I was just reading about a magic trick invented by Charles Peirce and described by Martin Gardner and I wanted to mention my own favorite in math magic tricks (also described by Martin Gardner). I’ve done it 8 times and it has received quite a reception each time. I have modified it slightly from the original to improve the presentation.

Supplies: deck of cards, calculator, envelope. Blindfold optional.

Turn your back to the audience, eyes closed. (The blindfold goes here if you have one.)

Have an audience member pick any row, column, or diagonal on the calculator.

789

456

123

Have them type those three digits in any order (so it could be for example 879, 528, or 195). Have them hit the “times” button on the calculator.

Have another audience member pick a different row, column or diagonal and type those three digits in any order, then hit equals.

Get a third audience member to then look at that number, say, 447795, and find cards from the deck that match the number. If a digit is repeated there should be multiple cards; so the number above means the audience member would find two 4s, two 7s, one 9, and one 5. Any zeros should be ignored (so for 56088 it should be one 5, one 6, and two 8s). The remaining deck of cards should be set aside.

Using the cards selected from the deck, have an audience member choose one at random and seal it in the envelope. Have the envelope handed to another person at the far corner of the room, and have the audience member now read the remaining cards to you.

You can now name the missing card and ask for the envelope to be opened. Remove the blindfold, turn, and wait for the cries of disbelief.

(No, really, you get cries of disbelief.)

So, how does the trick work?

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Mr. K, on November 15, 2008 at 10:47 am said:Each row, column, or diagonal is a multiple of 3. Therefore your product is a multiple of 9. A simple test for divisibility by 9 is whether the digits sum to 9. So if you sum the digits that were read out to you, (and sum them again until you only have a single digit) subtracting that number from 9 will give you the missing card. The special case is when the difference is 0, in which case the resulting card will be a 9.

I have a deck of single digit cards all ready to go.

John, on November 15, 2008 at 10:48 am said:That’s fun. I need to show that to my kids.

Both of the three-digit numbers being multiplied together is divisible by 3, so the product is divisible by 9. The missing digit in the product is whatever it has to be for the sum of the digits in the product to be a multiple of 9. Since you’ve discarded zeros, the missing card is uniquely determined.

jd2718, on November 15, 2008 at 11:08 am said:You guys beat me to the punch…

mathmom, on November 15, 2008 at 4:39 pm said:Beat me too, but I did figure out the same thing as John and Mr. K both did. 🙂

Awesome trick, though. Perhaps I will do that for my middle school math students.

Jason Dyer, on November 15, 2008 at 4:52 pm said:I still have a student in one class who swears I cheated (even after the explanation).

Carnival of Mathematics #44 « Maxwell’s Demon, on November 21, 2008 at 7:06 am said:[…] an elegant solution that is easy to explain, but hard to find. Another interesting puzzle from Jason Dyer, can be turned into a magic trick, or maybe that should be turned back into a magic trick. Less […]

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