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It's a Puzzlement
5 Card Magic
by John P. Robertson
Here’s a trick that requires two magicians working together. An audience member picks any five cards from a standard 52-card deck and hands them to Magician A. This magician looks through the five cards, picks one, hands it back to the audience member, and arranges the remaining four into a neat pile face down on a table. Magician B then looks at these four cards and announces the suit and denomination of the fifth card. Magician B has not seen any of the cards before inspecting the pile and there is no secret communication between the two magicians. The identity of the fifth card is deduced purely from a review of the pile of four. The puzzlement is to explain how the trick works. This magic trick is originally credited to magician and mathematician Fitch Cheney.
Comparing Ages
Boy did we ever get a variety of solutions to this one! Space does not permit printing the full solutions from everyone, but we’ll do the best we can. The puzzle is that two people know their ages (in years) differ but don’t know who is older. Neither is willing to divulge their age to the other or to a third party. How can they determine who is older?Jon Evans, Dave Skurnick, David Uhland, and Hank Youngerman (independently) suggest having each person put a number of ball bearings equal to their age in identical opaque jars. Put the jars on a balance scale, and the side that goes up indicates the younger person. Dave also suggests having each light a number of candles equal to their age on opposites sides of an opaque partition. A temperature sensor will report which side is hotter and, thus, which person is older. Steve Philbrick gives a variation; have one person prepare a container with their age in grams in hydrogen, and the other a container with their age in grams in anti-hydrogen. “Combine the contents of the two containers, duck, then observe what remains.” If there is hydrogen the first person is older, otherwise the second person is older.
Chris Yaure has each person prepare three cards, one with a date equal to their birthdays, one with a date earlier than their birthdays, and one with a date later than their birthdays. Prepared in such a way that you cannot tell who prepared which cards, these cards are given to a third party who shuffles them without looking, sorts them into order, and reports a date between the third and fourth earliest dates. You can work out that this is enough for each of the first two to determine who is older.
Bob Conger, Alex Kozmin, Richard Nichols, and Tom Struppeck (independently) have a third person pick a random number and whisper it to the first person. They add their age and whisper the sum to the second person. That person subtracts their age, and whispers the result to the third person. By comparing this last number to the original random number, the third person can announce who is older. (Tom suggests doing all of the arithmetic here modulo 1,000.)
David Uhland also gives a method whereby a calculator or computer is used to calculate (A - B)/|A - B|, and display the resulting ±1, from which one can determine whether A or B is larger. Input is suitably masked. Barry Zurbuchen submitted a similar solution.
Greg Cuzzi has each person pick a random number and give that to a third person. Then they add their ages to the random numbers and give those totals to a fourth person. The third and fourth person each announce the difference between the two numbers they have been given. From the relative magnitudes of these two differences the participants can determine which of the first two people is older.
Other honorable mention solvers include Pete Lindquist, Bret Shroyer, Jon Evans. David Westerberg submitted a solution to the Mobius strip puzzle, featured in the February 2004 issue of The Actuarial Review.
