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
<span>
</span>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
<span>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:</span>
(1/2) + (1/y) +
(1/z) = 1
Therefore
(1/y) + (1/z) =
1/2
<span>We also know that:</span>
(1/4) + (1/4) =
(1/2)
<span>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</span>
(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
<span>so {x,y,z}={2,3,6} </span>
<span> </span>
