computed.
Secret : Just add 7 to the total which finally shows.
Proof : Being opposites on a die, the two unknown numbers add up to 7.
waiting for our wits to grow sharper.
