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 $24$!

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
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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