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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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Solution the magician A coaches partner B | Comment 15 of 51 |

Listen, my friend B

I will teach you know how to encode the distribution of numbers a1 a2 a3 and a4 (this is a series of chosen numbers in increasing order) between partcipants C D E F.

first of all memorize(or write down or enter in your cellular phone /notebook ) the following table.:

CDEF   1

CDFE   2

CEDF  3

CEFD   4

CFDE   5

CFED   6

DCEF   7 

.......

........

FEDC   24

Assuming the men are seated CDEF  , sending a code representing #1 means that a1 belongs to C,  a2 belongs to D,a3 belongs to  E,a4 belongs to F

Sending a code representing #7 means that a1 belongs to D,  a2 belongs to C,a3 belongs to  E,a4 belongs to F

Sending a code representing #24 means that a1 belongs to F,  a2 belongs to E,a3 belongs to D,a4 belongs to C.

Lets assign letter Z to the actual code

Now you ask me how out of 4 given numbers will I create the designating number from which is possible to decode the right combination.

Write down the numbers in inreasing order a1, a2, a3, a4,100+a1 (circular extension)

If a1 belongs to C AND  AND a2 belongs to D AND a3 belongs to F AND a4 belongs to E -the combination's designating number is 2 .

Now in the series a1 a2 a3 a4 locate the lowest number  ak such that the numerical difference between ak and the number to the right of ak is 25 or more- thus enabling  enough space to send any number beween 1 and 24 .

This is always possible and unique(no way to invent three numbers   x u v  between any a1 and a1+100 such that :

x-a1<25

u-x<25

v-u<25

100+a1-v<25

== from  adding follows 100<100 contradiction!!)!!

Add to the series a new number Z+ak , shuffle and send to me.

To illustrate the point here are few examples to define ak 

1  14  58 70  101   ak=14    the lowest!!!

5  34  58 70  105   ak=5

7  14  37 61  107   ak=61

20 40 60 80 120     ak=80

13  14 33 97 113    ak=33

if you want to transmit designating number 2 for the series in the last row Z+ak=2+33=35  add another member i.e. 35 to the original  series 13  14 33 97 ==.> 13  14 33 97 35, shuffle

and forward to me(MasterA!!)

I receive 5 numbers e.g. 33 97 14 35 13, put them in order

 13  14 33  35   97   add another 2 members  (100+13 and 100+14 ) and out of new series 13  14 33  35 97   113 114 calculate for the 2nd,3rd ,4th, 5th and 6th (6th mod 100=1st) differences between the neighbors i.e. number to the right minus number to the left.

 In our example  FOR 14 33-13=20<25

FOR 33     35-14=21<25

FOR 35 97-13=84>25     OK    ak=33 Z=2

Z=2  Corresponds to CDFE

and the magician announces:

13 chosen by C

 14   chosen by D

33 chosen by F

 35  BY B

97 by E

I hope it makes it clear - Charlie will write for  you a program where the input will be 5 NUMBERS and 5 PROPER NAMES, Although there are 120 ways  of matching the names with the numbers the laptop will display the right table.

That is if you dont err, my dear friend!

ELEMENTARY, Dear Watson.

NB If there still ambiguites or ommissions(hope not)

please advise.

ATHERON- I like your puzzle!! Is it original or a trick someone actually performed?  If so, who and where...? 

 

Edited on May 12, 2007, 2:41 pm
  Posted by Ady TZIDON on 2007-05-12 13:16:50

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