Difference between revisions of "2021 AMC 12B Problems/Problem 20"
MRENTHUSIASM (talk | contribs) (→Solution 4 (Division Analysis Without Finding Q(z)): Made the solution more concise.) |
MRENTHUSIASM (talk | contribs) (1. Prioritized solutions based on easiness. I don't think finding Q(z) is the most efficient. 2. Deleted repetitive solutions while retained credits to the original authors. Let me know if you are unhappy.) |
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<math>\textbf{(A) }-z \qquad \textbf{(B) }-1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1</math> | <math>\textbf{(A) }-z \qquad \textbf{(B) }-1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1</math> | ||
− | ==Solution 1== | + | ==Solution 1 (Difference of Cubes)== |
− | ===Solution 1.1=== | + | Let <math>z=s</math> be a root of <math>z^2+z+1</math> so that <math>s^2+s+1=0.</math> It follows that <cmath>(s-1)\left(s^2+s+1\right)=s^3-1=0,</cmath> from which <math>s^3=1.</math> |
+ | |||
+ | Note that | ||
+ | <cmath>\begin{align*} | ||
+ | s^{2021}+1 &= s^{3\cdot673+2}+1 \\ | ||
+ | &= (s^3)^{673}\cdot s^2+1 \\ | ||
+ | &= s^2+1 \\ | ||
+ | &= \left(s^2+s+1\right)-s \\ | ||
+ | &= -s. | ||
+ | \end{align*}</cmath> | ||
+ | Since <math>z^{2021}+1=-z</math> for each root <math>z=s</math> of <math>z^2+z+1,</math> we conclude that the remainder when <math>z^{2021}+1</math> is divided by <math>z^2+z+1</math> is <math>R(z)=\boxed{\textbf{(A) }-z}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2 (Finds Q(z) Using Patterns)== | ||
+ | |||
+ | Note that the equation above is in the form of polynomial division, with <math>z^{2021}+1</math> being the dividend, <math>z^2+z+1</math> being the divisor, and <math>Q(x)</math> and <math>R(x)</math> being the quotient and remainder respectively. Since the degree of the dividend is <math>2021</math> and the degree of the divisor is <math>2</math>, that means the degree of the quotient is <math>2021-2 = 2019</math>. Note that <math>R(x)</math> can't influence the degree of the right hand side of this equation since its degree is either <math>1</math> or <math>0</math>. Since the coefficients of the leading term in the dividend and the divisor are both <math>1</math>, that means the coefficient of the leading term of the quotient is also <math>1</math>. Thus, the leading term of the quotient is <math>z^{2019}</math>. Multiplying <math>z^{2019}</math> by the divisor gives <math>z^{2021}+z^{2020}+z^{2019}</math>. We have our <math>z^{2021}</math> term but we have these unnecessary terms like <math>z^{2020}</math>. We can get rid of these terms by adding <math>-z^{2018}</math> to the quotient to cancel out these terms, but this then gives us <math>z^{2021}-z^{2018}</math>. Our first instinct will probably be to add <math>z^{2017}</math>, but we can't do this as although this will eliminate the <math>-z^{2018}</math> term, it will produce a <math>z^{2019}</math> term. Since no other term of the form <math>z^n</math> where <math>n</math> is an integer less than <math>2017</math> will produce a <math>z^{2019}</math> term when multiplied by the divisor, we can't add <math>z^{2017}</math> to the quotient. Instead, we can add <math>z^{2016}</math> to the coefficient to get rid of the <math>-z^{2018}</math> term. Continuing this pattern, we get the quotient as <cmath>z^{2019}-z^{2018}+z^{2016}-z^{2015}+\cdots-z^2+1.</cmath> | ||
+ | The last term when multiplied with the divisor gives <math>z^2+z+1</math>. This will get rid of the <math>-z^2</math> term but will produce the expression <math>z+1</math>, giving us the dividend as <math>z^{2021}+z+1</math>. Note that the dividend we want is of the form <math>z^{2021}+1</math>. Therefore, our remainder will have to be <math>-z</math> in order to get rid of the <math>z</math> term in the expression and give us <math>z^{2021}+1</math>, which is what we want. Therefore, the remainder is <math>\boxed{\textbf{(A) }-z}.</math> | ||
+ | |||
+ | ~ rohan.sp ~rocketsri | ||
