2021 AMC 12B Problems/Problem 20
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Complex numbers)
- 4 Solution 3 (similar to Solution 1.2)
- 5 Solution 4 (Division Analysis, Similar Approach but Different Explanation from Solution 1)
- 6 Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-solving)
- 7 Video Solution using long division(not brutal)
- 8 See Also
Problem
Let and
be the unique polynomials such that
and the degree of
is less than
What is
Solution 1
Solution 1.1
Note that
so if
is the remainder when dividing by
,
Now,
So
, and
The answer is
Solution 1.2 (More Thorough Version of Solution 1.1)
Instead of dealing with a nasty , we can instead deal with the nice
, as
is a factor of
. Then, we try to see what
is. Of course, we will need a
, getting
. Then, we've gotta get rid of the
term, so we add a
, to get
. This pattern continues, until we add a
to get rid of
, and end up with
. We can't add anything more to get rid of the
, so our factor is
. Then, to get rid of the
, we must have a remainder of
, and to get the
we have to also have a
in the remainder. So, our product is
Then, our remainder is
. The remainder when dividing by
must be the same when dividing by
, modulo
. So, we have that
, or
. This corresponds to answer choice
. ~rocketsri
Solution 2 (Complex numbers)
One thing to note is that takes the form of
for some constants A and B.
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;-;)
Solution 3 (similar to Solution 1.2)
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 R(x) 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
Solution 4 (Division Analysis, Similar Approach but Different Explanation from Solution 1)
By the difference of cubes or the short geometric series, we get
We rewrite
by the polynomial division algorithm:
where
and
are unique polynomials such that
Taking
in modulo
(in which
), we have
Substituting
back into
gives
which almost resembles to the original equation
Since we require that
the divisor
goes into the remaining
for one more time. Rewriting
produces
from which
~MRENTHUSIASM
Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-solving)
Video Solution using 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 |
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