Difference between revisions of "2021 Fall AMC 12A Problems/Problem 25"
MRENTHUSIASM (talk | contribs) |
MRENTHUSIASM (talk | contribs) m (→Solution 1 (Complete Residue System)) |
||
(3 intermediate revisions by the same user not shown) | |||
Line 12: | Line 12: | ||
Since <math>\gcd(m,4)=1,</math> we conclude that <cmath>\{k+4(0),k+4(1),k+4(2),\ldots,k+4(m-1)\}</cmath> is the complete residue system modulo <math>m</math> for all integers <math>k.</math> | Since <math>\gcd(m,4)=1,</math> we conclude that <cmath>\{k+4(0),k+4(1),k+4(2),\ldots,k+4(m-1)\}</cmath> is the complete residue system modulo <math>m</math> for all integers <math>k.</math> | ||
− | Given any ordered quadruple <math>(a'_1, a'_2, a'_3, a'_4)</math> in modulo <math>m,</math> it follows that exactly one of these <math>m</math> ordered quadruples | + | Given any ordered quadruple <math>(a'_1, a'_2, a'_3, a'_4)</math> in modulo <math>m,</math> it follows that exactly one of these <math>m</math> ordered quadruples has sum <math>0</math> modulo <math>m:</math> |
<cmath>\begin{array}{c|c} | <cmath>\begin{array}{c|c} | ||
& \\ [-2.5ex] | & \\ [-2.5ex] | ||
Line 18: | Line 18: | ||
\hline | \hline | ||
& \\ [-2ex] | & \\ [-2ex] | ||
− | (a'_1, a'_2, a'_3, a'_4) & S+4(0) \\ [0.5ex] | + | (a'_1, a'_2, a'_3, a'_4) & S'+4(0) \\ [0.5ex] |
− | (a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S+4(1) \\ [0.5ex] | + | (a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S'+4(1) \\ [0.5ex] |
− | (a'_1+2, a'_2+2, a'_3+2, a'_4+2) & S+4(2) \\ [0.5ex] | + | (a'_1+2, a'_2+2, a'_3+2, a'_4+2) & S'+4(2) \\ [0.5ex] |
\cdots & \cdots \\ [0.5ex] | \cdots & \cdots \\ [0.5ex] | ||
− | (a'_1+m-1, a'_2+m-1, a'_3+m-1, a'_4+m-1) & S+4(m-1) \\ [0.5ex] | + | (a'_1+m-1, a'_2+m-1, a'_3+m-1, a'_4+m-1) & S'+4(m-1) \\ [0.5ex] |
\end{array}</cmath> | \end{array}</cmath> | ||
We conclude that <math>q(m)=\frac1m\cdot[m(m-1)(m-2)(m-3)]=(m-1)(m-2)(m-3),</math> so <cmath>q(x)=(x-1)(x-2)(x-3)=c_3x^3+c_2x^2+c_1x+c_0.</cmath> | We conclude that <math>q(m)=\frac1m\cdot[m(m-1)(m-2)(m-3)]=(m-1)(m-2)(m-3),</math> so <cmath>q(x)=(x-1)(x-2)(x-3)=c_3x^3+c_2x^2+c_1x+c_0.</cmath> |
Latest revision as of 00:04, 29 April 2023
Contents
Problem
Let be an odd integer, and let
denote the number of quadruples
of distinct integers with
for all
such that
divides
. There is a polynomial
such that
for all odd integers
. What is
Solution 1 (Complete Residue System)
For a fixed value of there is a total of
possible ordered quadruples
Let We claim that exactly
of these
ordered quadruples satisfy that
divides
Since we conclude that
is the complete residue system modulo
for all integers
Given any ordered quadruple in modulo
it follows that exactly one of these
ordered quadruples has sum
modulo
We conclude that
so
By Vieta's Formulas, we get
~MRENTHUSIASM
Solution 2 (Symmetric Congruent Numbers and Interpolation)
Define
Hence, is a one-to-one and onto function of
, and the range of
is
.
Therefore, to solve this problem, it is equivalent for us to count the number of tuples that are all distinct and satisfy
.
Denote by the number of such tuples that are also subject to the constraint
.
Hence, .
We do the following casework analysis to compute .
: There is one
in
.
Denote by the number of tuples with
.
By symmetry, and
.
: There is no
in
.
Denote by the number of tuples with
positive entries.
By symmetry, and
.
Therefore,
Now, we compute for
.
.
We have .
:
We cannot have distinct positive integers. So
.
:
Because there are positive integers, we must have
,
.
Hence,
. However, this is out of the range of
.
Thus,
.
:
We cannot have distinct positive integers. So
.
:
We cannot have distinct positive integers. So
.
:
The only solution is . So
.
Therefore, .
.
We have .
:
We have no feasible solution. Thus, .
:
The only solution is .
Thus,
.
:
We cannot have distinct positive integers. So
.
:
To get distinct positive integers, we have
. This implies
. However, this is out of the range of
. So
.
:
We have .
Therefore, .
.
We have .
:
The only solution is .
Thus,
.
:
The feasible solutions are ,
.
Thus,
.
:
There is no feasible solution. So .
:
To get distinct positive integers, we have
. This implies
. However, this is out of the range of
. So
.
:
We have .
Therefore, .
.
We have .
:
The only solution is .
Thus,
.
:
The feasible solutions are ,
,
,
.
Thus,
.
:
The only feasible solution is . So
.
:
The only feasible solution is . So
.
:
We have .
Therefore, .
We know that for odd
.
Plugging into this equation, we get
Now, we solve this system of equations.
Taking ,
,
, we get
Taking ,
, we get
Taking , we get
.
Plugging into Equation (3.1), we get
.
Plugging and
into Equation (2.1), we get
.
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 3
As before, note that we have numbers we can choose as
From here, there is exactly one possible value of
that could make
divisible by
However, there is a
chance that this value of
has already been chosen as
or
. Thus our polynomial is
. By Vieta's,
.
~Dhillonr25
Video Solution
~MathProblemSolvingSkills.com
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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.