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Math Help - Is there a solution?

  1. #1
    Newbie
    Joined
    Aug 2009
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    Cool Is there a solution?

    A three-digit number 'xyz' is such that the number equals x!+y!+z!.Find the difference of the number formed by reversing its digits and the original.
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  2. #2
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517
    Interesting problem

    145 does the trick.

    1!=1
    4!=24
    5!=120
    1!+4!+5!=145

    if you just look 7! is more that 3 digits, so none of the numbers can be 7.
    6!=720 which means the number would have to have at least a 7 in the first place, which means it is too big by above.

    This narrows your choices to 1,2,3,4,5.
    If you use all 4 or lower its not enough to get up into the 3 digits.

    So you know there has to be a 5 in there somewhere. From there I just played around with it.
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