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

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{{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 19:08, 17 January 2008

Problem

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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