Art of Problem Solving

2006 AMC 8 Problems/Problem 9: Difference between revisions

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== 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 your 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

These problems are copyrighted © by the Mathematical Association of America.