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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
+1

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
+1

I got 21/2 but the answer is wrong, am I missing something?

Guest May 23, 2022
#4
+1

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
+1

Yes, I got 23/21 but it was wrong again.

Guest May 23, 2022
#6
+1

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
0

BuilderBoi, why can't you get this right?

Guest May 24, 2022