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

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is <math>2x+1</math>. There is a unique polynomial of least degree with these two properties. What is the sum of
 
is <math>2x+1</math>. 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?
 
the squares of the coefficients of that polynomial?
 +
 +
==Solution 1 (Experimentation)==
 +
 +
Given that all the answer choices and coefficients are integers, we hope that <math>P(x)</math> has positive integer coefficients.
 +
 +
Throughout this solution, we will express all polynomials in base <math>x</math>. E.g. <math>x^2 + x + 1 = 111_{x}</math>.
 +
 +
We are given:
 +
 +
<cmath>111a + 12 = 101b + 21 = P(x)</cmath>.
 +
 +
We add <math>111</math> and <math>101</math> to each side and balance respectively:
 +
 +
<cmath>111(a - 1) + 123 = 101(b - 1) + 122 = P(x)</cmath>
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 +
We make the units digits equal:
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 +
<cmath>111(a - 1) + 123 = 101(b - 2) + 223 = P(x)</cmath>
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 +
We now notice that:
 +
 +
<cmath>111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x)</cmath>.
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 +
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>
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P.S. The 4 computational steps can be deduced through quick experimentation.
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 +
~ numerophile
  
 
==Video Solutions==
 
==Video Solutions==

Revision as of 20:04, 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

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 (Problems • Answer Key • Resources)
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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