Difference between revisions of "2004 AMC 12A Problems/Problem 3"
(New page: ==Problem== {{empty}} ==Solution== {{solution}} ==See Also== {{AMC12 box|year=2004|ab=A|num-b=2|num-a=4}}) |
|||
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 19:08, 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 (Problems • Answer Key • Resources) | |
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 |
The answer is (B) since x can't be 0, which leaves only 49 different values for y.