Difference between revisions of "2023 AMC 10A Problems/Problem 21"
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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>? | 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>? | ||
| + | |||
| + | Solution 1 | ||
| + | From the problem statement, we know P(1)=1, P(2-2)=0, P(9)=0 and 4P(4)=0 therefore we know P(x) must at least have the factors x(x-9)(x-4) and we can assume the last factor to be (x-a) where a is the fractional factor. Then we can use the fact that P(1)=1 to obtain that a 1-a must be 1/24 and a is 23/24. The answer is then 47. | ||
| + | ~aiden22gao | ||
== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == | ||
https://youtu.be/aOL04sKGyfU | https://youtu.be/aOL04sKGyfU | ||
Revision as of 17:57, 9 November 2023
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 
.
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 know P(1)=1, P(2-2)=0, P(9)=0 and 4P(4)=0 therefore we know P(x) must at least have the factors x(x-9)(x-4) and we can assume the last factor to be (x-a) where a is the fractional factor. Then we can use the fact that P(1)=1 to obtain that a 1-a must be 1/24 and a is 23/24. The answer is then 47. ~aiden22gao
