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

Line 10: Line 10:
 
==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}}

Revision as of 01:13, 5 July 2013

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

By telescoping, it's easy to see the sum becomes $\frac{2006}{2}=\boxed{\textbf{(C)}\ 1003}$.

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