AMC 8 2019 Problem 17 | Value of Product

Content

Try out this beautiful algebra problem from AMC 8, 2019 based on finding the value of the product. You may use sequential hints to solve the problem.

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

But try the problem first...

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

Source

AMC 8, 2019, Problem 17

Try with Hints

First hint

We write

$\left(\frac{1.3}{2.2}\right)\left(\frac{2.4}{3.3}\right)\left(\frac{3.5}{4.4}\right) \ldots\left(\frac{97.99}{98.98}\right)\left(\frac{98.100}{99.99}\right)$

in a different form like

$\frac{1}{2} \cdot\left(\frac{3.2}{2.3}\right) \cdot\left(\frac{4.3}{3.4}\right) \cdots \cdots \left(\frac{99.98}{98.99}\right) \cdot \frac{100}{99}$

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.