Difference between revisions of "2005 AMC 12A Problems/Problem 24"
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− | Clearly, we could not have included any other constant functions. For any linear function, we have <math>2\cdot Q(2) = Q(1) + Q(3)</math>. | + | Clearly, we could not have included any other constant functions. For any linear function, we have <math>2\cdot Q(2) = Q(1) + Q(3)</math> because <math>2(2)=1+3</math>. So we have not included any other linear functions. Therefore, the desired answer is <math>27 - 5 = \boxed{\textbf{(B) }22}</math>. |
== See also == | == See also == |
Revision as of 19:12, 9 November 2013
Problem
Let . For how many polynomials does there exist a polynomial of degree 3 such that ?
Solution
Since has degree three, then has degree six. Thus, has degree six, so must have degree two, since has degree three.
Hence, we conclude , , and must each be , , or . Since a quadratic is uniquely determined by three points, there can be different quadratics after each of the values of , , and are chosen.
However, we have included which are not quadratics. Namely,
Clearly, we could not have included any other constant functions. For any linear function, we have because . So we have not included any other linear functions. Therefore, the desired answer is .
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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