A unit fractionis a fraction of the form 1/n for some nonzero integer n. Compute the number of ways we can write 1/60 as the sum of two distinct positive unit fractions. (The order of the fractions in the sum does not matter, so 1/2+1/3 would be considered the same sum as 1/3+1/2)
Each pair of distinct factors of 60 produce a pair of distinct unit fractions whose sum is \(\frac{1}{60}\).
This is done as follows:
assume m and n are distinct factors of 60;
the two fractions \(\frac{m}{(m+n)(60)}\ and \ \frac{n}{(m+n)(60)}\)can be then proven to be unit fractions whose sum is \(\frac{1}{60}\).
Instead of spending time on the proof (I will be glad to show the proof to anyone interested), I will do an example.
Let m=5 and n=15. Check to make sure they are factors of 60. The fractions
\(\frac{5}{(5+15)(60)}=\frac{5}{20(60)}=\frac{1}{240}\)
and
\(\frac{15}{(5+15)(60)}= \frac{1}{80}\)
are clearly unit fractions and if we add them together, the result would be what we expected:
\(\frac{1}{240}+\frac{1}{80}=\frac{1}{240}+\frac{3}{240}=\frac{4}{240}=\frac{1}{60}\)
Note that the above calculations show that the pair of factors (1, 3) produce the same unit fractions as (5, 15). This is not accidental and in the table displayed above the relatively prime pairs, that is the ones that have only 1 as a common factor, are separated and these are the only ones we count; each non-realtively-prime pair of factors is associated with a relatively prime pair that produces the same unit fractions. So if I am not mistaken( and there is a 0.99 probability that I am not) , there are a total of 21 pairs of distinct (in two different ways) unit fractions that add up to \(\frac{1}{60}\).
