Question

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?

NEVERMIND. ITS 11/10

Answer 1

Let's denote the first fraction Levans wrote as $\frac{a}{b}$, where $a$ is the numerator and $b$ is the denominator.

According to the given information, we know that $\frac{a}{b}$ is a positive fraction in which the numerator is $1$ greater than the denominator. Therefore, we can write the equation:

$a = b + 1$

We also know that Levans wrote a total of $20$ fractions, so we can set up an equation using the product of the fractions:

$\left(\frac{a}{b}\right) \cdot \left(\frac{a+1}{b+1}\right) \cdot \left(\frac{a+2}{b+2}\right) \cdot \ldots \cdot \left(\frac{a+19}{b+19}\right) = 3$

To simplify the equation, we can cancel out common factors between the numerator and denominator in each fraction:

$\frac{a(a+1)(a+2)\ldots(a+19)}{b(b+1)(b+2)\ldots(b+19)} = 3$

Now, substituting $a = b + 1$ into the equation:

$\frac{(b+1)(b+2)(b+3)\ldots(b+19)(b+20)}{b(b+1)(b+2)\ldots(b+19)} = 3$

We can see that all the terms in the numerator and denominator cancel out except for the term $(b+20)$ in the numerator and the term $b$ in the denominator:

$\frac{b+20}{b} = 3$

Cross-multiplying, we have:

$b + 20 = 3b$

Simplifying the equation, we get:

$2b = 20$

$b = 10$

Since $a = b + 1$, we have:

$a = 10 + 1 = 11$

Therefore, the value of the first fraction Levans wrote is $\frac{11}{10}$.