Difference between revisions of "2005 AMC 12A Problems/Problem 24"

Revision as of 22:16, 22 September 2007

Problem

Let $P(x)=(x-1)(x-2)(x-3)$ . For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P(Q(x))=P(x)* R(x)$ ?

Solution

Since $R(x)$ has degree three, then $P(x)\cdot R(x)$ has degree six. Thus, $P(Q(x))$ has degree six, so $Q(x)$ must have degree two, since $P(x)$ has degree three.

$P(Q(1))=(Q(1)-1)(Q(1)-2)(Q(1)-3)=P(1)\cdot R(1)=0,$
$P(Q(2))=(Q(2)-1)(Q(2)-2)(Q(2)-3)=P(2)\cdot R(2)=0,$
$P(Q(3))=(Q(3)-1)(Q(3)-2)(Q(3)-3)=P(3)\cdot R(3)=0.$

Hence, we conclude $Q(1)$ , $Q(2)$ , and $Q(3)$ must each be $1$ , $2$ , or $3$ . Since a quadratic is uniquely determined by three points, there can be $3*3*3 = 27$ different quadratics $Q(x)$ after each of the values of $Q(1)$ , $Q(2)$ , and $Q(3)$ are chosen.

However, we have included $Q(x)$ which are not quadratics. Namely,

$Q(1)=Q(2)=Q(3)=1 \Rightarrow Q(x)=1,$
$Q(1)=Q(2)=Q(3)=2 \Rightarrow Q(x)=2,$
$Q(1)=Q(2)=Q(3)=3 \Rightarrow Q(x)=3,$
$Q(1)=1, Q(2)=2, Q(3)=3 \Rightarrow Q(x)=x,$
$Q(1)=3, Q(2)=2, Q(3)=1 \Rightarrow Q(x)=4-x.$

Clearly, we could not have included any other constant functions. For any linear function, we have $2\cdot Q(2) = Q(1) + Q(3)$ . Again, it is pretty obvious that we have not included any other linear functions. Therefore, the desired answer is $27 - 5 = 22$ .

@@ Line 3: / Line 3: @@
 == 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 ==
-* [[2005 AMC 12A Problems/Problem 23 | Previous problem]]
+{{AMC12 box|year=2005|ab=A|num-b=23|num-a=25}}
-* [[2005 AMC 12A Problems/Problem 25 | Next problem]]
-* [[2005 AMC 12A Problems]]
+[[Category:Intermediate Algebra Problems]]

2005 AMC 12A (Problems • Answer Key • Resources)
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Difference between revisions of "2005 AMC 12A Problems/Problem 24"

Revision as of 22:16, 22 September 2007

Problem

Solution

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