Difference between revisions of "2023 AMC 10A Problems/Problem 21"

(Video Solution 1 by OmegaLearn)
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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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 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
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 +
https://www.youtube.com/watch?v=HEqewKGKrFE
  
 
==Video Solution 2==
 
==Video Solution 2==

Revision as of 20:32, 28 January 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 know $P(2-2)=0$, $P(9)=0$ 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 $1(-8)(-3)(1 - a) = 1$, therefore $24(1 - a) = 1 \implies 1 - a = 1/24 \implies a = 23/24$. Therefore, our answer is $23 + 24 =\boxed{\textbf{(D) }47}$

~aiden22gao

~cosinesine

~walmartbrian

~sravya_m18

~ESAOPS

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 1 by OmegaLearn

https://youtu.be/aOL04sKGyfU

Video Solution by CosineMethod [🔥Fast and Easy🔥]

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

Video Solution 2

https://youtu.be/Jan9feBsPEg

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

Video Solution 3 by EpicBird08

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

Video Solution 4 by MegaMath

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

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
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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