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Magic trick (Posted on 2007-05-11) Difficulty: 3 of 5
Two magicians A and B perform the following trick:

A leaves the room and B chooses 4 members from the audience at random. Each member chooses a card numbered from 1 to 100 (each chooses a different card) and after B has seen their cards he chooses a card from the remaining deck of cards. The 5 chosen cards are shuffled by an audience member and handed to A who just returned to the room. Prove that A is able to figure out which cards each member picked. Consider that the chosen members form a row and e.g. the leftmost member picks the first card and the rightmost member (B) picks the last card.

No Solution Yet Submitted by atheron    
Rating: 4.1667 (6 votes)

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re: Address time here it comes- my way | Comment 48 of 51 |
(In reply to Address time by brianjn)

Dear Brianjn,<o:p></o:p>

Having read your comment about the "magic trick" I realized that I have to reply.

I wanted to do it about a year ago, but had in mind a long explanation.<o:p></o:p>

Since that time I have successfully performed this trick in front of   crowds (or teams) of gifted<o:p></o:p>


All you need is a quick mind and a good   partner with the same ability.<o:p></o:p>

Let's sum-up the problems we faced, while trying to invent a full-proof solution (an impossible dream- as Charlie pointed out):<o:p></o:p>

a)       the magician has to devise a number within a range of permissible choices such that it defines the permutation of ABCD( four volunteers from the non-cooperative audience )<o:p></o:p>

b)       this number must be easily identified by magician's partner as the guide number, easily decoded into a number between 0 and 23 to denote the correct permutation e.g. 0 =ABCD,  1=ABDC, 2=ACBD…..23=DCBA<o:p></o:p>

c)       it is proven as not-attainable task using math only<o:p></o:p>

d)       we must find some "dirty  trick" to make it full-proof doable<o:p></o:p>

<o:p> </o:p>

I use a shuffled deck of 276 numbered cards,     (To my audience I declare:  Pick up a card – there are numbers from 1 to 300). The deck consists of all the integer numbers from 1 to 300 with the following numbers excluded:<o:p></o:p>

<o:p> </o:p>

2     11    23    31     43     53     61        71        83     97<o:p></o:p>

101   113    127   131    149   151    163    173    181    191 <o:p></o:p>

288   211    223   233    <o:p></o:p>

<o:p> </o:p>

Those are the guide numbers – used by the magician – and as you can see they are easily identifiable, all except 288 are defined as "the first prime number in the decade" i.e.  2>>0>>ABCD   11>>1>>ABDC….173>>17>>CDBA,…233>>23>>DCBA.<o:p></o:p>

<o:p> </o:p>

Since there is no prime to represent 20 (201,203,205,207,209 are composite)<o:p></o:p>

I choose 288 to represent 20(=DBAC), this is a special number, even, easily recognised, so the transformation is easily remembered.<o:p></o:p>

<o:p> </o:p>

Example: A selects 284, B 249, C 55, D 108<o:p></o:p>

<o:p> </o:p>

I, the magician realize that if placed in ascending order the permutation is CDBA.<o:p></o:p>

Now I associate it with number 17 (0-5 starts with A, 6-11  with B, 12-17 with C  , 18-23 starts with D….17 is the last in group C , ergo 17>> CDBA ).<o:p></o:p>

I try to find the lowest 3-digit prime number beginning with 17.<o:p></o:p>

Obviously only odd numbers, ending with 1 3 7 9 are candidates.<o:p></o:p>

171 –no, it is multiple of 3, 173 is ok<o:p></o:p>

Therefore I add  a number 173 to the list of 4 numbers (to make the dirty trick even dirtier  , one might pick up a card  from the deck , after deciding what is the number he is going to write, make a glimpse at the card and only then write 173—not related  to the card he saw-

I do not do it)<o:p></o:p>

  My accomplice gets the five numbers on a piece of paper in any order set by the<o:p></o:p>

Volunteers, say 249, 173, 108, 55, and 284<o:p></o:p>

He identifies the only prime number in the list i.e. 173(CDBA) and therefore decides that the original numbers in ascending order would be 55 108 249 284 and he announces<o:p></o:p>

 C has chosen 55,   D has chosen 108, B has chosen 249, A has chosen 284    and MAESTRO has chosen 173<o:p></o:p>

<o:p> </o:p>

<o:p> </o:p>

One more random sample (ABCD) >>>264, 54,177,281<o:p></o:p>

Magician adds   83<o:p></o:p>

The 5 numbers written in random sequence 264 281 83 54 177  <o:p></o:p>

<o:p> </o:p>

The partner does not consider the numbers 264 and 281, they are over 233, the last guide number and none of them is 288... 54 is even so the guide number is either 83 or 177. Since 83 is the first prime in the 8th decade no need to consider 177(even not a prime-=3*59).<o:p></o:p>

OK, reasons the partner 8 lies in the second 6-tuple of permutations, begins with a B, so   6=BACD, 7=BADC, 8=BCAD<o:p></o:p>

Ascending sequence    54  83 177 264 281, exclude 83 and get 54 177 264 281<o:p></o:p>

Now announce:<o:p></o:p>

B has chosen 54,   C has chosen 177,  A has chosen 264, D has chosen 281    and MAESTRO has chosen 83<o:p></o:p>

<o:p> </o:p>

As you see it takes a lot of words to explain it and I am not fully convinced that I succeeded . If anyone needs further explanation /  or wants to comment on my post – go ahead be my guest


Better late than never

Meglio tardi che mai

Mieux tard que jamais


  Posted by Ady TZIDON on 2008-01-26 07:47:01
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