Difference between revisions of "2023 AMC 10A Problems/Problem 21"
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− | + | ==Problem== | |
− | + | Let <math>P(x)</math> be the unique polynomial of minimal degree with the following properties: | |
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− | + | *<math>P(x)</math> has a leading coefficient <math>1</math>, | |
− | * | + | *<math>1</math> is a root of <math>P(x)-1</math>, |
− | + | *<math>2</math> is a root of <math>P(x-2)</math>, | |
− | + | *<math>3</math> is a root of <math>P(3x)</math>, and | |
− | + | *<math>4</math> is a root of <math>4P(x)</math>. | |
− | ~ | + | 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>? |
+ | |||
+ | <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 | ||
+ | |||
+ | ~Andrew_Lu | ||
+ | |||
+ | == 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 <math>a</math>, we get that <math>a = 23/24</math>. 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}} |
Latest revision as of 20:01, 10 August 2024
Contents
Problem
Let be the unique polynomial of minimal degree with the following properties:
has a leading coefficient
,
is a root of
,
is a root of
,
is a root of
, and
is a root of
.
The roots of are integers, with one exception. The root that is not an integer and can be written as
, where
and
are relatively prime integers. What is
?
Solution 1
From the problem statement, we know ,
and
. Therefore, we know that
,
, and
are roots. So, we can factor
as
, where
is the unknown root. Since
, we plug in
which gives
, therefore
. Therefore, our answer is
~aiden22gao
~cosinesine
~walmartbrian
~sravya_m18
~ESAOPS
~Andrew_Lu
Solution 2
We proceed similarly to solution one. We get that . Expanding, we get that
. We know that
, so the sum of the coefficients of the cubic expression is equal to one. Thus
. Solving for
, we get that
. Therefore, our answer is
~Aopsthedude
Video Solution 1 by OmegaLearn
Video Solution
Video Solution by CosineMethod
https://www.youtube.com/watch?v=HEqewKGKrFE
Video Solution 2
~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 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.