Difference between revisions of "2023 AIME I Problems/Problem 9"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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+ | ==Solution 3== | ||
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+ | <math>p(x)-p(2)</math> is a cubic with two integral real roots, therefore it has three real roots, which are all integers. There are exactly two distinct roots, so either <math>p(x)=p(2)+(x-2)^2(x-m)</math> or <math>p(x)=p(2)+(x-2)(x-m)^2</math> with <math>m\neq 2</math>. In the first case <math>p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)</math>, so <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1,3,4</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving <math>410</math> polynomials. In the second case <math>p(x)=x^3-(1+2m)x^2+(4m+m^2)x-2m^2+p(2)</math>, and <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving 328 polynomials. The total is <math>\boxed{738}</math>. | ||
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+ | ~EVIN- | ||
==See also== | ==See also== |
Revision as of 16:23, 8 February 2023
Contents
[hide]Problem 9
Find the number of cubic polynomials , where , , and are integers in , such that there is a unique integer with
Solution 1
Plugging into , we get . We can rewrite into , where can be any value in the range. Since must be . The problem also asks for unique integers, meaning can only be one value for each polynomial, so the discriminant must be . , and . Rewrite to be . must be even for to be an integer. because . However, plugging in result in . There are 8 pairs of and 41 integers for , giving ~chem1kall
Solution 2
Define q . Hence, for , beyond having a root 2, it has a unique integer root that is not equal to 2.
We have Thus, the polynomial has a unique integer root and it is not equal to 2.
Following from Vieta' formula, the sum of two roots of this polynomial is . Because is an integer, if a root is an integer, the other root is also an integer. Therefore, the only way to have a unique integer root is that the determinant of this polynomial is 0. Thus, In addition, because two identical roots are not 2, we have Equation (1) can be reorganized as Thus, . Denote . Thus, (2) can be written as Because , , and , we have .
Therefore, we have the following feasible solutions for : , , , , . Thus, the total number of is 8.
Because can take any value from , the number of feasible is 41.
Therefore, the number of is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3
is a cubic with two integral real roots, therefore it has three real roots, which are all integers. There are exactly two distinct roots, so either or with . In the first case , so can be and can be any integer from to , giving polynomials. In the second case , and can be and can be any integer from to , giving 328 polynomials. The total is .
~EVIN-
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.