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

Latest revision as of 22:42, 29 September 2024

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 $Q(x)-1, Q(x)-2,$ 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

Quickest Solution

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 ( $Q(1) =Q(2) = Q(3)$ ), and $2$ that are linear ( $\{1, 2, 3\} \rightarrow \{3, 2, 1\}$ and $\{1, 2, 3\} \rightarrow \{1, 2, 3\}$ ). Hence there are $3^3-3-2 = \boxed{22}$ solutions, so $\boxed{B}$ .

~Maximilian113

Guessing Solution(desperate),

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

rewrite it as $P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)(x-r1)(x-r2)(x-r3).$

say Q(x)= 2nd degree polymonial

that means (Q(x)-1) must equal to 2 factors of (R(x) times P(x))

we have 6 factors $(x-1)(x-2)(x-3)(x-r1)(x-r2)(x-r3).$

We need 2 factors,so it must be 6 choices, choose 2 or

6!/4!=30 none of choices are 30, so lets use the answers $\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32$

it cannot be E because it is above 30. Now we look for answers that are similar

so we see 22,27,32 which shows that we added or subtracted 5 from 27 in the actual solution. since it can't be greater than 30, the answer should be 22 or 27. Which means choose B or D if you are desperate.

only use if you are desperate Note: If you're desperate, please go back and review 1-20 instead of guessing on a 24

@@ Line 41: / Line 41: @@
 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>.
-Quicker Solution:
+==Quicker Solution==
 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 2.) 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>.
-<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>.
+~Williamgolly
+==Quickest Solution==
+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 (<math>Q(1) =Q(2) = Q(3)</math>), and <math>2</math> that are linear (<math>\{1, 2, 3\} \rightarrow \{3, 2, 1\}</math> and <math>\{1, 2, 3\} \rightarrow \{1, 2, 3\}</math>). Hence there are <math>3^3-3-2 = \boxed{22}</math> solutions, so <math>\boxed{B}</math>.
+~Maximilian113
+==Guessing Solution(desperate),==
+<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>
+rewrite it as
+<cmath>P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)(x-r1)(x-r2)(x-r3).</cmath>
+say Q(x)= 2nd degree polymonial
+that means (Q(x)-1) must equal to 2 factors of (R(x) times P(x))
+we have 6 factors <cmath>(x-1)(x-2)(x-3)(x-r1)(x-r2)(x-r3).</cmath>
+We need 2 factors,so it must be
+choices, choose 2 or
-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>.
+!/4!=30
+none of choices are 30, so lets use the answers
+<math>\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32</math>
-~Williamgolly
+it cannot be E because it is above 30. Now we look for answers that are similar
+so we see 22,27,32 which shows that we added or subtracted 5 from 27 in the actual solution.
+since it can't be greater than 30, the answer should be 22 or 27. Which means choose B or D if you are desperate.
+only use if you are desperate
+Note: If you're desperate, please go back and review 1-20 instead of guessing on a 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

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