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Difference between revisions of "2023 AMC 10A Problems/Problem 21"
(question options and see also)
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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>?
 
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>?
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<math>\textbf{(A) }41\qquad\textbf{(B) }43\qquad\textbf{(C) }45\qquad\textbf{(D) }47\qquad\textbf{(E) }49</math>
  
 
==Solution 1==
 
==Solution 1==
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== Video Solution 1 by OmegaLearn ==
 
== Video Solution 1 by OmegaLearn ==
 
https://youtu.be/aOL04sKGyfU
 
https://youtu.be/aOL04sKGyfU
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==See Also==
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{{AMC10 box|year=2023|ab=A|num-b=20|num-a=22}}
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{{MAA Notice}}

Revision as of 20:40, 9 November 2023

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 know $P(1)=1$, $P(2-2)=0$, $P(9)=0$ and $4P(4)=0$. Therefore, we know that $0$, $9$, and $4$ are roots. Because of this, we can factor $P(x)$ as $x(x - 9)(x - 4)(x - a)$, where $a$ is the unknown root. Plugging in $x = 1$ gives $1(-8)(-3)(1 - a) = 1$, so $24(1 - a) = 1/24 \implies 1 - a = 24 \implies a = 23/24$. Therefore, our answer is $23 + 24 =$ $47$, or $C$

~aiden22gao

~cosinesine

~walmartbrian


Video Solution 1 by OmegaLearn

https://youtu.be/aOL04sKGyfU

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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