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Math Help - 153

  1. #1
    Newbie
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    North Carolina
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    153

    Explain this:

    but first, notice this: if from 153 you take every digits (1,5,3) compute the cube of those digits and add them, you get back 153
    (1^3 + 5^3 + 3^3) = 1 +125 + 27 = 153

    -------------
    Then pick any 3 digits number that can be divided by 3, any one

    and do the same thing, then with the answer found, just repeat the same thing again, and again. You will always end up with 153

    Example
    take 342
    3^3 + 4^3 + 2^3 = 99
    9^3 + 9^3 =1458
    1^3 + 4^3 +5^3 + 8^3 = 702
    7^3 + 2^3 = 351
    3^3 + 5^3 + 1^3 = 153

    and other examples
    666 gives you the sequence 648, 792, 1080, 513, 153
    261 gives the sequence 225, 141, 66, 432, 99, 1458, 702, 351, 153
    369 gives the sequence 972, 1080, 513, 153

    ======================
    Last edited by LChenier; September 25th 2012 at 05:31 PM.
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  2. #2
    Super Member
    Joined
    Jun 2012
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    AZ
    Posts
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    97

    Re: 153

    Let f(x) be the sum of the cubes of the digits of x. It can be shown that x > f(x) for all x \ge 1999.

    Therefore, if x is at least 1999, then iterating f several times, we will arrive at some number less than 1999, in which f must cycle around some number (this is true due to a Pigeonhole argument, if we iterate f 2000 times, at least two of the outputs must be the same).

    It suffices to show that 153 is the only three-digit multiple of 3 that is equal to the sum of the cubes of its digits, which I'll leave that up to you. Apparently 370, 371, and 407 all share the same property, but none of them are multiples of 3.
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