Difference between revisions of "2022 AMC 10B Problems/Problem 21"
(→Solution 3) |
(→Solution 3) |
||
Line 45: | Line 45: | ||
then <math>P(x) = Q_1(x)(x^2+1) + Q_1(x)x + x + 2</math> | then <math>P(x) = Q_1(x)(x^2+1) + Q_1(x)x + x + 2</math> | ||
− | Also <math>P(x) = Q_2(x)(x^2+1) + 2x + 1</math> | + | Also <math>P(x) = Q_2(x)(x^2+1) + 2x + 1</math> |
+ | |||
+ | We infer that <math>Q_1(x)</math> and <math>Q_2(x)</math> have same degree | ||
Then we get: | Then we get: |
Revision as of 19:44, 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
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
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.