A simple combinatorics problem

To break the long hiatus, I’m going to solve a short combinatorics problem extracted from “A Walk Through Combinatorics”:

Find all positive integers a < b < c such that \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1.

As a is the smallest of the three integers, we know that \frac{1}{a} will be the bigger of the three fractions. The three fractions must add to one and they must be different, so \frac{1}{a} must be equal to \frac{1}{2}. Rewriting the restriction using this newly found knowledge:

\displaystyle \frac{1}{2} + \frac{1}{b} + \frac{1}{c} = 1

\displaystyle \frac{1}{b} + \frac{1}{c} = \frac{1}{2}.

b must be greater than 2, but it cannot be bigger than 3 because, otherwise,

\displaystyle \frac{1}{b} + \frac{1}{c} \leq \frac{1}{4} + \frac{1}{5} = \frac{9}{20} < \frac{1}{2}.

Then we have b = 3 and

\displaystyle c = \frac{1}{1 - \frac{1}{2} - \frac{1}{3}} = 6.

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