An amusing parlor trick is performed as follows. Ask spectator A to jot down any three-digit number and then to repeat the digits in the same order to make a six-digit number (e.g., 394,394). With your back turned so that you cannot see the number, ask A to pass the sheet of paper to spectator B, who is requested to divide the number by 7.
“Don’t worry about the remainder,” you say, “because there won’t be any.” B is surprised to discover that you are right (e.g., 394,394 divided by 7 is 56,342). Without telling you the result, B passes it on to spectator C, who is told to divide it by 11. Once again, you state that there will be no remainder, and this also proves correct (56,342 divided by 11 is 5,122).
With your back still turned, and no knowledge whatever of the figures obtained by these computations, you direct a fourth spectator, D, to divide the last result by 13. Again, the division comes out even (5,122 divided by 13 is 394). This final result is written on a slip of paper, which is folded and handed to you. Without opening it, you pass it on to spectator A. “Open this,” you say, “and you will find your original three-digit number.” Prove that the trick cannot fail to work regardless of the digits chosen by the first spectator.
Writing a three-digit number twice is the same as multiplying it by 1,001. This number has the factors 7, 11 and 13, so writing the chosen number twice is equivalent to multiplying it by 7, 11 and 13. Naturally when the product is successively divided by these same three numbers, the final remainder will be the original number.
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A version of this puzzle originally appeared in the August 1958 issue of Scientific American.