Difference between revisions of "2004 AMC 12A Problems/Problem 3"

(See Also)
Line 10: Line 10:
  
 
{{AMC12 box|year=2004|ab=A|num-b=2|num-a=4}}
 
{{AMC12 box|year=2004|ab=A|num-b=2|num-a=4}}
 
The answer is (B) since x can't be 0, which leaves only 49 different values for y.
 

Revision as of 20:08, 17 January 2008

Problem

This is an empty template page which needs to be filled. You can help us out by finding the needed content and editing it in. Thanks.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See Also

2004 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 AMC 12 Problems and Solutions