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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... ### Sum Equals Product

The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 ï¿½ 1 [1/3]. What other numbers have the sum equal to the product and can this be so for any whole numbers? ### 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.

# Keep it Simple

##### Age 11 to 14Challenge Level

Catherine and Poppy from Stoke by Nayland Middle School made a good start on this problem, and Kijung from Wind Point Elementary School found that:

Not all of Charlie's examples were right.

To be correct, one of the unit fractions must have a denominator which is 1 more than the denominator of the original unit fraction, and the other unit fraction must have a denominator which is the product of the other two denominators:

$$\frac{1}{n} = \frac{1}{n+1}+\frac{1}{n(n+1)}$$

Here are some other examples that work:

$\frac{1}{5} = \frac{1}{6}+\frac{1}{30}$
$\frac{1}{6} = \frac{1}{7}+\frac{1}{42}$
$\frac{1}{105} = \frac{1}{106}+\frac{1}{11130}$

$\frac{1}{8}$ can also be expressed as the sum of two unit fractions in several ways:

$\frac{1}{8} = \frac{1}{9} +\frac{1}{72}$
$\frac{1}{8} = \frac{1}{10} +\frac{1}{40}$
$\frac{1}{8} = \frac{1}{11} + \frac{1}{n}$ is not possible
$\frac{1}{8} = \frac{1}{12} +\frac{1}{24}$

Felix from Condover Primary acutely observed that unit fractions with denominators which are prime numbers can only be written in one way as the sum of two distinct unit fractions.

Rose, from Claremont Primary School in Tunbridge Wells, Kent worked out a general formula:

$\frac{1}{z} = \frac{1}{y}+\frac{1}{x}$ (where $z$, $y$ and $x$ are positive integers and $y < x$)

Using $\frac{1}{10}$ as an example:

$\frac{1}{10} = \frac{1}{11}+\frac{1}{110}$

$\frac{1}{10} = \frac{1}{12}+\frac{1}{60}$

$\frac{1}{10} = \frac{1}{14}+\frac{1}{35}$

$\frac{1}{10} = \frac{1}{15}+\frac{1}{30}$

I listed the values of $y-z$ that provide solutions:

$1$, $2$, $4$ and $5$

These are also the factors of $z^ 2$ (i.e. $100$) that are smaller than its square root: $1\times100$

$2\times50$

$4\times25$

$5\times20$

$10\times10$

This pattern also occurred for $\frac{1}{12}$:

$\frac{1}{12} = \frac{1}{13}+\frac{1}{156}$

$\frac{1}{12} = \frac{1}{14}+\frac{1}{84}$

$\frac{1}{12} = \frac{1}{15}+\frac{1}{60}$

$\frac{1}{12} = \frac{1}{16}+\frac{1}{48}$

$\frac{1}{12} = \frac{1}{18}+\frac{1}{36}$

$\frac{1}{12} = \frac{1}{20}+\frac{1}{30}$

$\frac{1}{12} = \frac{1}{21}+\frac{1}{28}$

Here $y - z = 1, 2, 3, 4, 6, 8, 9$

and the factors of $z ^ 2$ (i.e.$144$) are:

$1\times144$

$2\times72$

$3\times48$

$4\times36$

$6\times24$

$8\times18$

$9\times16$

$12\times12$

$\frac{1}{10}$ can be written as the sum of two different unit fractions in $4$ ways.

In this case $z ^ 2$ has $9$ factors and $y-z = 4$

$\frac{9-1}{2}=4$

$\frac{1}{12}$ can be written as the sum of two different unit fractions in $7$ ways.

In this case $z ^ 2$ has $15$ factors and $y-z = 7$

$\frac{15-1}{2}=7$

Conclusion:

If $n$ is the number of factors of $z ^ 2$,

$\frac{1}{z}$ can be written as the sum of two different unit fractions in $\frac{n -1}{2}$ ways.

Rose's conclusion draws on her two examples, but when we generalise in mathematics, we need to be sure that what we have noticed will be true in all other cases.

Can anyone provide a convincing explanation for why Rose's conclusion is, or is not, correct?