Fracmax

There were a number of good solutions to this problem and I have picked several to illustrate the different approaches adopted.

First Fred and Matt from Albion Heights sent in a solution, which I rather liked. They have not explained why they did not try 1/3 as the first fraction or give full reasoning for why they knew they had a largest value, but the argument as far as it goes is a good one. I have added some examples below to illustrate what they said in a little more detail. I have included a solution based on the one from Tom or STRS, which was similar to that of Curt from Reigate School . Andrei, of Tudor Vianu College, also looked for solutions using a spreadsheet and sent in a program in C++ to search for soultions. I have included this below.

Well done to all of you.

First Fred and Matt's approach to get us started::

To make a fraction as large as possible you need to make the denominator as small as possible. The smallest a denominator can be is $2$. All the denominators cannot be $2$ because:

$$ \frac{1}{2} + \frac{1}{2} + \frac{1}{2}= \frac{3}{2} \mbox{ which is }> 1 $$

The next we can try then is

$$ \frac{1}{2} + \frac{1}{3} + \frac{1}{2}= \frac{4}{3} \mbox{ which is }> 1 $$

Then try the next largest value, which is:

$$ \frac{1}{2} + \frac{1}{3} + \frac{1}{3}= \frac{7}{6} \mbox{ which is }> 1 $$

This is getting closer to $1$ but is still greater than $1$

Try denominators of $2$, $3$ and $4$ giving:

$$ \frac{1}{2} + \frac{1}{3} + \frac{1}{4}= \frac{13}{12} \mbox{ which is }> 1 $$

Continuing in this way, the first set of fractions that give a total less than $1$ and with the smallest possible denominators (and largest sum) is:

$$ \frac{1}{2} + \frac{1}{3} + \frac{1}{7}= \frac{41}{42} \mbox{ which is }< 1$$

Andrei's program (I haven't tested it!):