Difference between revisions of "2023 AIME I Problems/Problem 9"
(→Solution 1) |
(→Solution 2) |
||
Line 9: | Line 9: | ||
~chem1kall | ~chem1kall | ||
− | |||
− | |||
− | |||
==Solution== | ==Solution== |
Revision as of 15:24, 8 February 2023
Contents
Problem (Unofficial, please update when official one comes out):
is a polynomial with integer coefficients in the range
, inclusive. There is exactly one integer
such that
. How many possible values are there for the ordered triple
?
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)