2019 AMC 8

Complete problem set with solutions and individual problem pages

Problem 17 Hard

What is the value of the product

\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?

  • A.

    \frac{1}{2}

  • B.

    \frac{50}{99}

  • C.

    \frac{9800}{9801}

  • D.

    \frac{100}{99}

  • E.

    50

Answer:B

Solution 1

We rewrite:

\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}

The middle terms cancel, leaving us with

\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}

 

Solution 2

If you calculate the first few values of the equation, all of the values tend to close to \frac{1}{2}, but are not equal to it. The answer closest to \frac{1}{2} but not equal to it is \boxed{\textbf{(B)}\frac{50}{99}}.

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