2022 AMC 10B Problems/Problem 21

Revision as of 20:04, 18 November 2022 by Numerophile (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?

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

Video Solutions

https://youtu.be/yGUur4vP_6k

~ ThePuzzlr

https://youtu.be/ELdhkqVyB9E

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

See Also

2022 AMC 10B (ProblemsAnswer KeyResources)
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
All AMC 10 Problems and Solutions

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