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Thread: reciprocals haha good show. and done

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    Addicted Member dolor's Avatar
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    reciprocals haha good show. and done

    The sum of two positive integers is 9. What is the least possible sum of their reciprocals?
    Last edited by dolor; Apr 21st, 2004 at 04:54 PM.
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    Frenzied Member Acidic's Avatar
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    the two integers would have to be 4 and 5. hence the sum being 9/20, which is the smallest.
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    Fanatic Member davebat's Avatar
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    i dont know anything about maths and dont know what a reciprocals is but why cant the 2 numbers be 2+7 = 9?
    why do they have to be 4 + 5

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    Only Slightly Obsessive jemidiah's Avatar
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    1 + 8 = 1/1 + 1/8 = 9/8
    2 + 7 = 1/2 + 1/7 = 9/14
    3 + 6 = 1/3 + 1/6 = 9/18
    4 + 5 = 1/4 + 1/5 = 9/20

    Smallest is one with largest denominator, hence 4 + 5.
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    Frenzied Member Acidic's Avatar
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    Originally posted by davebat
    ...dont know what a reciprocals is
    a reciprocal of a number is that number to the power -1
    so the reciprocal of 2 is 1/2, of 10 is 1/10 of n is 1/n or n-1
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    Generalization

    In general we should say that:

    1/x + 1/y = (x+y)/xy

    Now minimising 1/x + 1/y is equivalent to minimising (x+y)/xy

    As (x+y) is constant, the problem reduces to maximizing xy.

    We know (x+y)^2 = (x-y)^2 + 4xy

    Thus if (x+y) is given, we can get xy to me maximum if x=y because in that case only (x-y)^2 will be zero, which is least for a square.

    In short, if sum of two numbers is constant, their product is maximum when they are equal.

    So making x and y towards equal, we will get 4 and 5.

    This concept can be generalised however big the number is.


    Similarly, vice versa, if product of two numbers is constant, their sum is minimum when they are equal.

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