Difference between revisions of "2004 AMC 12A Problems/Problem 3"
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==Solution== | ==Solution== | ||
| − | + | The answer is (B) since x can't be 0, which leaves only 49 different values for y. | |
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2004|ab=A|num-b=2|num-a=4}} | {{AMC12 box|year=2004|ab=A|num-b=2|num-a=4}} | ||
Revision as of 20:08, 17 January 2008
Problem
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Solution
The answer is (B) since x can't be 0, which leaves only 49 different values for y.
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 | |
