2021 Fall AMC 12A Problems/Problem 25
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[hide]Problem
Let be an odd integer, and let denote the number of quadruples of distinct integers with for all such that divides . There is a polynomial such that for all odd integers . What is
Solution 1 (Complete Residue System)
For a fixed value of there is a total of possible ordered quadruples
Let We claim that exactly of these ordered quadruples satisfy that divides
Since we conclude that is the complete residue system modulo for all integers
Given any ordered quadruple in modulo it follows that exactly one of these ordered quadruples satisfy that divides We conclude that so By Vieta's Formulas, we get
~MRENTHUSIASM
Solution 2 (Symmetric Congruent Numbers and Interpolation)
Define
Hence, is a one-to-one and onto function of , and the range of is .
Therefore, to solve this problem, it is equivalent for us to count the number of tuples that are all distinct and satisfy .
Denote by the number of such tuples that are also subject to the constraint .
Hence, .
We do the following casework analysis to compute .
: There is one 0 in .
Denote by the number of tuples with .
By symmetry, and .
: There is no 0 in .
Denote by the number of tuples with positive entries.
By symmetry, and .
Therefore,
Now, we compute for .
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We have .
:
We cannot have 3 distinct positive integers. So .
:
Because there are 2 positive integers, we must have , . Hence, . However, this is out of the range of . Thus, .
:
We cannot have 4 distinct positive integers. So .
:
We cannot have 3 distinct positive integers. So .
:
The only solution is . So .
Therefore, .
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We have .
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We have no feasible solution. Thus, .
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The only solution is . Thus, .
:
We cannot have 4 distinct positive integers. So .
:
To get 3 distinct positive integers, we have . This implies . However, this is out of the range of . So .
:
We have .
Therefore, .
.
We have .
:
The only solution is . Thus, .
:
The feasible solutions are , . Thus, .
:
There is no feasible solution. So .
:
To get 3 distinct positive integers, we have . This implies . However, this is out of the range of . So .
:
We have .
Therefore, .
.
We have .
:
The only solution is . Thus, .
:
The feasible solutions are , , , . Thus, .
:
The only feasible solution is . So .
:
The only feasible solution is . So .
:
We have .
Therefore, .
We know that for odd .
Plugging into this equation, we get
Now, we solve this system of equations. Taking , , , we get
Taking , , we get
Taking , we get .
Plugging into Equation (3.1), we get .
Plugging and into Equation (2.1), we get .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 3 (Educated Guess)
Note that you see numbers with absolute value and in the answer choices. What is special about those numbers? Well, you should notice that they are the coefficients of the polynomial when expanded (if you've already memorized this). Then, you can probably guess the polynomial is some form of whether negative or positive. Since is asked, the answer should be reasoned out as
Furthermore, you can gain confidence in your guess since that is the only answer choice with absolute value
~fidgetboss_4000
Video Solution
~MathProblemSolvingSkills.com
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
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