### Tweedle Dum and Tweedle Dee

Two brothers were left some money, amounting to an exact number of pounds, to divide between them. DEE undertook the division. "But your heap is larger than mine!" cried DUM...

### Special Sums and Products

Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.

### 3388

Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.

# Sum Equals Product

##### Age 11 to 14 Challenge Level:

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.