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Topic: How do they perform this magic trick? (Read 2568 times) |
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wonderful
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How do they perform this magic trick?
« on: Mar 28th, 2008, 12:45am » |
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A magician and his assistant perform the following show:
In the beginning, there are only the audiences and assistant on the stage. An audience then writes a 101 digit number on the board. The assistant uses two small plate to cover some two adjacent digits. Next, the magician appears on the stage and can guess exacly these two covered digits as well as their order in the orginal 101 digit number.
How do the magician and his assistant coordinate to perform this trick? Can 101 be reduced to a smaller number?
Hint: think of a simple example with binary numbers.
Have A Great Day!
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| « Last Edit: Mar 28th, 2008, 1:23am by wonderful » |
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Hippo
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Re: How do they perform this magic trick?
« Reply #1 on: Mar 28th, 2008, 4:06am » |
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What order do you mean?
The one I can imagine was not difficult to guess or I do not understand what the magican can see.
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towr
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Re: How do they perform this magic trick?
« Reply #2 on: Mar 28th, 2008, 4:35am » |
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on Mar 28th, 2008, 4:06am, Hippo wrote:It think he means the (mathe)magician can tell which covered number is which, rather then just deduce what pair of number it is; so rather than say e.g. "the numbers are 4 and 5" he'd say "the first number is 5 and the second number is 4".
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Wikipedia, Google, Mathworld, Integer sequence DB
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SMQ
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Re: How do they perform this magic trick?
« Reply #3 on: Mar 28th, 2008, 5:04am » |
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There are 100 possibilities for the two covered digits: 00 - 99. There are 100 possible selections of adjacent digits.
However, the magician and his assistant would have to agree on a correspondence before hand, and given any chosen correspondence it is possible to construct many 101 digit numbers which have no suitable adjacent digits... Hmm.
[edit]In fact, using this simple method, a little over a third (close to 1/e, not coincidentally) of 101-digit numbers will be unsuitable for the trick. I don't think most magicians would take those odds.[/edit]
--SMQ
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| « Last Edit: Mar 28th, 2008, 5:19am by SMQ » |
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Obob
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Re: How do they perform this magic trick?
« Reply #4 on: Mar 28th, 2008, 8:47am » |
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The magician must be awfully speedy at arithmetic! Suppose the digits of the number are abcdefg...z, and have the assistant compute 10a+b+10c+d+...+10z modulo 100...
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| « Last Edit: Mar 28th, 2008, 9:06am by Obob » |
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SMQ
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Re: How do they perform this magic trick?
« Reply #5 on: Mar 28th, 2008, 9:18am » |
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Good insight, Obob! I think that process can be simplified somewhat, however, using only roughly half the number of additions and no division:
[edit]Seems you thought of roughly the same simplification I did.[/edit]
The assistant computes, instead, the modulo 10 sums of a + c + e + ... and b + d + f + ..., concatenates the sums to form a two-digit number, and covers the numbers in that position. The magician does the same, substituting 0 for the covered digits. The two single-digit differences will be the covered digits, each missing from its respective sum.
As to whether or not any algorithm can perform the trick with fewer than 101 cards, I believe the hidden observation in my first post is sufficient to show that it is not possible.
--SMQ
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| « Last Edit: Mar 28th, 2008, 9:20am by SMQ » |
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rmsgrey
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Re: How do they perform this magic trick?
« Reply #6 on: Mar 28th, 2008, 10:01am » |
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This reminds me of the similar puzzle about identifying one of 5 playing cards chosen from an ordinary pack of 52 by being shown the other 4 in a particular order.
There's a thread around here somewhere
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wonderful
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Re: How do they perform this magic trick?
« Reply #8 on: Mar 28th, 2008, 10:07pm » |
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SMQ and Obob proposed an intersting solution. There is a bit subtle point here: what happened if one or both sums are divided by 10; thus the suggested position is 0n or 00. For instance, a+c+e+...=120, b+d+f+..= 250. Thus, mod 10 of a+c+e+..= mod 10 of b+d+f +...= 0. As such, which two numbers the assistant should cover?
Also, the second question can be clarified: can they perform the same trick for some number with less than 101 digits?
Have A Great Day!
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Obob
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Re: How do they perform this magic trick?
« Reply #9 on: Mar 28th, 2008, 11:19pm » |
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I don't think there is any subtlety, wonderful. There are 100 residues modulo 100, and 100 pairs of digits of the 101 digit number.
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wonderful
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Re: How do they perform this magic trick?
« Reply #10 on: Mar 29th, 2008, 7:44pm » |
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Quote:
"concatenates the sums to form a two-digit number, and covers the numbers in that position."
Suppose, after concatenating one has the number 09, so what positions the assistant should cover? Are they the numbers in position 9 and 10 or what other two adjacent positions?
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Obob
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Re: How do they perform this magic trick?
« Reply #11 on: Mar 30th, 2008, 12:46am » |
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Sure, she could cover digits 9 and 10 in that case. The magician and assistant have to agree on a convention. With that convention, if the sum 10a+b+10c+d+... is 0 mod 100 she would cover digits 100 and 101. Or she could cover the first and second, but then cover the 10th and 11th in the case where the sum is 9. Or the two could agree that they label the digits using the numbers 0 through 100, and then cover the digit corresponding to the residue mod 100 and the next digit. So there isn't any subtlety; they just have to agree on a set of representatives for the congruence classes mod 100.
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wonderful
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Re: How do they perform this magic trick?
« Reply #12 on: Mar 30th, 2008, 4:57pm » |
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Excellent Obob! That's exactly the answer I look for. Your solution is very neat.
For practical purpose, the magical and his assistant have to perfrom the show fast. The calculation of a+10b+c + ... would take sometime. can anyone suggest another solution?
Thanks and Have A Great Day!
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| « Last Edit: Mar 30th, 2008, 8:12pm by wonderful » |
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