# 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

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