Question

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

Answer 1

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}