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19 January, 20:39

Melissa has three different positive integers. she adds their reciprocals together and gets a sum of 1. what is the product of her integers?

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  1. 19 January, 21:16
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    First let us assign the three positive integers to be x, y, and z.

    From the given problem statement, we know that:

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

    Without loss of generality we can assume x < y < z.

    We know that:

    1 = (1/3) + (1/3) + (1/3)

    Where x = y = z = 3 would be a solution

    However this could not be true because x, y, and z must all be different integers. And x, y, and z cannot all be 3 or bigger than 3 because the sum would then be less than 1. So let us say that x is a denominator that is less than 3. So x = 2, and we have:

    (1/2) + (1/y) + (1/z) = 1

    Therefore

    (1/y) + (1/z) = 1/2

    We also know that:

    (1/4) + (1/4) = (1/2)

    and y = z = 4 would be a solution, however this is also not true because y and z must also be different. And y and z cannot be larger than 4, so y=3, therefore

    (1/2) + (1/3) + (1/z) = 1

    Now we are left by 1 variable so we calculate for z. Multiply both sides by 6z:

    3z + 2z + 6 = 6z

    z = 6

    Therefore:

    (1/2) + (1/3) + (1/6) = 1

    so {x, y, z}={2,3,6}
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