2019 AMC 8 Problems/Problem 17


What is the value of the product


$\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50$

Solution 1 (Telescoping)

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 $\boxed{\textbf{(B)}\frac{50}{99}}$.

Solution 3

Rewriting the numerator and the denominator, we get $\frac{\frac{100! \cdot 98!}{2}}{\left(99!\right)^2}$. We can simplify by canceling 99! on both sides, leaving us with: $\frac{100 \cdot 98!}{2 \cdot 99!}$ We rewrite $99!$ as $99 \cdot 98!$ and cancel $98!$, which gets $\boxed{\textbf{(B)}\frac{50}{99}}$.

Solution 4

All of the terms have the form $\frac{k^2-1}{k^2}$, which is $<1$, so the product is $<1$, so we eliminate options (D) and (E). (C) is too close to 1 to be possible. The partial products seem to be approaching 1/2, so we guess that 1/2 is the limit/asymptote, and so any finite product would be slightly larger than 1/2. Therefore, by process of elimination and a small guess, we get that the answer is $\boxed{\textbf{(B)}\frac{50}{99}}$.

Video Solution

Video Solution by Math-X (First fully understand the problem!!!)




Associated video


~ MathEx

Video Solution 2

Solution detailing how to solve the problem:


Video Solution 3



Video Solution 3(an Elegant way)


Video Solution 4 by OmegaLearn


~ pi_is_3.14

Video Solution


~Education, the Study of Everything

Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)



See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AJHSME/AMC 8 Problems and Solutions

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