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:
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|
1
2
|
+ |
1
2
|
+ |
1
2
|
= |
3
2
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which is > 1 |
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The next we can try then is
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1
2
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+ |
1
3
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+ |
1
2
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= |
4
3
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which is > 1 |
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Then try the next largest value, which is:
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1
2
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+ |
1
3
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+ |
1
3
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= |
7
6
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which is > 1 |
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This is getting closer to 1 but is still greater than 1 Try denominators of 2, 3 and 4 giving:
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|
1
2
|
+ |
1
3
|
+ |
1
4
|
= |
13
12
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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:
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 ³ 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:
If there is a solution better than that of (2,3,7), the following
inequality must hold true:
So it is established that
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(2r)/(r-2) < q < (21r)/(10r-21) |
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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
(2r)/(r-2) tends to 2 and (21r)/(10r-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 ³ 3.
Therefore the inequality (2r)/(r-2) < q < (21r)/(10r-21) has no
integer solutions for q ³ 3, r ³ 3.
Therefore as shown earlier there is no case for p = 2 that yields a
greater sum than the case (2,3,7).
So 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; }