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Levans writes a positive fraction in which the numerator and denominator are integers, and the numerator is 1  greater than the denominator. He then writes several more fractions. To make each new fraction, he increases both the numerator and the denominator of the previous fraction by 1 . He then multiplies all his fractions together. He has 20 fractions, and their product equals 3 . What is the value of the first fraction he wrote?

 

I realize you need to do cancellation of some sort, but dont exactly know how to do it. 

 

Thank you! 

 May 23, 2022
 #1
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The first fraction is 6/5.

 May 23, 2022
 #2
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We have the equation: \({{x + 1} \over {x}} \times {x + 2 \over x+1} \times {x + 3 \over x + 2} \times \cdot \cdot \cdot \times {x+21 \over x+20} = 3\).

 

Note that the \(x+1\) all the way to the \(x+20\) terms cancel out. 


This gives us: \({{x + 21} \over {x}} = 3\).

 

Can you take it from here?

 May 23, 2022
 #3
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I got 21/2 but the answer is wrong, am I missing something?

Guest May 23, 2022
 #4
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Close... 


We have the equation: \({{x + 21} \over x} = 3\)

 

Cross multiplying gives us: \(x + 21 = 3x \).

 

Subtracting x from both sides gives us: \(21 = 2x \)

 

This means \(x = {21 \over 2}\).

 

Now, we subsitute this in, and get: \(\large{{21 \over 2} + 1 \over {21 \over 2}}\).

 

Can you take it from here?

BuilderBoi  May 23, 2022
 #5
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Yes, I got 23/21 but it was wrong again. 

Guest May 23, 2022
 #6
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Wait... my bad... I made an error. In my original answer, there are 21 terms, not 20. 

 

It should be \({x+1 \over x} \times {x +2 \over x+1} \times {x +3 \over x +2 } \times \cdot \cdot \cdot \times {x + 20 \over x +19} = 3\)


This means we have the equation: \(x + 20 = 3x\), where x = 10, meaning the original fraction was \(\color{brown}\boxed{11 \over 10}\)

BuilderBoi  May 23, 2022
 #7
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BuilderBoi, why can't you get this right?

Guest May 24, 2022

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