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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> because <math>Q(2)</math> is y-value of the midpoint of <math>(1, Q(1))</math> and <math>(3, Q(3))</math>. So we have not included any other linear functions. Therefore, the desired answer is <math>27 - 5 = \boxed{\textbf{(B) }22}</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>Q(2)</math> is y-value of the midpoint of <math>(1, Q(1))</math> and <math>(3, Q(3))</math>. So we have not included any other linear functions. Therefore, the desired answer is <math>27 - 5 = \boxed{\textbf{(B) }22}</math>.
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Quicker Solution:
 +
We see that
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<math>P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)\cdot R(x)</math>.
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Therefore, <math>P(x) | P(Q(x))</math>.  Since <math>deg Q = 2,</math> we must have <math>x-1, x-2, x-3</math> divide <math>P(Q(x))</math> so we pair them off with one of <math>Q(x)-1, Q(x)-2, Q(x)-3</math> to see that there are <math>3!+3 \cdot 2 \cdot \binom{3}{2}</math> = 24 without restrictions (the count was made by pairing off the linear factors of <math>P(x)</math> with <math>Q(x)-1, Q(x)-2, Q(x)-3</math> and notice that the degree of <math>Q</math> is 2).  However, we have two functions which are constant, which are <math>Q(x) = x</math> and <math>Q(x) = 4-x.</math>  The answer is <math>22</math>.
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~Williamgolly
  
 
== See also ==
 
== See also ==

Revision as of 23:01, 2 March 2021

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 $P(Q(x))=P(x) \cdot R(x)$?


$\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32$

Solution

We can write the problem as

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


Since $\deg P(x) = 3$ and $\deg R(x) = 3$, $\deg P(x)\cdot R(x) = 6$. Thus, $\deg P(Q(x)) = 6$, so $\deg Q(x) = 2$.

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

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: lines. 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 $2\cdot Q(2) = Q(1) + Q(3)$ because $Q(2)$ is y-value of the midpoint of $(1, Q(1))$ and $(3, Q(3))$. So we have not included any other linear functions. Therefore, the desired answer is $27 - 5 = \boxed{\textbf{(B) }22}$.

Quicker Solution: We see that

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

Therefore, $P(x) | P(Q(x))$. Since $deg Q = 2,$ we must have $x-1, x-2, x-3$ divide $P(Q(x))$ so we pair them off with one of $Q(x)-1, Q(x)-2, Q(x)-3$ to see that there are $3!+3 \cdot 2 \cdot \binom{3}{2}$ = 24 without restrictions (the count was made by pairing off the linear factors of $P(x)$ with $Q(x)-1, Q(x)-2, Q(x)-3$ and notice that the degree of $Q$ is 2). However, we have two functions which are constant, which are $Q(x) = x$ and $Q(x) = 4-x.$ The answer is $22$.

~Williamgolly

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

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