Difference between revisions of "2022 AMC 12B Problems/Problem 20"

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== Problem ==
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#redirect [[2022 AMC 10B Problems/Problem 21]]
 
 
Let <math>P(x)</math> be a polynomial with rational coefficients such that when <math>P(x)</math> is divided by the polynomial <math>x^2+x+1</math>, the remainder is <math>x+2</math>, and when <math>P(x)</math> is divided by the polynomial <math>x^2+1</math>, the remainder 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?
 
 
 
<math>\textbf{(A)}\ 10 \qquad
 
\textbf{(B)}\ 13 \qquad
 
\textbf{(C)}\ 19 \qquad
 
\textbf{(D)}\ 20 \qquad
 
\textbf{(E)}\ 23 \qquad</math>
 
 
 
== Solution 1 ==
 
It is easy to see that <math>P(x)</math> has a degree of at least 2.
 
 
 
Suppose that it has degree <math>2</math>, so let <math>P(x) = a(x^2 + x + 1) + (x + 2) = b(x^2 + 1) + (2x + 1)</math>. Then comparing coefficients of <math>x^2</math> gives <math>a = b</math>, and comparing coefficients of <math>x^0</math> gives <math>a + 2 = b + 1</math>, a contradiction.
 
 
 
Now suppose it has degree <math>3</math>. Let <math>P(x) = (ax + b)(x^2 + x + 1) + (x + 2) = (cx + d)(x^2 + 1) + (2x + 1)</math>. Equating coefficients of <math>x^3</math> gives <math>a = c</math>, so <math>(ax + b)(x^2 + x + 1) + (x + 2) = (ax + d)(x^2 + 1) + (2x + 1)</math>.
 
 
 
Equating coefficients of <math>x^0</math> gives <math>b + 2 = d + 1</math>, so <math>d = b + 1</math> and <math>(ax + b)(x^2 + x + 1) + (x + 2) = (ax + b + 1)(x^2 + 1) + (2x + 1)</math>.
 
 
 
Now equating coefficients of <math>x^2</math> gives <math>b + a = b + 1</math> and hence <math>a = 1</math>. Hence <math>(x + b)(x^2 + x + 1) + (x + 2) = (x + b + 1)(x^2 + 1) + (2x + 1)</math>.
 
 
 
Then, we equate coefficients of <math>x</math> to get <math>1 + b + 1 = 1 + 2</math>, so <math>b = 1</math>.
 
 
 
Hence, <math>P(x) = (x + 1)(x^2 + x + 1) + (x + 2) = x^3 + 2x^2 + 3x + 3</math> and the sum of the squares of coefficients is <math>1^2 + 2^2 + 3^2 + 3^2 = \fbox{(E)23}</math>, and we're done!
 
 
 
~[[User:Bxiao31415 | Bxiao31415]]
 
 
 
== Video Solution by OmegaLearn Using Polynomial Remainders ==
 
https://youtu.be/HdrbPiZHim0
 
 
 
~ pi_is_3.14
 
 
 
 
 
== See Also ==
 
{{AMC12 box|year=2022|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 

Latest revision as of 23:21, 4 January 2023