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Math Help - Number system Problem

  1. #1
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    Number system Problem

    If N= 144 How many sets of two values (a,b) are there for which the LCM (a,b) is 144 ??

    What the general formula for this (with proof) ??
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  2. #2
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    Re: Number system Problem

    Note that 144 = 2^43^2. Therefore one of a or b must be divisible by 2^4, and one of a or b must be divisible by 3^2. How many ordered pairs (a,b) satisfy these conditions?

    This method definitely generalizes, but I doubt there is a general formula since it becomes one giant counting problem when N has more prime divisors.
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  3. #3
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    Re: Number system Problem

    Why a or b will be divisible by 2^4 or 3^2 ????


    . Very confused
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  4. #4
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    Re: Number system Problem

    In order for LCM(a,b) to be 2^43^2, a and b must have the form, respectively, 2^x3^u and 2^y3^v for some nonnegative integers x, y, u and v. Further, \text{LCM}(2^x3^u,2^y3^v)=2^{\max(x,y)}3^{\max(u,v  ))}. Therefore, \max(x,y)=4 and \max(u,v)=2.
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  5. #5
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    Re: Number system Problem

    Yeah dude You are awesome thanks twice
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  6. #6
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    Re: Number system Problem

    @emakarov : We can't say intuitively that ia or b will be divisible by 2^4 or 3^2 ,right ? (can we ?)
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  7. #7
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    Re: Number system Problem

    Quote Originally Posted by timkuc View Post
    We can't say intuitively that ia or b will be divisible by 2^4 or 3^2 ,right ?
    We can. If the maximum power of 2 that divides either a or b is, say, 3, then there is no reason to include 2^4 as a factor of LCM(a, b). In the notation of post #4, since 2^4 is the maximum power of 2 that divides 144, we have max(x, y) = 4, so x = 4 or y = 4 or both. This means that 2^4 divides a or b or both. Similarly for 3^2.
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  8. #8
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    Re: Number system Problem

    Last edited by timkuc; July 19th 2012 at 06:04 AM.
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