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# Egyptian Fractions

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### Chocolate

### Tweedle Dum and Tweedle Dee

### Sum Equals Product

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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?

There are three tables in a room with blocks of chocolate on each. Where would be the best place for each child in the class to sit if they came in one at a time?

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...

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?