2005 AMC 12A Problems/Problem 24

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 $24$ !

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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