2022 AMC 10B Problems/Problem 21
- The following problem is from both the 2022 AMC 10B #21 and 2022 AMC 12B #20, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Experimentation)
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (Undetermined Coefficients)
- 6 Solution 5 (Quick, but Not Quicker Than 2)
- 7 Solution 6 (Similar to Solution 3)
- 8 Video Solutions
- 9 Video Solution by OmegaLearn Using Polynomial Remainders
- 10 Video Solution by The Power of Logic(#20-#21)
- 11 See Also
Problem
Let be a polynomial with rational coefficients such that when is divided by the polynomial , the remainder is , and when is divided by the polynomial , the remainder is . There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?
Solution 1 (Experimentation)
Given that all the answer choices and coefficients are integers, we hope that has positive integer coefficients.
Throughout this solution, we will express all polynomials in base . E.g. .
We are given: We add and to each side and balance respectively: We make the units digits equal: We now notice that: Therefore , , and . is the minimal degree of since there is no way to influence the ‘s digit in when is an integer. The desired sum is
P.S. The four computational steps can be deduced through quick experimentation.
~ numerophile
Solution 2
Let , then , therefore , or . Clearly the minimum is when , and expanding gives . Summing the squares of coefficients gives
~mathfan2020
Solution 3
Let , then
Also
We infer that and have same degree, we can assume , and , since has least degree. If this cannot work, we will try quadratic, etc.
Then we get:
The constant term gives us:
So
Substituting this in gives:
Solving this equation, we get
Plugging this into our original equation we get
Verify this works with
Therefore the answer is
~qgcui
Solution 4 (Undetermined Coefficients)
Notice that we cannot have the quotients equal to some constants, since the same constant will yield different constant terms for (which is bad) and different constants will yield different first coefficients (also bad). Thus, we try setting the quotients equal to linear terms (for minimizing degree).
Let and . The quotients have the same coefficient, since must have the same coefficient in both cases. Expanding, we get and
Equating coefficients, we get , , and . From the second equation, we get , then substituting into the first, . Finally, from , we have . Now, and our answer is
~MathHayden
Solution 5 (Quick, but Not Quicker Than 2)
We construct the following equations in terms of and the information given by the problem: Upon inspection, and cannot be constant, so the smallest possible degree of is and both and are linear.
Let and We know there will be values for and that make the below equation hold, so we can assume that has a leading coefficient of .
Substituting these values in, and setting and equal to each other, We plug in , yielding Substituting this value into the above equation, Letting we conclude that so Therefore, The requested sum is
-Benedict T (countmath1)
Solution 6 (Similar to Solution 3)
By remainder theorem, the polynomial can be written as follows.
This is a timed exam, we can use the information given by answer choices. The answer choices tell us this is the polynomial with integer coefficients, and we need to find the polynomial with the least degree so we can assume both and are linear (the coefficient of x should be same).
Then we can write as a cubic polynomial.
Substituting to determine the value of and .
We have:
We can solve the simultaneous equations:
Hence, . The answer is
~PythZhou
Video Solutions
~ ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn Using Polynomial Remainders
~ pi_is_3.14
Video Solution by The Power of Logic(#20-#21)
See Also
2022 AMC 10B (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 |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.