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# Fractions

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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 3 fractions, and their product equals 5. What is the value of the first fraction he wrote?

Feb 15, 2024

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This at first feels like it will be a really nasty equation, but lets just start writing it out.

Suppose the original fraction is $$\frac{a+1}{a}$$. The next one is $$\frac{a+2}{a+1}$$. The third one is $$\frac{a+3}{a+2}$$. Multiplying these, we get $$\frac{a+1}{a}*\frac{a+2}{a+1}*\frac{a+3}{a+2}=\frac{a+3}{a}$$. We see these cancel out!

This equals five, so $$\frac{a+3}{a}=5$$, 5a = a+3, 4a=3. a = 3/4. The original fraction written is $$\frac{\frac{3}{4}+1}{\frac{3}{4}}=\frac{7}{3}$$. BUT NOTE!! A better answer might be better LEAVING IT IN THE FORM $$\frac{\frac{3}{4}+1}{\frac{3}{4}}$$ or $$\frac{\frac{7}{4}}{\frac{3}{4}}$$, BECAUSE $$\frac{7+1}{3+1} \neq \frac{\frac{7}{4}+1}{\frac{3}{4}+1}$$.

Feb 15, 2024
edited by hairyberry  Feb 15, 2024

#1
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Suppose the original fraction is $$\frac{a+1}{a}$$. The next one is $$\frac{a+2}{a+1}$$. The third one is $$\frac{a+3}{a+2}$$. Multiplying these, we get $$\frac{a+1}{a}*\frac{a+2}{a+1}*\frac{a+3}{a+2}=\frac{a+3}{a}$$. We see these cancel out!
This equals five, so $$\frac{a+3}{a}=5$$, 5a = a+3, 4a=3. a = 3/4. The original fraction written is $$\frac{\frac{3}{4}+1}{\frac{3}{4}}=\frac{7}{3}$$. BUT NOTE!! A better answer might be better LEAVING IT IN THE FORM $$\frac{\frac{3}{4}+1}{\frac{3}{4}}$$ or $$\frac{\frac{7}{4}}{\frac{3}{4}}$$, BECAUSE $$\frac{7+1}{3+1} \neq \frac{\frac{7}{4}+1}{\frac{3}{4}+1}$$.