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 Magic trick (Posted on 2007-05-11)
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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Ken,

Thank you for your counter example.

...."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...."

While decoding, A looks for the first available(starting from a2 )gap>25 between each number's neighbors and stops after finding the first occurence

In your case for the series 20 40 60 80 100

60-20   stop ==.> no further hesitationS  40-20=20

...and the 20th alphabetically sorted comb. is FCDE (not DFEC, BUT THAT IS  BESIDE THE ISSUE)

I challenge you to find a valid counter -example,  God knows
I have tried...

I challenge  Charlie to write a program and to test it for
10 millions randomly chosen 4-tuples  in search for a failure- reward granted in case of finding a failing set.

Remember : the series of 5 numbers is not a random set-
one of its members is a powerful indicator ,immediately and uniquely identifiable and providing information how to sort the
remaining (and known) four- not how to find the right  combination out of all existing 4-tuples  (not the millions existing possibilities just 24!!).

 Posted by Ady TZIDON on 2007-05-13 03:20:55

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