Difference between revisions of "2022 AMC 10B Problems/Problem 21"
(→Solution 3) |
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<math>P(x) = x^3 + 2x^2 + 3x + 3 </math> | <math>P(x) = x^3 + 2x^2 + 3x + 3 </math> | ||
− | Verify this works with | + | Verify this works with <math>P(x) = (x^2+1)Q_2(x) + 2x + 1</math> |
Therefore the answer is <math>1^2 + 2^2 + 3^2 + 3^2 = \boxed{\textbf{(E)} \ 23}</math> | Therefore the answer is <math>1^2 + 2^2 + 3^2 + 3^2 = \boxed{\textbf{(E)} \ 23}</math> |
Revision as of 18:56, 20 November 2022
Contents
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 4 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
Video Solutions
~ ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn using Circular Tangency
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.