Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 8 Problems/Problem 17"
m (→‎Solution 3)
Line 18: Line 18:
  
 
==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}{99!}</math> We rewrite <math>99!</math> as <math>99 \cdot 98!</math> and cancel <math>98!</math>, which gets <math>\boxed{\frac{50}{99}}</math>. Answer B.
+
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{\frac{50}{99}}</math>. Answer B.
 +
 
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2019|num-b=16|num-a=18}}
 
{{AMC8 box|year=2019|num-b=16|num-a=18}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:37, 3 February 2020

Problem 17

What is the value of the product \[\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?\]

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

Solution 1

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

\[\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}\]

~phoenixfire

Solution 2

If you calculate the first few values of the equation, all of the values tend to $\frac{1}{2}$, but are not equal to it. The answer closest to $\frac{1}{2}$ but not equal to it is $\boxed{\textbf{(B)}\frac{50}{99}}$.~heeeeeeeheeeee

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 $\boxed{\frac{50}{99}}$. Answer B.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_17&oldid=116773"