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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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Some Thoughts A new approach | Comment 34 of 51 |

This may also end up failing as well, but I though I'd share.

B's plan will be to always pick and even card (from 2 to 48) based only on the even cards that have been picked.  C=code number.  B=B's card

Case 1.  All cards are odd (O1,O2,O3,O4).  C=B.  A will see a single even card and know C=B

Case 2.  One even (O1,O2,O3,E1).  B=E1+C mod 100.  A will see two even cards, E1 and B.  E1-B is too big, but B-E1=C.

Case 3.  Two Even (O1,O2,E1,E2).  if E1>E2 then B=(E1-E2)+C if E2>E1 then B=(E2-E1)+C.  A will see three even cards, E1, E2, and B and will try to decode to find C.  There are three choices:
B-(E1-E2) = B-E1+E2 = C
E1-(B-E2) = -B+E1+E2 = not C
E2-(B-E1) = -B+E1+E2 = not C
The unique solution is the correct one, the two incorrect ones come out the same.  [Is there an example that doesn't work?]

Case 4.  Three even (O1,E1,E2,E3).  I haven't tried this yet but I'm thinking if E1>E2>E3 then B=E1-E2+E3+C may work but I haven't checked it yet.

Case 5.   Four even (O1,O2,O3,O4).  Again I haven't tried this yet but I'm thinking if E1>E2>E3>E4 then B = E1-E2+E3-E4+C may work but I haven't tried it yet.


  Posted by Jer on 2007-05-14 12:59:05
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