Congratulations to Wei Zhang, age 18 from Merchiston Castle School for the superb solution given below. Andrei Lazanu, age 14, School No. 205, Bucharest, Romania submitted a similar solution.

Suppose there is a fraction $1 / n$ , where $n$ belongs to ${1, 2, 3, 4, 5, 6}$ .

$\begin{matrix} 1 / 1 : 1 x & = 1 so x = 1 \\ 1 / 2 : 2 x & = 1 so x = 4 \\ 1 / 3 : 3 x & = 1 so x = 5 \\ 1 / 4 : 4 x & = 1 so x = 2 \\ 1 / 5 : 5 x & = 1 so x = 3 \\ 1 / 6 : 6 x & = 1 so x = 6 \end{matrix}$

As I showed above, $1 / n$ is in the set ${1, 2, 3, 4, 5, 6}$ .

Now we have $a / n = a \times 1 / n$ , where $a$ belongs to ${0, 1, 2, 3, 4, 5, 6}$ , this is a number from the set ${0, 1, 2, 3, 4, 5, 6}$ times a number from the set of ${1, 2, 3, 4, 5, 6}$ . Therefore we will get a set of numbers for $a / n$ , which is ${0, 1, 2, 3, 4, 5, 6}$ . Solving the equation

$\frac{1}{x} + \frac{1}{y} = \frac{1}{x + y}$

is equivalent to solving $x^{2} + xy + y^{2} = 0$ .

when $x = 1$ $y^{2} + y + 1 = 0$ so $y^{2} + y = 6$ $y = 2$ or $y = 4$

when $x = 2$ $y^{2} + 2 y + 4 = 0$ so $y^{2} + 2 y = 3$ $y = 1$ or $y = 4$

when $x = 3$ $y^{2} + 3 y + 2 = 0$ so $y^{2} + 3 y = 5$ $y = 6$ or $y = 5$

when $x = 4$ $y^{2} + 4 y + 2 = 0$ so $y^{2} + 4 y = 5$ $y = 1$ or $y = 2$

when $x = 5$ $y^{2} + 5 y + 4 = 0$ so $y^{2} + 5 y = 3$ $y = 3$ or $y = 6$

when $x = 6$ $y^{2} + 6 y + 1 = 0$ so $y^{2} + 6 y = 6$ $y = 3$ or $y = 5 .$

We have found the six solution pairs $(1, 2)$ , $(1, 4)$ , $(2, 4)$ , $(3, 6)$ , $(3, 5)$ , $(5, 6)$ . The equation is symmetric in $x$ and $y$ so there are six corresponding solutions when we exchange $x$ and $y$ . We don't allow negative solutions (e.g. $x = 1, y = - 3$ ) because we are working entirely in the set ${0, 1, 2, 3, 4, 5, 6}$ . If we work with real numbers, we think of solving the quadratic equation $x^{2} + yx + y^{2} = 0$ as an equation in $x$ . The formula for the solution of this quadratic equation involves $y^{2} - 4 y^{2} = - 3 y^{2} < 0$ which has no real values, therefore there is no real solutions for $x$ .

when	$x = 1$	$y^{2} + y + 1 = 0$	so	$y^{2} + y = 6$	$y = 2$ or $y = 4$
when	$x = 2$	$y^{2} + 2 y + 4 = 0$	so	$y^{2} + 2 y = 3$	$y = 1$ or $y = 4$
when	$x = 3$	$y^{2} + 3 y + 2 = 0$	so	$y^{2} + 3 y = 5$	$y = 6$ or $y = 5$
when	$x = 4$	$y^{2} + 4 y + 2 = 0$	so	$y^{2} + 4 y = 5$	$y = 1$ or $y = 2$
when	$x = 5$	$y^{2} + 5 y + 4 = 0$	so	$y^{2} + 5 y = 3$	$y = 3$ or $y = 6$
when	$x = 6$	$y^{2} + 6 y + 1 = 0$	so	$y^{2} + 6 y = 6$	$y = 3$ or $y = 5 .$