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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. \par
Suppose there is a fraction $1/n$ , where $n$ belongs to
$\{1,2,3,4,5,6\}$.
$$\eqalign{ 1/1: 1x &=1\ {\rm so}\ x=1 \cr 1/2: 2x &=1\
{\rm so}\ x=4 \cr 1/3: 3x &=1\ {\rm so}\ x=5 \cr 1/4: 4x
&=1\ {\rm so}\ x=2 \cr 1/5: 5x &=1\ {\rm so}\ x=3 \cr 1/6:
6x &=1\ {\rm so}\ x=6 }$$
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
$${1\over x} + {1\over y}= {1\over{x+y}}$$
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 |
$x=2$ |
$y^2+2y+4=0$ |
so |
$y^2+2y=3$ |
$y=1$ |
or |
$y=4$ |
when |
$x=3$ |
$y^2+3y+2=0$ |
so |
$y^2+3y=5$ |
$y=6$ |
or |
$y=5$ |
when |
$x=4$ |
$y^2+4y+2=0$ |
so |
$y^2+4y=5$ |
$y=1$ |
or |
$y=2$ |
when |
$x=5$ |
$y^2+5y+4=0$ |
so |
$y^2+5y=3$ |
$y=3$ |
or |
$y=6$ |
when |
$x=6$ |
$y^2+6y+1=0$ |
so |
$y^2+6y=6$ |
$y=3$ |
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 $y^2-4y^2=-3y^2< 0$
which has no real values, therefore there is no real solution for
$x$.