Difference between revisions of "2022 AMC 12B Problems/Problem 20"
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\textbf{(D)}\ 20 \qquad | \textbf{(D)}\ 20 \qquad | ||
\textbf{(E)}\ 23 \qquad</math> | \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 == | == Video Solution by OmegaLearn Using Polynomial Remainders == | ||
Revision as of 05:10, 18 November 2022
Contents
Problem
Let 
be a polynomial with rational coefficients such that when is divided by the polynomial 
, the remainder is 
, and when is divided by the polynomial 
, the remainder is 
. 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
It is easy to see that has a degree of at least 2.
Suppose that it has degree 
, so let 
. Then comparing coefficients of 
gives 
, and comparing coefficients of 
gives 
, a contradiction.
Now suppose it has degree 
. Let 
. Equating coefficients of 
gives 
, so 
.
Equating coefficients of gives 
, so 
and 
.
Now equating coefficients of gives 
and hence 
. Hence 
.
Then, we equate coefficients of 
to get 
, so 
.
Hence, 
and the sum of the squares of coefficients is 
, and we're done!
Video Solution by OmegaLearn Using Polynomial Remainders
~ 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
