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Difference between revisions of "2023 AMC 10A Problems/Problem 21"
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*<math>4</math> is a root of <math>4P(x)</math>.
 
*<math>4</math> is a root of <math>4P(x)</math>.
  
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The roots of <math>P(x)</math> are integers, with one exception. The root that is not an integer and can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime integers. What is <math>m+n</math>?
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The roots of <math>P(x)</math> are integers, with one exception. The root that is not an integer can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime integers. What is <math>m+n</math>?
  
 
<math>\textbf{(A) }41\qquad\textbf{(B) }43\qquad\textbf{(C) }45\qquad\textbf{(D) }47\qquad\textbf{(E) }49</math>
 
<math>\textbf{(A) }41\qquad\textbf{(B) }43\qquad\textbf{(C) }45\qquad\textbf{(D) }47\qquad\textbf{(E) }49</math>
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==Solution 1==
 
==Solution 1==
  
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From the problem statement, we know <math>P(2-2)=0</math>, <math>P(9)=0</math> and <math>4P(4)=0</math>. Therefore, we know that <math>0</math>, <math>9</math>, and <math>4</math> are roots. So, we can factor <math>P(x)</math> as <math>x(x - 9)(x - 4)(x - a)</math>, where <math>a</math> is the unknown root. Since <math>P(x) - 1 = 0</math>, we plug in <math>x = 1</math> which gives <math>1(-8)(-3)(1 - a) = 1</math>, therefore <math>24(1 - a) = 1 \implies 1 - a = 1/24 \implies a = 23/24</math>. Therefore, our answer is <math>23 + 24 =\boxed{\textbf{(D) }47}</math>
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From the problem statement, we find <math>P(2-2)=0</math>, <math>P(9)=0</math> and <math>4P(4)=0</math>. Therefore, we know that <math>0</math>, <math>9</math>, and <math>4</math> are roots. So, we can factor <math>P(x)</math> as <math>x(x - 9)(x - 4)(x - a)</math>, where <math>a</math> is the unknown root. Since <math>P(x) - 1 = 0</math>, we plug in <math>x = 1</math> which gives <math>1(-8)(-3)(1 - a) = 1</math>, so <math>24(1 - a) = 1 \implies 1 - a = 1/24 \implies a = 23/24</math>. Therefore, our answer is <math>23 + 24 =\boxed{\textbf{(D) }47}</math>
  
 
~aiden22gao  
 
~aiden22gao  
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~Andrew_Lu
 
~Andrew_Lu
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 +
~sl_hc
  
 
== Solution 2==
 
== Solution 2==
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~Aopsthedude
 
~Aopsthedude
  
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==Video Solution by Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=oAdscLjoWqHtVcUX&t=4673
 +
~little-fermat
 
==Video Solution 1 by Math-X (First fully understand the problem!!!)==
 
==Video Solution 1 by Math-X (First fully understand the problem!!!)==
 
https://youtu.be/GP-DYudh5qU?si=HDaV3kGwwywXP2J5&t=7475  
 
https://youtu.be/GP-DYudh5qU?si=HDaV3kGwwywXP2J5&t=7475  

Latest revision as of 06:36, 5 November 2024

Problem

Let $P(x)$ be the unique polynomial of minimal degree with the following properties:

  • $P(x)$ has a leading coefficient $1$,
  • $1$ is a root of $P(x)-1$,
  • $2$ is a root of $P(x-2)$,
  • $3$ is a root of $P(3x)$, and
  • $4$ is a root of $4P(x)$.

The roots of $P(x)$ are integers, with one exception. The root that is not an integer can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime integers. What is $m+n$?

$\textbf{(A) }41\qquad\textbf{(B) }43\qquad\textbf{(C) }45\qquad\textbf{(D) }47\qquad\textbf{(E) }49$

Solution 1

From the problem statement, we find $P(2-2)=0$, $P(9)=0$ and $4P(4)=0$. Therefore, we know that $0$, $9$, and $4$ are roots. So, we can factor $P(x)$ as $x(x - 9)(x - 4)(x - a)$, where $a$ is the unknown root. Since $P(x) - 1 = 0$, we plug in $x = 1$ which gives $1(-8)(-3)(1 - a) = 1$, so $24(1 - a) = 1 \implies 1 - a = 1/24 \implies a = 23/24$. Therefore, our answer is $23 + 24 =\boxed{\textbf{(D) }47}$

~aiden22gao

~cosinesine

~walmartbrian

~sravya_m18

~ESAOPS

~Andrew_Lu

~sl_hc

Solution 2

We proceed similarly to solution one. We get that $x(x-9)(x-4)(x-a)=1$. Expanding, we get that $x(x-9)(x-4)(x-a)=x^4-(a+13)x^3+(13a+36)x^2-36ax$. We know that $P(1)=1$, so the sum of the coefficients of the cubic expression is equal to one. Thus $1-(a+13)+(13a+36)-36a=1$. Solving for $a$, we get that $a = 23/24$. Therefore, our answer is $23 + 24 =\boxed{\textbf{(D) }47}$

~Aopsthedude

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=oAdscLjoWqHtVcUX&t=4673 ~little-fermat

Video Solution 1 by Math-X (First fully understand the problem!!!)

https://youtu.be/GP-DYudh5qU?si=HDaV3kGwwywXP2J5&t=7475

~Math-X

Video Solution 2 by OmegaLearn

https://youtu.be/aOL04sKGyfU

Video Solution

https://youtu.be/jIqF_dhczsY

Video Solution by CosineMethod

https://www.youtube.com/watch?v=HEqewKGKrFE

Video Solution 3

https://youtu.be/Jan9feBsPEg

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution 4 by EpicBird08

https://www.youtube.com/watch?v=D4GWjJmpqEU&t=25s

Video Solution 5 by MegaMath

https://www.youtube.com/watch?v=4Hwt3f1bi1c&t=1s

See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 10 Problems and Solutions

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