Difference between revisions of "2006 AMC 8 Problems/Problem 9"

m (Solution)
 
(5 intermediate revisions by 4 users not shown)
Line 6: Line 6:
  
 
== Solution ==
 
== Solution ==
After looking at the problem, we immediately notice that terms cancel out, leaving us with
+
The numerator in each fraction cancels out with the denominator of the next fraction. There are only two numbers that didn't cancel: <math>\frac{2006}{2}=\boxed{\textbf{(C)}\ 1003} </math>.
: <math> \frac{2006}{2}=\boxed{\textbf{(C)}\ 1003} </math>. And that is the answer.
+
 
 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=HrlLDNc4u34  ~David
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/yStRsLftX_0
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2006|num-b=8|num-a=10}}
 
{{AMC8 box|year=2006|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:22, 29 October 2024

Problem

What is the product of $\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{2006}{2005}$ ?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1002\qquad\textbf{(C)}\ 1003\qquad\textbf{(D)}\ 2005\qquad\textbf{(E)}\ 2006$

Solution

The numerator in each fraction cancels out with the denominator of the next fraction. There are only two numbers that didn't cancel: $\frac{2006}{2}=\boxed{\textbf{(C)}\ 1003}$.

Video Solution

https://www.youtube.com/watch?v=HrlLDNc4u34 ~David

Video Solution by WhyMath

https://youtu.be/yStRsLftX_0

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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