Difference between revisions of "2019 AMC 8 Problems/Problem 17"
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==Solution 3== | ==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>. | 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>. | ||
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| + | ==Solution 4== | ||
| + | Obviously all of the terms are less than one by difference of squares, so we eliminate options <math>(D)</math> and <math>(E)</math>. <math>(C)</math> is too close to one to be possible, and as stated in Solution 2 it gets close to but never actually becomes <math>1/2</math>. Therefore, by process of elimination we get that the answer is <math>B</math>. | ||
==Video Solution 1== | ==Video Solution 1== | ||
Revision as of 16:20, 20 December 2022
Contents
Problem
What is the value of the product
![\[\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?\]](http://latex.artofproblemsolving.com/9/c/a/9ca55aef127eee6fc870173ae1b23e1abd56a02b.png)
Solution 1 (Telescoping)
We rewrite: ![\[\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}\]](http://latex.artofproblemsolving.com/7/2/4/724d2c4b0da8e0a1d6b0f9ba10f8269651695790.png)
The middle terms cancel, leaving us with
![\[\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}\]](http://latex.artofproblemsolving.com/e/c/8/ec85e9681a2b497df4971e3bc0a27d4515b654e8.png)
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
Obviously all of the terms are less than one by difference of squares, so we eliminate options 
and 
. 
is too close to one to be possible, and as stated in Solution 2 it gets close to but never actually becomes 
. Therefore, by process of elimination we get that the answer is 
.
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
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. 
