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

Latest revision as of 22:38, 10 May 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

https://youtu.be/jIqF_dhczsY

Video Solution by CosineMethod

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

@@ Line 1: / Line 1: @@
-.
+==Problem==
-The bond dissociation energies for F2 and Cl2 are
+Let <math>P(x)</math> be the unique polynomial of minimal degree with the following properties:
-approximately 158 and 242 kJ mol−1, respectively.
-Given that the enthalpy change for the reaction below
-is ΔH = −54 kJ mol−1, what is the bond dissociation
-energy for the F-Cl bond?
-½ F2(g) + ½ Cl2(g) → FCl(g)
-A 200 kJ mol−1
+*<math>P(x)</math> has a leading coefficient <math>1</math>,
-B 254 kJ mol−1
+*<math>1</math> is a root of <math>P(x)-1</math>,
-C 146 kJ mol−1
+*<math>2</math> is a root of <math>P(x-2)</math>,
-D 454 kJ mol−1
+*<math>3</math> is a root of <math>P(3x)</math>, and
-E 346 kJ mol−1
+*<math>4</math> is a root of <math>4P(x)</math>.
-~kaisssssgao
+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>
+==Solution 1==
+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>
+~aiden22gao
+~cosinesine
+~walmartbrian
+~sravya_m18
+~ESAOPS
+== Solution 2==
+We proceed similarly to solution one. We get that <math>x(x-9)(x-4)(x-a)=1</math>. Expanding, we get that <math>x(x-9)(x-4)(x-a)=x^4-(a+13)x^3+(13a+36)x^2-36ax</math>. We know that <math>P(1)=1</math>, so the sum of the coefficients of the cubic expression is equal to one. Thus <math>1+(a+13)+(13a+36)-36a=1</math>. Solving for a, we get that a=23/24. Therefore, our answer is <math>23 + 24 =\boxed{\textbf{(D) }47}</math>
+~Aopsthedude
+== Video Solution 1 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 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==
+{{AMC10 box|year=2023|ab=A|num-b=20|num-a=22}}
+{{MAA Notice}}

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
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