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2022 AMC 12B Problems/Problem 20

Revision as of 04:10, 18 November 2022 by Gracelimi (talk | contribs) (→‎Problem)

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?

$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 19 \qquad \textbf{(D)}\ 20 \qquad \textbf{(E)}\ 23 \qquad$

Solution 1

It is easy to see that $P(x)$ has a degree of at least 2.

Suppose that it has degree $2$, so let $P(x) = a(x^2 + x + 1) + (x + 2) = b(x^2 + 1) + (2x + 1)$. Then comparing coefficients of $x^2$ gives $a = b$, and comparing coefficients of $x^0$ gives $a + 2 = b + 1$, a contradiction.

Now suppose it has degree $3$. Let $P(x) = (ax + b)(x^2 + x + 1) + (x + 2) = (cx + d)(x^2 + 1) + (2x + 1)$. Equating coefficients of $x^3$ gives $a = c$, so $(ax + b)(x^2 + x + 1) + (x + 2) = (ax + d)(x^2 + 1) + (2x + 1)$.

Equating coefficients of $x^0$ gives $b + 2 = d + 1$, so $d = b + 1$ and $(ax + b)(x^2 + x + 1) + (x + 2) = (ax + b + 1)(x^2 + 1) + (2x + 1)$.

Now equating coefficients of $x^2$ gives $b + a = b + 1$ and hence $a = 1$. Hence $(x + b)(x^2 + x + 1) + (x + 2) = (x + b + 1)(x^2 + 1) + (2x + 1)$.

Then, we equate coefficients of $x$ to get $1 + b + 1 = 1 + 2$, so $b = 1$.

Hence, $P(x) = (x + 1)(x^2 + x + 1) + (x + 2) = x^3 + 2x^2 + 3x + 3$ and the sum of the squares of coefficients is $1^2 + 2^2 + 3^2 + 3^2 = \fbox{(E)23}$, and we're done!

~ Bxiao31415

Video Solution by OmegaLearn Using Polynomial Remainders

https://youtu.be/HdrbPiZHim0

~ pi_is_3.14


See Also

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 12 Problems and Solutions

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