The 2 is going to be in the denominator of the simplified fraction in all cases.

For the other numbers consider the following: the fraction is A/B assembled from the first n-1 numbers.  Then we are to add in the division for n.  There are two ways to do this; either (A/B)/n or A/(B/n).  In the first case n gets pulled into the denominator forming A/(B*n) or in the second case n gets pulled into the numerator forming (A*n)/B.  So n can be either in the numerator denominator.

Then each of 3, 4, 5, and 6 could occur in the numerator or denominator of the simplified fraction.

5 must be in the numerator for the fraction to simplify to an integer.

There can be at most two factors of 2 in the denominator.  But because 2 is already in the denominator, then 4 must be in the numerator.

Then at most one of 3 and 6 can occur in the denominator.

This leaves us with three integer values: (1*3*4*5*6)/2=180, (1*3*4*5)/(2*6)=5, and (1*4*5*6)/(2*3)=20.

For completeness, the original divisions of the integer values are 1/((((2/3)/4)/5)/6)=180, (1/((((2/3)/4)/5))/6=5, and (1/2)/(((3/4)/5)/6)=20.