Question

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?

$\ 10$

$\ 13$

$\ 19$

$\ 20$

$\ 23$

Answer: E

Concepts:polynomial system of equations

Difficulty rating: 2150

Solution:

The first remainder condition gives $\begin{aligned}P(x)&=(x^2+x+1)Q(x)\\&\quad+x+2.\end{aligned}$ for some polynomial $Q.$ Modulo $x^2+1,$ we have $x^2\equiv-1,$ so $P(x)\equiv xQ(x)+x+2.$

If $Q(x)=c$ is constant, this remainder is $(c+1)x+2,$ which cannot equal $2x+1.$ Thus $Q$ must have degree at least $1.$

Now let $Q(x)=ax+b.$ Reducing modulo $x^2+1$ gives $P(x)\equiv(b+1)x+(2-a).$ Matching this with $2x+1$ yields $a=b=1.$ This constructs a degree- $3$ polynomial, and the failed constant case proves that degree $3$ is minimal.

Therefore, $\begin{aligned}P(x)&=(x+1)(x^2+x+1)\\&\quad+x+2\\&=x^3+2x^2+3x+3.\end{aligned}$ The sum of the squares of its coefficients is $1^2+2^2+3^2+3^2=23.$

Thus, the answer is E .

2022 AMC 10B Problem 21

Answer: E

Difficulty rating: 2150

Solution:

Problem 21 in Other Years