# Sum Equals Product

I have been practising arithmetic with fractions.

I worked out $4 + 1 \frac{1}{3}$ but then realised that I had misread the question!

I was supposed to work out $4 \times 1 \frac{1}{3}$

When I worked out the multiplication, I was surprised to find I got the same answer to both calculations!

Can you find other examples of calculations where replacing the multiplication sign by an addition sign does not alter the result of the calculation?

Here are some examples you could try:

$4 + 1 \frac{1}{4}$ and $4 \times 1 \frac{1}{4}$

$5 + 1 \frac{1}{4}$ and $5 \times 1 \frac{1}{4}$

$5 + 1 \frac{1}{5}$ and $5 \times 1 \frac{1}{5}$

$6 + 1 \frac{1}{5}$ and $6 \times 1 \frac{1}{5}$**Which of these give the same answer for both calculations?Can you use these examples to help you find some more pairs that have the same sum and product?**

There was a good world-wide response to this question. James Bollard of St. Peter's College, Australia, answered the first question.Only the whole numbers 0 and 2 will have their sum equal to their product.

The second question about the relationship between the numbers, where one of the numbers is an integer, was successfully answered by a number of people: Kate Lister and Katherine Taylor students at The Emmbrook School,Wokingham,England; Ling Xiang Ning of Tao Nan School, Singapore; Jonathan Bloxham of the Royal Grammar School, Newcastle, England; and Joyce Fu and Sheila Luk both students at the Mount School, York, England. They all come to the correct solution that numbers of the form $n$ and $n/(n-1)$ will have sums equal to their product. Joyce Fu and Sheila Luk also point out this will be true for negative numbers.

Claire Kruithof and Cinde Lau of Madras College, Scotland, go further to investigate and find other pairs of numbers for which the same relationship holds.

\begin{equation*} \left(\frac{(2n + 1)}{n}\right) , \left(\frac{(2n + 1)}{(n + 1)}\right) \end{equation*}

and

\begin{equation*} \left(\frac{(3n + 1)}{n}\right) , \left(\frac{(3n + 1)}{(2n + 1)} \right) \end{equation*}

Catherine Aitken and Elisabeth Brewster, also of Madras College, found these two related pairs and another family of solutions (where $x$ and $y$ are whole numbers and $y \geq x$):

\begin{equation*} \left(\frac{(y + 1)}{x}\right), \left(\frac{(y + 1)}{(y - x + 1)}\right). \end{equation*}

Well done, Catherine and Elisabeth. Their proof is as follows:

$$\begin{eqnarray} \\ \left(\frac{(y + 1)}{x} \right) + \left(\frac{(y + 1)}{(y - x + 1)}\right) &=& \left(\frac{(y + 1)(y - x + 1 + x)}{x(y - x + 1)}\right) \\ &=& \left(\frac{(y + 1)^2}{x(y - x + 1)}\right). \end{eqnarray}$$

To find other families of solutions like this you simply take two algebraic fractions with the same numerator and any two denominators that add up to give the numerator.