Difference between revisions of "2023 AIME I Problems/Problem 9"
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− | + | Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> | |
− | + | are integers in <math>{−20, −10, −18, . . . , 18, 19, 20}</math>, such that there is a unique integer | |
− | <math> | + | <math>m \neq 2</math> with <math>p(m) = p(2)</math>. |
==Solution== | ==Solution== |
Revision as of 15:49, 8 February 2023
Find the number of cubic polynomials , where
,
, and
are integers in ${−20, −10, −18, . . . , 18, 19, 20}$ (Error compiling LaTeX. Unknown error_msg), such that there is a unique integer
with
.
Contents
[hide]Solution
Solution 1
Plugging into
, we get
. We can rewrite into
, where
can be any value in the range. Since
must be
. The problem also asks for unique integers, meaning
can only be one value for each polynomial, so the discriminant must be
.
, and
. Rewrite to be
.
must be even for
to be an integer.
because
. However, plugging in
result in
. There are 8 pairs of
and 41 integers for
, giving
~chem1kall
Solution
Define .
Hence, for
, beyond having a root 2, it has a unique integer root that is not equal to 2.
We have
Thus, the polynomial has a unique integer root and it is not equal to 2.
Following from Vieta' formula, the sum of two roots of this polynomial is .
Because
is an integer, if a root is an integer, the other root is also an integer.
Therefore, the only way to have a unique integer root is that the determinant of this polynomial is 0.
Thus,
In addition, because two identical roots are not 2, we have
Equation (1) can be reorganized as
Thus, . Denote
.
Thus, (2) can be written as
Because ,
, and
, we have
.
Therefore, we have the following feasible solutions for :
,
,
,
,
.
Thus, the total number of
is 8.
Because can take any value from
, the number of feasible
is 41.
Therefore, the number of is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution
It can be easily noticed that is independent of the condition
, and can thus safely take all
possible values between
and
.
There are two possible ways for to be the only integer satisfying
:
has a double root at
or a double root at
.
Case 1: has a double root at
:
In this case, , or
. Thus
ranges from
to
. One of these values,
corresponds to a triple root at
, which means
. Thus there are
possible values of
. (It can be verified that
is an integer).
Case 2: has a double root at
:
See the above solution. This yields possible combinations of
and
.
Thus, in total we have combinations of
.
-Alex_Z