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 <math> \frac{2006}{2}=\boxed{\textbf{(C)}\ 1003} </math>. And that is the answer.
We notice that the numerator in each fraction cancels out with the denominator of the one following it. This leaves us with only two numbers that didn't cancel: <math> \frac{2006}{2}=\boxed{\textbf{(C)}\ 1003} </math>.


==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 02:40, 14 June 2022

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

We notice that the numerator in each fraction cancels out with the denominator of the one following it. This leaves us with only two numbers that didn't cancel: $\frac{2006}{2}=\boxed{\textbf{(C)}\ 1003}$.

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

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