A correct solution came from Bradley University students Dan Keshen, Ryan Collins.  Another correct solution was submitted by Lou Cairoli. 

There are 16 such simple fractions.  

This is one of those situations in mathematics where it is easier to solve the more general problem:   There are 2^n-2 simple fractions that can be built from the n primes a₁, ... , a_n.  Let's show this by induction, using the trivial case n = 2 to anchor the induction. 

First note that whenever you simplify a complex fraction like the ones in the statement of the problem a₁ will end up in the numerator and a₂ in the denominator; this is easy to see but the skeptics can verify it by induction.  Since a₁ ends up in the numerator and a₂ ends up in the denominator, and each of the other n - 2 primes can end up in (at most) one of these, it follows that you can realize at most 2^n-2 simple fractions as complex fractions.  Let's show that, in fact, any simple fraction with a₁ in the numerator and a₂ in the denominator can be so realized.  

Let N  and D  be the numerator and denominator of the desired simple fraction.  Let N = a₁ a_i+1 N ' and D = a₂ ... a_i D ' so that i is the smallest index with i > 1 for which a_i+1 appears in the numerator.  By induction, you can realize the simple fraction F₁ = a₁ /(a₂ ... a_i-1); similarly, you can realize the simple fraction F₂ = (a_i D ') /(a_i+1 N ').  Therefore, the complex fraction F₁ / F₂ realizes N /D.

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©2005 Alberto L. Delgado