+ | |||
+ | ==Solution 3 (Modular Arithmetic in Polynomials)== | ||
Note that | Note that | ||
<cmath>z^3-1\equiv 0\pmod{z^2+z+1}</cmath> | <cmath>z^3-1\equiv 0\pmod{z^2+z+1}</cmath> | ||
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The answer is <math>\boxed{\textbf{(A) }-z}.</math> | The answer is <math>\boxed{\textbf{(A) }-z}.</math> | ||
− | + | ==Solution 4 (Complex Numbers)== | |
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One thing to note is that <math>R(z)</math> takes the form of <math>Az + B</math> for some constants <math>A</math> and <math>B.</math> | One thing to note is that <math>R(z)</math> takes the form of <math>Az + B</math> for some constants <math>A</math> and <math>B.</math> | ||
Note that the roots of <math>z^2 + z + 1</math> are part of the solutions of <math>z^3 -1 = 0.</math> | Note that the roots of <math>z^2 + z + 1</math> are part of the solutions of <math>z^3 -1 = 0.</math> | ||
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~Jamess2022 (burntTacos ;-)) | ~Jamess2022 (burntTacos ;-)) | ||
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== Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-Solving) == | == Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-Solving) == |
Revision as of 02:37, 2 October 2021
Contents
Problem
Let and
be the unique polynomials such that
and the degree of
is less than
What is
Solution 1 (Difference of Cubes)
Let be a root of
so that
It follows that
from which
Note that
Since
for each root
of
we conclude that the remainder when
is divided by
is
~MRENTHUSIASM
Solution 2 (Finds Q(z) Using Patterns)
Note that the equation above is in the form of polynomial division, with being the dividend,
being the divisor, and
and
being the quotient and remainder respectively. Since the degree of the dividend is
and the degree of the divisor is
, that means the degree of the quotient is
. Note that
can't influence the degree of the right hand side of this equation since its degree is either
or
. Since the coefficients of the leading term in the dividend and the divisor are both
, that means the coefficient of the leading term of the quotient is also
. Thus, the leading term of the quotient is
. Multiplying
by the divisor gives
. We have our
term but we have these unnecessary terms like
. We can get rid of these terms by adding
to the quotient to cancel out these terms, but this then gives us
. Our first instinct will probably be to add
, but we can't do this as although this will eliminate the
term, it will produce a
term. Since no other term of the form
where
is an integer less than
will produce a
term when multiplied by the divisor, we can't add
to the quotient. Instead, we can add
to the coefficient to get rid of the
term. Continuing this pattern, we get the quotient as
The last term when multiplied with the divisor gives
. This will get rid of the
term but will produce the expression
, giving us the dividend as
. Note that the dividend we want is of the form
. Therefore, our remainder will have to be
in order to get rid of the
term in the expression and give us
, which is what we want. Therefore, the remainder is
~ rohan.sp ~rocketsri
Solution 3 (Modular Arithmetic in Polynomials)
Note that
so if
is the remainder when dividing by
,
Now,
So
, and
The answer is
Solution 4 (Complex Numbers)
One thing to note is that takes the form of
for some constants
and
Note that the roots of
are part of the solutions of
They can be easily solved with roots of unity:
Obviously the right two solutions are the roots of
We substitute
into the original equation, and
becomes 0. Using De Moivre's theorem, we get:
Expanding into rectangular complex number form:
Comparing the real and imaginary parts, we get:
The answer is
~Jamess2022 (burntTacos ;-))
Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-Solving)
Video Solution (Long Division, Not Brutal)
https://youtu.be/kxPDeQRGLEg ~hippopotamus1
~ pi_is_3.14
See Also
2021 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.