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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts a good start to go on ( no spoiler!!) | Comment 6 of 51 |

To a set of 4 numbers B adds a card easily recognized

as a 'guide"  :  the number that changes the average

of the other four more than any other card (out of final 5)

would do  .

For instance  : if 1,2,3,4 were selected - clearly any card

greater than 5 would be recognized as B's card.

Now out of 95   possible numbers a subset of 24

should be defined and used with a predefined

correspondence table: 1st ABCD,2nd ABDC ,3rd ACBD,

...24th DCBA. in our case 6==>ABCD ,7==>ABDC 8==>ACBD


For  set (10,15,37,99) B should use a number   between 75 and 98

75 as ABCD 98 as DCBA / etc

When A sees a set 10,15,37, 77, 99   he might hesitate

between 77, 99    as to decision which is the "guide" number

but finally(wise guy) he will understand that 77 is the right

decision , because for the set 10,15,37, 77 ther are no 24

consecutive choices between 78 and 98 ,  and B would choose a

number between 38 and 61 (10,15,37, 77  average 34.75).

It seems to me that there will be always a 24- number- long

uniquelly defined subset to enable coding and decoding with

no ambiguity.

It is very late here,and although I like this puzzle

and feel that I can perform the trick- I leave it to someone else

to finalize a formal proof.

Hope it can be done.

  Posted by Ady TZIDON on 2007-05-11 20:21:39
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