Challenge Level

The examples below give a feel for the range of solutions we received. They are interesting in their variety of approaches and detail. Well done to all of you.

Firstly, Rachael and Katy of Ardingly College Junior School sent a solution obtained through trial and improvement. Some people undervalue this valid mathematical method and it can often lead us to a complete answer or, at the very least, throw light on what to be looking for to find a fuller solution to the problem.

For this problem we have found more than one solution.

To work this out we used trial and error.

We started with $0.5$ giving $0.5 \times 0.5 + 0.5 = 0.25 + 0.5 = 0.75$

The answer was too low so we tried 1 giving $1 \times 1 + 1 = 1 + 1 = 2$. This is a correct solution.

The numbers are the same and so equal the same whichever way round they go.

Then I remembered that if you square a negative number the answer is positive so I tried $-2$ giving $-2\ \times -2 + (-2) = 4 + (-2) = 2 $.

This is another possible solution.

James sent us this solution.

From the question we can form $3$ simultaneous equations.

\begin{eqnarray} ab+ c&=&2 \\ bc+ a&=&2 \\ ca+ b&=&2. \end{eqnarray}

So, $$ab + c= ac+ b. $$ Take $b$ and $c$ from each side: $$ a(b-1) = a(c-1) $$ So, if $a$ is not equal to zero, then $b = c$. Since any two of the three equations could have been chosen, by symmetry we must have $a = b= c$. This gives us the quadratic:
\begin{eqnarray} a^2 + a&=& 2 \\ a^2 + a- 2 &=& 0 \\ (a-1)(a+2) &=& 0 \end{eqnarray}

So the only 2 solutions are: $a =b =c = 1$ and $-2 $.A complete solution, which explains the equality of the three unknowns more fully was supplied by Derek of Tin College.

We seek numbers $a, b$ and $c$ such that:

$a + b c = 2 \quad (1)$

$b + c a = 2 \quad (2)$

$c + a b = 2 \quad (3)$

By subtracting (1)-(2) and simplifying we get

$(a-b)(c-1) = 0 \quad (4)$

and similarly

$(b - c)(a - 1)=0 \quad (5)$

$(c - a)(b - 1)=0 \quad (6)$

Notice that if $a=1$ then (from (4)) we have either $b=1$ or $c=1$. Then using (3) or (2) respectively, we deduce that $a=b=c=1$. Similarly if $b=1$ or $c=1$ then it follows that $a=b=c=1$.

Now suppose for the moment that none of $a$, $b$ and $c$ are 1 because we are looking for other solutions. Then from (4), (5) and (6), we deduce $a = b = c$. From (1), $a^2 + a - 2 = 0$ and factorising this we get $(a - 1)(a + 2) = 0$. As $a \ne 1$, this gives $a = -2$ and hence $a = b = c = -2$.