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AMC 8, 2019 Problem 17 | Smallest Positive Integer

Try out this beautiful algebra problem from AMC 8, 2019 based on the smallest positive integer.

AMC 8, 2019: Problem 17


What is the value of the product

$$\left(\frac{1 \cdot 3}{2 \cdot 2}\right)\left(\frac{2 \cdot 4}{3 \cdot 3}\right)\left(\frac{3 \cdot 5}{4 \cdot 4}\right) \cdots\left(\frac{97 \cdot 99}{98 \cdot 98}\right)\left(\frac{98 \cdot 100}{99 \cdot 99}\right) ?$$

(A) $\frac{1}{2}$


(B) $\frac{50}{99}$


(C) $\frac{9800}{9801}$


(D) $\frac{100}{99}$


(E) $50$


Key Concepts

Algebra

Value

Telescoping


Check the Answer


Answer: is $\frac{50}{99}$

AMC 8, 2019, Problem 17

Try with Hints


First hint

We write $\left(\frac{1.3}{2.2}\right)\left(\frac{2.4}{3.3}\right)\left(\frac{3.5}{4.4}\right) \ldots \ldots . .\left(\frac{97.99}{98.98}\right)\left(\frac{98.100}{99.99}\right)$ in a different form like
$\frac{1}{2} \cdot\left(\frac{3.2}{2.3}\right) \cdot\left(\frac{4.3}{3.4}\right) \cdots \cdots \cdots \cdots\left(\frac{99.98}{98.99}\right) \cdot \frac{100}{99}$

Second Hint

All of the middle terms eliminate each other, and only the first and last term remains i.e.

$\frac{1}{2} \cdot \frac{100}{99}$

Final Step

$\frac{1}{2} \cdot \frac{100}{99}=\frac{50}{99}$

and that is the final answer.

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