Difference between revisions of "2023 AMC 10A Problems/Problem 21"
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==Solution 1== | ==Solution 1== | ||
− | From the problem statement, we | + | 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== |
Revision as of 07:50, 10 October 2024
Contents
[hide]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 find , and . Therefore, we know that , , and are roots. So, we can factor as , where is the unknown root. Since , we plug in which gives , so . Therefore, our answer is
~aiden22gao
~cosinesine
~walmartbrian
~sravya_m18
~ESAOPS
~Andrew_Lu
~sl_hc
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 Math-X (First fully understand the problem!!!)
https://youtu.be/GP-DYudh5qU?si=HDaV3kGwwywXP2J5&t=7475
~Math-X
Video Solution 2 by OmegaLearn
Video Solution
Video Solution by CosineMethod
https://www.youtube.com/watch?v=HEqewKGKrFE
Video Solution 3
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.