2019 AMC 8 Problems/Problem 17
Contents
- 1 Problem
- 2 Solution 1 (Telescoping)
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution 1
- 7 Video Solution 2
- 8 Video Solution 3
- 9 Video Solution 3(an Elegant way)
- 10 Video Solution 4 by OmegaLearn
- 11 Video Solution
- 12 Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 13 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 (B) .
Video Solution 1
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.