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\(\dfrac{a}{b}+\dfrac{c}{d}+\dfrac{e}{f}+\dfrac{g}{h}<2, \)all 8 of the variables are different single digit numbers, what values for each make the expression on the left the biggest possible.

Guest Feb 2, 2018
edited by Guest  Feb 2, 2018
 #1
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+1

Just to clarify. There are 8 variables, half are divided by the other half, and they are all different single digit numbers. Together, in the previously mention format, they need to be less than 2, but as big as possible.

Guest Feb 2, 2018
 #2
avatar+2075 
+2

It appears as if \(\frac{1}{2}+\frac{3}{7}+\frac{4}{9}+\frac{5}{8}\) is the best combination.

 

By the way, \(\frac{1}{2}+\frac{3}{7}+\frac{4}{9}+\frac{5}{8}\approx1.99802\)

TheXSquaredFactor  Feb 3, 2018
edited by TheXSquaredFactor  Feb 3, 2018
 #3
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Thank you! Could you explain how you got there though? I'm a little confused.

Guest Feb 3, 2018
 #4
avatar+2075 
+2

Being confused is quite natural. I am not going to say that my method is completely sound, but it seemed to get the job done. 

 

I broke up each fraction into individual terms, and I knew that each one would, on average, be \(\frac{1}{2}\) since \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\), so my goal was to make every fraction as close as possible to the value 1/2. If I can do that, then I should be reasonably close to 2. 

 

I filled in the first a/b with 1/2 because that is exactly 1/2, so I have already solved one fraction, but I have used up the 1 and the 2. 

 

Now, a=1 and b=2. That seems pretty good. I then tried this with the other fractions. c/d=3/6 and e/f equals 4/8. I have one fraction left. We now have a few variables defined. They are the following:
 

a=1

b=2

c=3

d=6

e=4

f=8

 

The numbers I have left are 5,7, and 9, and there is unfortunately no way of making a fraction that is less than 1/2, so I had to backtrack. 

 

I then realized that I had to make the fractions' average to 1/2, so I decided to take a different approach. I still left a/b=1/2, so a=1 and b=2. 

 

This means that I should have denominators 7,8, and 9. This way, I do not affect the average as much. Let me explain. If I have a fraction with a denominator of 4, for example, then the halfway point is 2. However, the closest possibilities such as 3 and 1 affect the fraction's value by 25%, so I realized that, the larger the denominator, the less the affect is if I increment the numerator by 1.
 

a=1

b=2

 

 

I have now reduced the possibilities to \(\frac{1}{2}+\frac{c}{7}+\frac{e}{8}+\frac{g}{9}\). I knew that none of the fractions should have a 6 because that would be too large for the average to work, so I just tested a few possibilities and \(\frac{1}{2}+\frac{3}{7}+\frac{4}{9}+\frac{5}{8}\) is the one that worked the best.

TheXSquaredFactor  Feb 3, 2018

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