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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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Second attempt | Comment 42 of 51 |
(In reply to Oops... by John zadeh)

What Magician B has to do is take the sum of the audience's numbers and add this sum to the highest of these numbers.
(and if the sum is over 100 return and resume the count from 1)
Then, from that number Magician B has to add a number (1-24) that represents the relative magnitude of the permutation.

For example, if the numbers and order chosen was:
1,4,9,54, Magician B adds 68 (1+4+9+54=68) to 54, which = 122, and return the count to 22 (because the sum is over 100). Then
the magician adds 2 to 22 (because 1-4-9-54  represents the 2nd smallest permutation) to get a total of 24.  24 is the number that Magician B should choose.

In the case that magician B has a arrangement such as 95,1,2,3 (the sum being 101), where the sum value will be a number already chosen (in this case 1) then the magician should move to
the next available value. In this case they should move to 4. And add 18 to get 22. 22 will be Magician B's cue number.

In this way, Magician A should not have a problem determining Magician B's cue number.  In the example above with 95,1,2,3,10
as the numbers, it is obvious that 10 was not included as one of the audiences numbers (10 added to 95, in itself, would mean that the sum value will be at least 5 - which the 1, 2, or 3 are less than).

Am I on to something, or am I just confused.  I have a feeling that the answer should be represented with a more mathematical approach.

  Posted by John zadeh on 2007-06-27 01:48:30

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