Difference between revisions of "2019 AMC 8 Problems/Problem 17"
(→Solution 1) |
|||
(22 intermediate revisions by 12 users not shown) | |||
Line 1: | Line 1: | ||
− | ==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> | |
+ | |||
+ | 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 <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{(B)\frac{50}{99}}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | Associated video - https://www.youtube.com/watch?v=yPQmvyVyvaM | ||
+ | |||
+ | https://www.youtube.com/watch?v=ffHl1dAjs7g&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=1 ~ MathEx | ||
+ | |||
+ | == Video Solution == | ||
− | + | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=VezsRMJvGPs&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=18 | |
− | + | ==Video Solution== | |
+ | https://youtu.be/e1EJNZu-jxM | ||
− | ~ | + | ~savannahsolver |
==See Also== | ==See Also== |
Revision as of 12:32, 7 February 2022
Contents
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 , 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 .
Video Solution
Associated video - https://www.youtube.com/watch?v=yPQmvyVyvaM
https://www.youtube.com/watch?v=ffHl1dAjs7g&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=1 ~ MathEx
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=VezsRMJvGPs&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=18
Video Solution
~savannahsolver
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.