Difference between revisions of "2023 AMC 10A Problems/Problem 21"
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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> | 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> | ||
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== Solution 2== | == Solution 2== | ||
Latest revision as of 07:26, 25 October 2025
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution by Little Fermat
- 5 Video Solution 1 by Math-X (First fully understand the problem!!!)
- 6 Video Solution 2 by OmegaLearn
- 7 Video Solution
- 8 Video Solution by CosineMethod
- 9 Video Solution 3
- 10 Video Solution 4 by EpicBird08
- 11 Video Solution 5 by MegaMath
- 12 See Also
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 
,
is a root of 
, and
is a root of 
, and
is a root of 
.
.The roots of are integers, with one exception. The root that is not an integer 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 by Little Fermat
https://youtu.be/h2Pf2hvF1wE?si=oAdscLjoWqHtVcUX&t=4673 ~little-fermat
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
