Difference between revisions of "2005 AMC 12A Problems/Problem 24"
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== Solution == | == Solution == | ||
− | + | Since <math>R(x)</math> has degree three, then <math>P(x)\cdot R(x)</math> has degree six. Thus, <math>P(Q(x))</math> has degree six, so <math>Q(x)</math> must have degree two, since <math>P(x)</math> has degree three. | |
+ | <div style="text-align:center;"> | ||
+ | <math> | ||
+ | P(Q(1))=(Q(1)-1)(Q(1)-2)(Q(1)-3)=P(1)\cdot R(1)=0, | ||
+ | </math><br /><math> | ||
+ | P(Q(2))=(Q(2)-1)(Q(2)-2)(Q(2)-3)=P(2)\cdot R(2)=0, | ||
+ | </math><br /><math> | ||
+ | P(Q(3))=(Q(3)-1)(Q(3)-2)(Q(3)-3)=P(3)\cdot R(3)=0. | ||
+ | </math> | ||
+ | </div> | ||
+ | Hence, we conclude <math>Q(1)</math>, <math>Q(2)</math>, and <math>Q(3)</math> must each be <math>1</math>, <math>2</math>, or <math>3</math>. Since a [[quadratic equation|quadratic]] is uniquely determined by three points, there can be <math>3*3*3 = 27</math> different quadratics <math>Q(x)</math> after each of the values of <math>Q(1)</math>, <math>Q(2)</math>, and <math>Q(3)</math> are chosen. | ||
+ | |||
+ | |||
+ | However, we have included <math>Q(x)</math> which are not quadratics. Namely, | ||
+ | <div style="text-align:center;"> | ||
+ | <math> | ||
+ | Q(1)=Q(2)=Q(3)=1 \Rightarrow Q(x)=1, | ||
+ | </math><br /><math> | ||
+ | Q(1)=Q(2)=Q(3)=2 \Rightarrow Q(x)=2, | ||
+ | </math><br /><math> | ||
+ | Q(1)=Q(2)=Q(3)=3 \Rightarrow Q(x)=3, | ||
+ | </math><br /><math> | ||
+ | Q(1)=1, Q(2)=2, Q(3)=3 \Rightarrow Q(x)=x, | ||
+ | </math><br /><math> | ||
+ | Q(1)=3, Q(2)=2, Q(3)=1 \Rightarrow Q(x)=4-x. | ||
+ | </math> | ||
+ | </div> | ||
+ | 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>. Again, it is pretty obvious that we have not included any other linear functions. Therefore, the desired answer is <math>27 - 5 = 22</math>. | ||
+ | |||
== See also == | == See also == | ||
− | + | {{AMC12 box|year=2005|ab=A|num-b=23|num-a=25}} | |
− | + | ||
− | + | [[Category:Intermediate Algebra Problems]] |
Revision as of 22:16, 22 September 2007
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 . Again, it is pretty obvious that 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 |