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Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo On Vacation
    Strange Egg
    19 May '08 00:34
    start with any 3 digit number, with a stipulation that the digits must be descending

    321

    reverse the order of the digits and subtract it from the original number

    321 - 123 = 198

    reverse the order of the new number and add it to the number first produced

    198 + 891 = 1089

    this will always process always produces 1089

    algebraiclly

    100a + 10b + c -(100c + 10b + a )

    99 ( a - c ) is the first produced number, but how do we show that this number reversed added to itself always produces 1089

    I understand that the multiples of 99 are the key, but is there algebraically a insight into this peculiar property?
  2. 19 May '08 04:03 / 1 edit
    (a - c) is a 1-digit number from 2 to 8, call it x. 99x = 100(x-1) + 10 (9) + (10-x). Reversing this gives 100(10-x) + 10(9) + (x-1). Adding the two clearly gives 1089.

    In fact, the digits don't have to be descending; it only matters that the first digit must be greater than the last. If the first and last are equal, you'll obviously get zero. If the first is smaller than the last, you'll simply get -1089. And it makes no difference what the middle digit is.
  3. 19 May '08 05:19
    Almost right: The first and last digit have to differ by at least 2. If they differ by 1, the difference would only be 99, so unless 099 counts as a three-digit number, this wouldn't work.