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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.

# Egyptian Fractions

##### Age 11 to 14Challenge Level

This was a tough one, well done to lots of you who sent in lots of different examples, but only Rosie gave a reason and general rule for her findings:

I used Keep It Simple first to find that all unit fractions can be expressed as the sum of two other unit fractions like this,
$\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}$
I noticed that if $n$ was odd, then the numbers $n+1$ and $n(n+1)$ were both even. So if $n$ is odd, say for $m$ a positive integer, $n=2m+1$.
$\frac{2}{n}=\frac{2}{n+1}+\frac{2}{n(n+1)}$
$\frac{2}{n}=\frac{2}{2m+1}=\frac{2}{2m+2}+\frac{2}{(2m+1)(2m+2)}$
$\frac{2}{n}=\frac{2}{2m+1}=\frac{1}{m+1}+\frac{1}{(2m+1)(m+1)}$
Which are unit fractions.

Now if $n$ is even, then $n+1$ is odd, and so will not cancel like above. However, as $n$ is even than half of it is still a whole number. So if $n=2p$, $\frac{2}{n}$ is going to cancel to $\frac{1}{p}$, and we know that $\frac{1}{p}=\frac{1}{p+1}+\frac{1}{p(p+1)}$ from above.
So
$\frac{2}{n}=\frac{1}{p+1}+\frac{1}{p(p+1)}$
which we can write in terms of $n$
$\frac{2}{n}=\frac{2}{n+2}+\frac{4}{n(n+2)}$
These are unit fractions, and so we're done.

Great, can anyone use this to find $\frac{3}{n},\frac{4}{n},\frac{5}{n}$ and so on?