Difference between revisions of "2019 AMC 8 Problems/Problem 17"
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− | ==Problem | + | ==Problem== |
What is the value of the product | What is the value of the product | ||
+ | |||
<cmath>\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)?</cmath> | <cmath>\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)?</cmath> | ||
<math>\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</math> | <math>\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</math> | ||
− | ==Solution 1== | + | ==Solution 1 (Telescoping)== |
We rewrite: <cmath>\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}</cmath> | We rewrite: <cmath>\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}</cmath> | ||
− | If | + | The middle terms cancel, leaving us with |
+ | |||
+ | <cmath>\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | If you calculate the first few values of the equation, all of the values tend to close to <math>\frac{1}{2}</math>, but are not equal to it. The answer closest to <math>\frac{1}{2}</math> but not equal to it is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Rewriting the numerator and the denominator, we get <math>\frac{\frac{100! \cdot 98!}{2}}{\left(99!\right)^2}</math>. We can simplify by canceling 99! on both sides, leaving us with: <math>\frac{100 \cdot 98!}{2 \cdot 99!}</math> We rewrite <math>99!</math> as <math>99 \cdot 98!</math> and cancel <math>98!</math>, which gets <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | All of the terms have the form <math>\frac{k^2-1}{k^2}</math>, which is <math><1</math>, so the product is <math><1</math>, 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 <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>. | ||
+ | |||
+ | ==Video Solution == | ||
+ | |||
+ | ==Video Solution by Math-X (First fully understand the problem!!!)== | ||
+ | https://youtu.be/IgpayYB48C4?si=UJVe2zopeqT-4rLM&t=5256 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | https://www.youtube.com/watch?v=yPQmvyVyvaM | ||
− | + | Associated video | |
− | + | https://www.youtube.com/watch?v=ffHl1dAjs7g&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=1 | |
− | ==Solution 2== | + | ~ MathEx |
− | + | ||
+ | == Video Solution 2== | ||
+ | |||
+ | Solution detailing how to solve the problem: | ||
+ | |||
+ | https://www.youtube.com/watch?v=VezsRMJvGPs&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=18 | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | https://youtu.be/e1EJNZu-jxM | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution 3(an Elegant way)== | ||
+ | https://www.youtube.com/watch?v=la3en2tgBN0 | ||
+ | |||
+ | ==Video Solution 4 by OmegaLearn== | ||
+ | https://youtu.be/TkZvMa30Juo?t=3326 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution == | ||
+ | https://youtu.be/wUvi7tzxuTk | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)== | ||
+ | https://youtu.be/Xm4ZGND9WoY | ||
+ | |||
+ | ~Hayabusa1 | ||
==See Also== | ==See Also== |
Latest revision as of 21:37, 16 January 2024
Contents
- 1 Problem
- 2 Solution 1 (Telescoping)
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution
- 7 Video Solution by Math-X (First fully understand the problem!!!)
- 8 Video Solution 2
- 9 Video Solution 3
- 10 Video Solution 3(an Elegant way)
- 11 Video Solution 4 by OmegaLearn
- 12 Video Solution
- 13 Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 14 See Also
Problem
What is the value of the product
Solution 1 (Telescoping)
We rewrite:
The middle terms cancel, leaving us with
Solution 2
If you calculate the first few values of the equation, all of the values tend to close to , but are not equal to it. The answer closest to
but not equal to it is
.
Solution 3
Rewriting the numerator and the denominator, we get . We can simplify by canceling 99! on both sides, leaving us with:
We rewrite
as
and cancel
, which gets
.
Solution 4
All of the terms have the form , which is
, so the product is
, 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
.
Video Solution
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/IgpayYB48C4?si=UJVe2zopeqT-4rLM&t=5256
~Math-X
https://www.youtube.com/watch?v=yPQmvyVyvaM
Associated video
https://www.youtube.com/watch?v=ffHl1dAjs7g&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=1
~ MathEx
Video Solution 2
Solution detailing how to solve the problem:
https://www.youtube.com/watch?v=VezsRMJvGPs&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=18
Video Solution 3
~savannahsolver
Video Solution 3(an Elegant way)
https://www.youtube.com/watch?v=la3en2tgBN0
Video Solution 4 by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=3326
~ pi_is_3.14
Video Solution
~Education, the Study of Everything
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~Hayabusa1
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.