Difference between revisions of "2023 AIME I Problems/Problem 9"
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<math>m \neq 2</math> with <math>p(m) = p(2).</math> | <math>m \neq 2</math> with <math>p(m) = p(2).</math> | ||
− | Solution 1 | + | ==Solution 1== |
Plugging <math>2</math> into <math>P(x)</math>, we get <math>8+4a+2b+c = m^3+am^2+bm+c</math>. We can rewrite into <math>(2-m)(m^2+2m+4+a(2+m)+b)=0</math>, where <math>c</math> can be any value in the range. Since <math>m\neq2, m^2+2m+4+a(2+m)+b</math> must be <math>0</math>. The problem also asks for unique integers, meaning <math>m</math> can only be one value for each polynomial, so the discriminant must be <math>0</math>. <math>m^2+2m+4+a(2+m)+b = m^2+m(2+a)+(2a+b+4)= 0</math>, and <math>(2+a)^2-4(2a+b+4)=0</math>. Rewrite to be <math>a(a-4)=4(b+3)</math>. <math>a</math> must be even for <math>4(b+3)</math> to be an integer. <math>-6<=a<=10</math> because <math>4(20+3) = 92</math>. However, plugging in <math>a=-6, b=12</math> result in <math>m=2</math>. There are 8 pairs of <math>(a,b)</math> and 41 integers for <math>c</math>, giving<cmath>41\cdot8 = \boxed{328}</cmath> | Plugging <math>2</math> into <math>P(x)</math>, we get <math>8+4a+2b+c = m^3+am^2+bm+c</math>. We can rewrite into <math>(2-m)(m^2+2m+4+a(2+m)+b)=0</math>, where <math>c</math> can be any value in the range. Since <math>m\neq2, m^2+2m+4+a(2+m)+b</math> must be <math>0</math>. The problem also asks for unique integers, meaning <math>m</math> can only be one value for each polynomial, so the discriminant must be <math>0</math>. <math>m^2+2m+4+a(2+m)+b = m^2+m(2+a)+(2a+b+4)= 0</math>, and <math>(2+a)^2-4(2a+b+4)=0</math>. Rewrite to be <math>a(a-4)=4(b+3)</math>. <math>a</math> must be even for <math>4(b+3)</math> to be an integer. <math>-6<=a<=10</math> because <math>4(20+3) = 92</math>. However, plugging in <math>a=-6, b=12</math> result in <math>m=2</math>. There are 8 pairs of <math>(a,b)</math> and 41 integers for <math>c</math>, giving<cmath>41\cdot8 = \boxed{328}</cmath> | ||
~chem1kall | ~chem1kall | ||
− | Solution | + | ==Solution 2== |
Define <math>q \left( x \right) = p \left( x \right) - p \left( 2 \right)</math>. Hence, for <math>q \left( x \right)</math>, beyond having a root 2, it has a unique integer root that is not equal to 2. | Define <math>q \left( x \right) = p \left( x \right) - p \left( 2 \right)</math>. Hence, for <math>q \left( x \right)</math>, beyond having a root 2, it has a unique integer root that is not equal to 2. | ||
− | We have\begin{align*} q \left( x \right) & = p \left( x \right) - p \left( 2 \right) \\ & = \left( x - 2 \right) \left( x^2 + \left( 2 + a \right) x + 4 + 2a + b \right) . \end{align*} | + | We have <math>\begin{align*} q \left( x \right) & = p \left( x \right) - p \left( 2 \right) \\ & = \left( x - 2 \right) \left( x^2 + \left( 2 + a \right) x + 4 + 2a + b \right) . \end{align*}</math> |
Thus, the polynomial <math>x^2 + \left( 2 + a \right) x + 4 + 2a + b</math> has a unique integer root and it is not equal to 2. | Thus, the polynomial <math>x^2 + \left( 2 + a \right) x + 4 + 2a + b</math> has a unique integer root and it is not equal to 2. | ||
Revision as of 17:15, 8 February 2023
Find the number of cubic polynomials , where
,
, and
are integers in
, such that there is a unique integer
with
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 2
Define . Hence, for
, beyond having a root 2, it has a unique integer root that is not equal to 2.
We have $\begin{align*} q \left( x \right) & = p \left( x \right) - p \left( 2 \right) \\ & = \left( x - 2 \right) \left( x^2 + \left( 2 + a \right) x + 4 + 2a + b \right) . \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
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)