Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
(Improved formatting of problem statement and solutions)
 
(18 intermediate revisions by 11 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Let <math>P(x)=(x-1)(x-2)(x-3)</math>. For how many [[polynomial]]s <math>Q(x)</math> does there exist a polynomial <math>R(x)</math> of degree 3 such that <math>P(Q(x))=P(x)* R(x)</math>?
+
Let <math>P(x)=(x-1)(x-2)(x-3)</math>. For how many polynomials <math>Q(x)</math> does there exist a polynomial <math>R(x)</math> of degree <math>3</math> such that <math>P(Q(x)) = P(x) \cdot R(x)</math>?
  
 +
<math>\mathrm {(A) } \ 19 \qquad \mathrm {(B) } \ 22 \qquad \mathrm {(C) } \ 24 \qquad \mathrm {(D) } \ 27 \qquad \mathrm {(E) } \ 32</math>
 +
 +
== Solution 1==
 +
We can write the problem as
 +
<div style="text-align:center;">
 +
<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>.
 +
</div>
  
−
<math>\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32</math>
 
  
−
== Solution ==
+
Since <math>\deg P(x) = 3</math> and <math>\deg R(x) = 3</math>, <math>\deg P(x)\cdot R(x) = 6</math>. Thus, <math>\deg P(Q(x)) = 6</math>, so <math>\deg Q(x) = 2</math>.
−
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;">
 
<div style="text-align:center;">
 
<math>
 
<math>
Line 16: Line 21:
 
</math>
 
</math>
 
</div>
 
</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.
+
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 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,
+
However, we have included polynomials <math>Q(x)</math> which are linear, rather than quadratic. Namely,
 
<div style="text-align:center;">
 
<div style="text-align:center;">
 
<math>
 
<math>
Line 33: Line 38:
 
</math>
 
</math>
 
</div>
 
</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 \mathrm{(B)}</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 the <math>y</math>-coordinate 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>.
 +
 
 +
==Solution 2==
 +
We see that
 +
<cmath>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).</cmath>
 +
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,</math> and <math>Q(x)-3</math> to see that there are <math>3!+3 \cdot 2 \cdot \binom{3}{2} = 24</math> without restrictions. (Note that this count was made by pairing off linear factors of <math>P(x)</math> with <math>Q(x)-1, Q(x)-2,</math> and <math>Q(x)-3</math>, and also note that the degree of <math>Q</math> is <math>2</math>.) However, we have two functions which are constant, which are <math>Q(x) = x</math> and <math>Q(x) = 4-x</math>. So we subtract <math>2</math> to get a final answer of <math>\boxed{22} \implies \boxed{B}</math>.
 +
 
 +
~Williamgolly
 +
 
 +
==Solution 3 (fastest)==
 +
Obviously, <math>\deg Q = 2</math> and the RHS has roots <math>1, 2, 3</math>. Thus, the only requirement is that <math>Q</math> is a quadratic that maps each of the numbers <math>1, 2, 3</math> to one of <math>{1, 2, 3}.</math> It suffices to count the number of such mappings, as this uniquely determines <math>Q</math> through polynomial interpolation. Thus there are <math>3^3</math> possibilities. But we have three functions that are constant (i.e. that give <math>Q(1) =Q(2) = Q(3)</math>), and <math>2</math> that are linear (i.e. that map <math>\{1, 2, 3\} \mapsto \{3, 2, 1\}</math> or <math>\{1, 2, 3\} \mapsto \{1, 2, 3\}</math>). Hence there are <math>3^3-3-2 = \boxed{22}</math> solutions, so <math>\boxed{B}</math>.
 +
 
 +
~Maximilian113
 +
 
 +
==Non-Solution 4 (guessing, only if desperate!)==
 +
As <math>R(x)</math> must have degree <math>3</math>, let its roots be <math>r_1</math>, <math>r_2</math>, and <math>r_3</math>, giving
 +
<cmath>\begin{align*}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) \\ &= \left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-r_1\right)\left(x-r_2\right)\left(x-r_3\right).\end{align*}</cmath>
 +
 
