Difference between revisions of "2022 AMC 10B Problems/Problem 21"
Mathhayden (talk | contribs) (→Solution 4 (undetermined coefficients)) |
Mathhayden (talk | contribs) (→Solution 2) |
||
Line 36: | Line 36: | ||
==Solution 2== | ==Solution 2== | ||
− | Let <math>P(x) = Q(x)(x^2+x+1) + x + 2</math>, then <math>P(x) = Q(x)(x^2+1) + xQ(x) + x + 2</math>, therefore <math>xQ(x) + x + 2 \equiv 2x + 1 \pmod{x^2+1}</math>, or <math>xQ(x) \equiv x-1 \pmod{x^2+1}</math>. Clearly the minimum is when <math>Q(x) = x | + | Let <math>P(x) = Q(x)(x^2+x+1) + x + 2</math>, then <math>P(x) = Q(x)(x^2+1) + xQ(x) + x + 2</math>, therefore <math>xQ(x) + x + 2 \equiv 2x + 1 \pmod{x^2+1}</math>, or <math>xQ(x) \equiv x-1 \pmod{x^2+1}</math>. Clearly the minimum is when <math>Q(x) = x+1</math>, and expanding gives <math>P(x) = x^3+2x^2+3x+3</math>. Summing the squares of coefficients gives <math>\boxed{\textbf{(E)} \ 23}</math> |
~mathfan2020 | ~mathfan2020 |
Revision as of 13:58, 22 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
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
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.