Difference between revisions of "2023 AMC 10A Problems/Problem 21"
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== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == | ||
https://youtu.be/aOL04sKGyfU | https://youtu.be/aOL04sKGyfU | ||
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+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/Jan9feBsPEg | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=20|num-a=22}} | {{AMC10 box|year=2023|ab=A|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:27, 11 November 2023
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 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
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.