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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 � 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 14
Challenge 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?



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