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

(Solution)
m (Solution)
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== Solution ==
 
== Solution ==
 
After looking at the problem, we immediately notice that terms cancel out, leaving us with
 
After looking at the problem, we immediately notice that terms cancel out, leaving us with
: <math> \frac{2006}{2}=\boxed{\textbf{(C)}\ 1003} </math>. And that is your answer.
+
: <math> \frac{2006}{2}=\boxed{\textbf{(C)}\ 1003} </math>. And that is the answer.
  
 
==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}}

Revision as of 22:56, 8 August 2018

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

After looking at the problem, we immediately notice that terms cancel out, leaving us with

$\frac{2006}{2}=\boxed{\textbf{(C)}\ 1003}$. And that is the answer.

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

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