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

Revision as of 18:46, 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 $P(x) = (x^2+x+1)Q_1(x) + x + 2$ , then $P(x) = (x^2+1)Q_1(x) + xQ_1(x) + x + 2$

Also $P(x) = (x^2+1)Q_2(x) + 2x + 1$

We infer that $Q_1(x)$ and $Q_2(x)$ have same degree

Then we get: $(x^2+1)(Q_1(x) - Q_2(x)) + xQ_1(x) - x + 1 = 0$

The constant term gives us: $(Q_1(x) - Q_2(x)) + 1 = 0$

So $Q_1(x) - Q_2(x) = -1$

Substituting this in gives: $-(x^2+1) + xQ_1(x) - x + 1 = 0$

Solving this equation, we get $Q_1(x) = x + 1$

Plugging this into our original equation we get $P(x) = x^3 + 2x^2 + 3x + 3$

Therefore the answer is $1^2 + 2^2 + 3^2 + 3^2 = \boxed{\textbf{(E)} \ 23}$

~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) = Q_1(x)(x^2+x+1) + x + 2</math>,
+Let <math>P(x) = (x^2+x+1)Q_1(x) + x + 2</math>,
-then <math>P(x) = Q_1(x)(x^2+1) + Q_1(x)x + x + 2</math>
+then <math>P(x) = (x^2+1)Q_1(x) + xQ_1(x) + x + 2</math>
-Also <math>P(x) = Q_2(x)(x^2+1) + 2x + 1</math>
+Also <math>P(x) = (x^2+1)Q_2(x) + 2x + 1</math>
 We infer that <math>Q_1(x)</math> and <math>Q_2(x)</math> have same degree
 Then we get:
-<math>( Q_1(x) - Q_2(x)) (x^2+1) + Q_1(x)x - x + 1 = 0 </math>
+<math>(x^2+1)(Q_1(x) - Q_2(x)) + xQ_1(x) - x + 1 = 0 </math>
 The constant term gives us:

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