# An interesting problem...?

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Stephen is making "the Stephen fractions". He chooses a set of two whole numbers (could be identical) greater than 1. He multiplies them, squares the result, and finds the reciprocal of the answer. (e.g. #(3,4)# =>#1/(3*4)^2=>1/144)# Given that the order of the numbers do not matter (#(3,4)# and #(4,3)# are the same), what is the sum of all Stephen fractions if there is a sum? Note that #1/(3*4)^2# and #1/(2*6)^2# are counted as two different fractions.

I think I have found a solution, and I would like to know how you would solve this problem!

Stephen is making "the Stephen fractions". He chooses a set of two whole numbers (could be identical) greater than 1. He multiplies them, squares the result, and finds the reciprocal of the answer. (e.g.

I think I have found a solution, and I would like to know how you would solve this problem!

##### 1 Answer

#### Answer:

The sum of all Stephen fractions is

#### Explanation:

Evaluating the problem, we know that Stephen fractions sum

I replace in each sum

\0/ Here's our answer !