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Math Help - a!b! = a! + b! + c!

  1. #1
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    a!b! = a! + b! + c!

    This is a BMO 1 question from 2003 that has me completely stumped.

    a!b! = a! + b! + c!

    Any help on basic operations on factorials would be appreciated as I have no idea how they can be manipulated.

    Thanks.

    EDIT:

    Apogies, I forgot to state that all possible values of a b and c are to be found
    Last edited by jj451; November 30th 2011 at 09:20 AM.
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  2. #2
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    Re: a!b! = a! + b! + c!

    Quote Originally Posted by jj451 View Post
    a!b! = a! + b! + c!
    What is to be done here?
    What are the instructions?
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  3. #3
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    Re: a!b! = a! + b! + c!

    Quote Originally Posted by jj451 View Post
    This is a BMO 1 question from 2003 that has me completely stumped.

    a!b! = a! + b! + c!
    There is one obvious solution (a,b,c) = (3,3,4). I suspect that there are no others, but I don't see how to prove it.

    You could start by rewriting the equation as (a!-1)(b!-1) = c!+1, but then what?
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    Re: a!b! = a! + b! + c!

    A quick thought (I don't know if it's useful). You can try independent looking at the possible orderings of {a,b,c}. For example, what happens if a\leq b\leq c? We can safely assume that a\leq b because of the symmetry in the equation.

    With them being ordered, you will know how much you can "divide out" (sorry, don't know the English word) and where.
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  5. #5
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    Re: a!b! = a! + b! + c!

    If we assume that a \le b say, then we can rewrite a!b! as (a!)^2 \times \frac{b!}{a!} and a! + b! can be rewritten as 2a! + (b! - a!) (I know this is obvious but it might help to see something else).

    No other ideas at the moment; I'm not too good with factorials.
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  6. #6
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    Re: a!b! = a! + b! + c!

    I found a solution here.
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    Re: a!b! = a! + b! + c!

    Quote Originally Posted by alexmahone View Post
    I found a solution here.
    Sehr gut. Marking as solved.
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