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"

(→‎Guessing Solution(incorrect),)
Line 48: Line 48:
 
~Williamgolly
 
~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 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
  
  

Revision as of 17:17, 8 April 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 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_24&oldid=217982"