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

Revision as of 21:35, 18 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

Video Solutions

https://youtu.be/yGUur4vP_6k

~ ThePuzzlr

https://youtu.be/ELdhkqVyB9E

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

@@ Line 28: / Line 28: @@
 <cmath>111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x)</cmath>.
-Therefore <math>a = 11_{x} = x + 1</math>, <math>b = 12_{x} = x + 2</math>, and <math>P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3</math>. <math>3</math> is the minimal degree of <math>P(x)</math> since there is no way to influence the <math>x</math>‘s digit in <math>101b + 21</math> when <math>b</math> is an integer. The desired sum is <math>1^2 + 2^2 + 3^2 + 3^2 = \boxed{\textbf{(E) 23}}</math>
+Therefore <math>a = 11_{x} = x + 1</math>, <math>b = 12_{x} = x + 2</math>, and <math>P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3</math>. <math>3</math> is the minimal degree of <math>P(x)</math> since there is no way to influence the <math>x</math>‘s digit in <math>101b + 21</math> when <math>b</math> is an integer. The desired sum is <math>1^2 + 2^2 + 3^2 + 3^2 = \boxed{\textbf{(E)} \ 23}</math>
 P.S. The 4 computational steps can be deduced through quick experimentation.
 ~ numerophile
+==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-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
 ==Video Solutions==

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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