Difference between revisions of "2022 AMC 10B Problems/Problem 21"

Revision as of 17:47, 20 November 2022

Problem

Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder is $2x+1$ . 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 $P(x)$ has positive integer coefficients.

Throughout this solution, we will express all polynomials in base $x$ . E.g. $x^2 + x + 1 = 111_{x}$ .

We are given:

$111a + 12 = 101b + 21 = P(x)$ .

We add $111$ and $101$ to each side and balance respectively:

$111(a - 1) + 123 = 101(b - 1) + 122 = P(x)$

We make the units digits equal:

$111(a - 1) + 123 = 101(b - 2) + 223 = P(x)$

We now notice that:

$111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x)$ .

Therefore $a = 11_{x} = x + 1$ , $b = 12_{x} = x + 2$ , and $P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3$ . $3$ is the minimal degree of $P(x)$ since there is no way to influence the $x$ ‘s digit in $101b + 21$ when $b$ is an integer. The desired sum is $1^2 + 2^2 + 3^2 + 3^2 = \boxed{\textbf{(E)} \ 23}$

P.S. The 4 computational steps can be deduced through quick experimentation.

~ numerophile

Solution 2

Let $P(x) = Q(x)(x^2+x+1) + x + 2$ , then $P(x) = Q(x)(x^2+1) + xQ(x) + x + 2$ , therefore $xQ(x) + x + 2 \equiv 2x + 1 \pmod{x^2+1}$ , or $xQ(x) \equiv x-1 \pmod{x^2+1}$ . Clearly the minimum is when $Q(x) = x-1$ , and expanding gives $P(x) = x^3+2x^2+3x+3$ . Summing the squares of coefficients gives $\boxed{\textbf{(E)} \ 23}$

~mathfan2020

Solution 3

Let $P(x) = Q_1(x)(x^2+x+1) + x + 2$ , then $P(x) = Q1(x)(x^2+1) + xQ1(x) + x + 2$ ,

$P(x) = Q2(x)(x^2+x+1) + x + 2$ , then $P(x) = Q2(x)(x^2+1) + 2x + 1$ ,

Then we get: $( Q1(x) - Q2(x)) (x^2+1) + xQ1(x) - x + 1 = 0$

$( Q1(x) - Q2(x)) + 1 = 0$

$-(x^2+1) + xQ1(x) - x + 1 = 0$

$Q1(x) = x + 1$

$P(x) = x^3 + 2x^2 + 3x + 3$

~qgcui

Video Solutions

https://youtu.be/yGUur4vP_6k

~ ThePuzzlr

https://youtu.be/ELdhkqVyB9E

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by OmegaLearn using Circular Tangency

https://youtu.be/HdrbPiZHim0

~ pi_is_3.14

@@ Line 42: / Line 42: @@
 ==Solution 3==
-Let <math>P(x) = Q1(x)(x^2+x+1) + x + 2</math>, then <math>P(x) = Q1(x)(x^2+1) + xQ1(x) + x + 2</math>,
+Let <math>P(x) = Q_1(x)(x^2+x+1) + x + 2</math>, then <math>P(x) = Q1(x)(x^2+1) + xQ1(x) + x + 2</math>,
 <math>P(x) = Q2(x)(x^2+x+1) + x + 2</math>, then <math>P(x) = Q2(x)(x^2+1) + 2x + 1</math>,
@@ Line 58: / Line 59: @@
 ~qgcui
 ==Video Solutions==

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
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