 +
If we now guess that <math>Q(x)</math> is a quadratic polynomial, it follows that e.g. <math>\left(Q(x)-1\right)</math> must be equal to a product of <math>2</math> factors from the expression <math>\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-r_1\right)\left(x-r_2\right)\left(x-r_3\right)</math>. As this expression contains <math>6</math> factors in total, the number of ways of choosing <math>2</math> factors from it, where the order matters, would be <math>6!/4! = 30</math>.
 +
 
 +
While none of the answer choices are <math>30</math>, we can now eliminate <math>\mathrm {(E) }</math> as <math>32 > 30</math>. Looking for answers that are similar, we further observe that <math>22</math>, <math>27</math>, or <math>32</math> would have been derived from a <math>27</math> in the actual solution, to which <math>5</math> was possibly added or subtracted. We can therefore guess, if sufficiently desperate, that the answer should be either <math>\mathrm {(B) } \ 22</math> or <math>\mathrm {(D) } \ 27</math>.
 +
 
 +
On the other hand, if you're desperate for time, it's probably a better idea to go back and review Problems <math>1</math> to <math>20</math>, rather than guessing on Problem <math>24</math>!
  
 
== See also ==
 
== See also ==
Line 39: Line 66:
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 15:24, 1 July 2025

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 1

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 polynomials $Q(x)$ which are linear, rather than quadratic. 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 the $y$-coordinate 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 $27 - 5 = \boxed{\textbf{(B) }22}$.

Solution 2

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 $Q(x)-1, Q(x)-2,$ and $Q(x)-3$ to see that there are $3!+3 \cdot 2 \cdot \binom{3}{2} = 24$ without restrictions. (Note that this count was made by pairing off linear factors of $P(x)$ with $Q(x)-1, Q(x)-2,$ and $Q(x)-3$, and also note that the degree of $Q$ is $2$.) However, we have two functions which are constant, which are $Q(x) = x$ and $Q(x) = 4-x$. So we subtract $2$ to get a final answer of $\boxed{22} \implies \boxed{B}$.

~Williamgolly

Solution 3 (fastest)

Obviously, $\deg Q = 2$ and the RHS has roots $1, 2, 3$. Thus, the only requirement is that $Q$ is a quadratic that maps each of the numbers $1, 2, 3$ to one of ${1, 2, 3}.$ It suffices to count the number of such mappings, as this uniquely determines $Q$ through polynomial interpolation. Thus there are $3^3$ possibilities. But we have three functions that are constant (i.e. that give $Q(1) =Q(2) = Q(3)$), and $2$ that are linear (i.e. that map $\{1, 2, 3\} \mapsto \{3, 2, 1\}$ or $\{1, 2, 3\} \mapsto \{1, 2, 3\}$). Hence there are $3^3-3-2 = \boxed{22}$ solutions, so $\boxed{B}$.

~Maximilian113

Non-Solution 4 (guessing, only if desperate!)

As $R(x)$ must have degree $3$, let its roots be $r_1$, $r_2$, and $r_3$, giving \begin{align*}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) \\ &= \left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-r_1\right)\left(x-r_2\right)\left(x-r_3\right).\end{align*}

If we now guess that $Q(x)$ is a quadratic polynomial, it follows that e.g. $\left(Q(x)-1\right)$ must be equal to a product of $2$ factors from the expression $\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-r_1\right)\left(x-r_2\right)\left(x-r_3\right)$. As this expression contains $6$ factors in total, the number of ways of choosing $2$ factors from it, where the order matters, would be $6!/4! = 30$.

While none of the answer choices are $30$, we can now eliminate $\mathrm {(E) }$ as $32 > 30$. Looking for answers that are similar, we further observe that $22$, $27$, or $32$ would have been derived from a $27$ in the actual solution, to which $5$ was possibly added or subtracted. We can therefore guess, if sufficiently desperate, that the answer should be either $\mathrm {(B) } \ 22$ or $\mathrm {(D) } \ 27$.

On the other hand, if you're desperate for time, it's probably a better idea to go back and review Problems $1$ to $20$, rather than guessing on Problem $24$!

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_24&oldid=253038"