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 crack the puzzle we tried using algebra but it did not work so we took this problem from a different approach.

Our answer to this puzzle is p=2 q=3 and r=7 and this would end up like this 1/2+1/3+1/7=41/42 .

We also tried a number of other combination but none of them was as close as this one and this is how we were thinking:

If one of the fractions is 1/2 then this puzzle would be slightly easier. Then we had to try the other two numbers. Of couse the total of the other two should be less than 1 otherwise this puzzle would equal or be more than one so the second number has to be 1/3 1/4 1/5 .... and so on.

If you take1/3 for one of the fractions, the last number has to be 1/7 for the closest result (the bigger the denominator is the smaller the number is).

If you take 1/4 then the last fraction would be 1/3 for the closest result.

Then we figure out when we use 1/2 then all we have to do is find two other numbers that add up to less than 1/2 and we know that 1/3 and 1/6 add to 1/2 so instead of 6 we used 7 and this made the answer smaller because the denonminator is bigger.

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} which is > 1

The next we can try then is

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

Then try the next largest value, which is:

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

This is getting closer to

1

but is still greater than

1

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

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

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: I will use the notation $(p, q, r)$ when referring to set of values for the variables.

First I will eliminate the '3rd dimension' to this problem by proving that p = 2. To simplify the explanation I will always have p as the lowest variable, this is OK as p q and r are interchangeable without affecting the sum.

If $p = 3, q > = 3$ and $r > = 3$ as $p$ is the lowest variable. The greatest sum in this case is:

$(3, 3, 4) sum = 11 / 12$ as this is the minimum value each variable can take, therefore maximising the reciprocals. $(3, 3, 3)$ gives $1$ exactly which is not permissible as the sum must be $< 1$ .

For p > = 4, the greatest sum will be given by $p, q$ and $r$ all being equal ie $(p, p, p)$ , as this minimises them (they cannot be higher than $p$ as that is the lowest). The sum will be:

$3 * (1 / p) = 3 / p$

As $p > = 4$ , this cannot exceed $3 / 4$ . Now consider the case $(2, 3, 7)$ . The sum here is $41 / 42$ , which is greater than the maximum established for $p = 3$ and $p \geq 4$ . As $p$ is a positive integer, and $p = 0, p = 1$ both invalidate the condition that the sum must be lower than $1$ , it is proven that $p = 2$ .

The solution is $(2, 3, 7)$ , or if there is a one greater still it too must occur where $p = 2$ .

Now I will prove that $(2, 3, 7)$ in fact gives the maximum.

It is proven that $p = 2$ , therefore:

$\begin{matrix} 1 / 2 + 1 / q + 1 / r & < & 1 \\ 1 / q + 1 / r & < & 1 / 2 \\ q + r & < & qr / 2 \\ 2 q + 2 r - qr & < & 0 \\ qr - 2 q - 2 r & > & 0 \\ q (r - 2) - 2 r & > & 0 \\ q (r - 2) & > & 2 r \\ q & > & (2 r) / (r - 2) \end{matrix}$

If there is a solution better than that of (2,3,7), the following inequality must hold true:

$\begin{matrix} 1 / 2 + 1 / q + 1 / r & > & 41 / 42 \\ 1 / q + 1 / r & > & 10 / 21 \\ 21 q + 21 r & > & 10 qr \\ 21 q + 21 r - 10 qr & > & 0 \\ 10 qr - 21 q - 21 r & < & 0 \\ q (10 r - 21) - 21 r & < & 0 \\ q (10 r - 21) & < & 21 r \\ q & < & (21 r) / (10 r - 21) \end{matrix}$

So it is established that

$(2 r) / (r - 2) < q < (21 r) / (10 r - 21)$

Now I will show there are no solutions that will match this inequality. Here are some individual cases:

r = 3: 6 < q < 7

r = 4: 4 < q < 4.421...

r = 5: 3.333... < q < 3.621...

r = 6: 3 < q < 3.231...

r = 7: 2.8 < q < 3 q > = 3 !

r = 8: 2.666... < q < 2.739 q > = 3 !

None of these inequalities has an integer solution. As $r$ tends to infinity $(2 r) / (r - 2)$ tends to $2$ and $(21 r) / (10 r - 21)$ tends to $2.1$ . Therefore the upper bound will not exceed $3$ for $r > 8$ , so the inequality will continue to fail as $q \geq 3$ .

Therefore the inequality $(2 r) / (r - 2) < q < (21 r) / (10 r - 21)$ has no integer solutions for $q \geq 3, r \geq 3$ .

Therefore as shown earlier there is no case for $p = 2$ that yields a greater sum than the case $(2, 3, 7)$ .

$p = 2$

$q = 3$

$r = 7$

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

//fracmax

#include< iostream.h>

long double p,q,r,p1,q1,r1,max1;

void main() {

max1=0;

for(p=1;p< =20;p++) {

for(q=1;q< =20;q++) {

for(r=1;r< =20;r++) {

if((1/p)+(1/q)+(1/r)< 1 && 1/p+1/q+1/r> max1) {

max1=(1/p)+(1/q)+(1/r);p1=p;q1=q;r1=r;

}

cout< < "p= « < p1< < "; q= « < q1< < "; r= « < r1< < "; maximum= « < max1;

}