Difference between revisions of "2021 Fall AMC 12A Problems/Problem 25"

Latest revision as of 00:04, 29 April 2023

Problem

Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial $q(x) = c_3x^3+c_2x^2+c_1x+c_0$ such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $c_1?$

$\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11$

Solution 1 (Complete Residue System)

For a fixed value of $m,$ there is a total of $m(m-1)(m-2)(m-3)$ possible ordered quadruples $(a_1, a_2, a_3, a_4).$

Let $S=a_1+a_2+a_3+a_4.$ We claim that exactly $\frac1m$ of these $m(m-1)(m-2)(m-3)$ ordered quadruples satisfy that $m$ divides $S:$

Since $\gcd(m,4)=1,$ we conclude that $\{k+4(0),k+4(1),k+4(2),\ldots,k+4(m-1)\}$ is the complete residue system modulo $m$ for all integers $k.$

Given any ordered quadruple $(a'_1, a'_2, a'_3, a'_4)$ in modulo $m,$ it follows that exactly one of these $m$ ordered quadruples has sum $0$ modulo $m:$ $\begin{array}{c|c} & \\ [-2.5ex] \textbf{Ordered Quadruple} & \textbf{Sum Modulo }\boldsymbol{m} \\ [0.5ex] \hline & \\ [-2ex] (a'_1, a'_2, a'_3, a'_4) & S'+4(0) \\ [0.5ex] (a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S'+4(1) \\ [0.5ex] (a'_1+2, a'_2+2, a'_3+2, a'_4+2) & S'+4(2) \\ [0.5ex] \cdots & \cdots \\ [0.5ex] (a'_1+m-1, a'_2+m-1, a'_3+m-1, a'_4+m-1) & S'+4(m-1) \\ [0.5ex] \end{array}$ We conclude that $q(m)=\frac1m\cdot[m(m-1)(m-2)(m-3)]=(m-1)(m-2)(m-3),$ so $q(x)=(x-1)(x-2)(x-3)=c_3x^3+c_2x^2+c_1x+c_0.$ By Vieta's Formulas, we get $c_1=1\cdot2+1\cdot3+2\cdot3=\boxed{\textbf{(E)}\ 11}.$

~MRENTHUSIASM

Solution 2 (Symmetric Congruent Numbers and Interpolation)

Define $b_i = \left\{ \begin{array}{ll} a_i & \mbox{ if } 1 \leq a_i \leq \frac{m-1}{2} \\ a_i - m & \mbox{ if } \frac{m-1}{2} + 1 \leq a_i \leq m - 1 \\ 0 & \mbox{ if } a_i = m \end{array} \right..$

Hence, $b_i$ is a one-to-one and onto function of $a_i$ , and the range of $b_i$ is $\left\{- \frac{m-1}{2} , \cdots , \frac{m-1}{2} \right\}$ .

Therefore, to solve this problem, it is equivalent for us to count the number of tuples $\left( b_1 , b_2 , b_3 , b_4 \right)$ that are all distinct and satisfy $m | b_1 + b_2 + b_3 + b_4$ .

Denote by $d \left( m \right)$ the number of such tuples that are also subject to the constraint $b_1 < b_2 < b_3 < b_4$ .

Hence, $D \left( m \right) = 4! d \left( m \right) = 24 d \left( m \right)$ .

We do the following casework analysis to compute $d \left( m \right)$ .

$\textbf{Case 1}$ : There is one $0$ in $\left( b_1 , b_2 , b_3 , b_4 \right)$ .

Denote by $d_{1i} \left( m \right)$ the number of tuples with $b_i = 0$ .

By symmetry, $d_{11} \left( m \right)= d_{14} \left( m \right)$ and $d_{12} \left( m \right)= d_{13} \left( m \right)$ .

$\textbf{Case 2}$ : There is no $0$ in $\left( b_1 , b_2 , b_3 , b_4 \right)$ .

Denote by $d_{2i} \left( m \right)$ the number of tuples with $i$ positive entries.

By symmetry, $d_{20} \left( m \right) = d_{24} \left( m \right)$ and $d_{21} \left( m \right) = d_{23} \left( m \right)$ .

Therefore, $\begin{align*} D \left( m \right) & = 24 d \left( m \right) \\ & = 24 \left( \sum_{i=1}^4 d_{1i} \left( m \right) + \sum_{i=0}^4 d_{2i} \left( m \right) \right) \\ & = 24 \left( 2 d_{11} \left( m \right) + 2 d_{12} \left( m \right) + 2 d_{24} \left( m \right) + 2 d_{23} \left( m \right) + d_{22} \left( m \right) \right) . \end{align*}$

Now, we compute $D \left( m \right)$ for $m = 5 , 7 , 9 , 11$ .

$\underline{\textbf{SCENARIO}}$ $m = 5$ .

We have $b_i \in \left\{ - 2 , \cdots , 2 \right\}$ .

$\textbf{Case 1}$

$\textbf{Case 1.1}$ : $b_1 = 0$

We cannot have $3$ distinct positive integers. So $d_{11} \left( 5 \right) = 0$ .

$\textbf{Case 1.2}$ : $b_2 = 0$

Because there are $2$ positive integers, we must have $b_3 = 1$ , $b_4 = 2$ . Hence, $b_1 = - 3$ . However, this is out of the range of $b_i$ . Thus, $d_{12} \left( 5 \right) = 0$ .

$\textbf{Case 2}$

$\textbf{Case 2.1}$ : $b_1 > 0$

We cannot have $4$ distinct positive integers. So $d_{24} \left( 5 \right) = 0$ .

$\textbf{Case 2.2}$ : $b_1 < 0 < b_2$

We cannot have $3$ distinct positive integers. So $d_{23} \left( 5 \right) = 0$ .

$\textbf{Case 2.3}$ : $b_2 < 0 < b_3$

The only solution is $\left( b_1 , b_2 , b_3 , b_4 \right) = \left( - 2 , - 1 , 1 , 2 \right)$ . So $d_{22} \left( 5 \right) = 1$ .

Therefore, $D \left( 5 \right) = 24$ .

$\underline{\textbf{SCENARIO}}$ $m = 7$ .

We have $b_i \in \left\{ - 3 , \cdots , 3 \right\}$ .

$\textbf{Case 1}$

$\textbf{Case 1.1}$ : $b_1 = 0$

We have no feasible solution. Thus, $d_{11} \left( 7 \right) = 0$ .

$\textbf{Case 1.2}$ : $b_2 = 0$

The only solution is $\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)$ . Thus, $d_{12} \left( 7 \right) = 1$ .

$\textbf{Case 2}$

$\textbf{Case 2.1}$ : $b_1 > 0$

We cannot have $4$ distinct positive integers. So $d_{24} \left( 7 \right) = 0$ .

$\textbf{Case 2.2}$ : $b_1 < 0 < b_2$

To get $3$ distinct positive integers, we have $\left( b_2 , b_3 , b_4 \right) = \left( 1 , 2 , 3 \right)$ . This implies $b_1 = - 6$ . However, this is out of the range of $b_1$ . So $d_{23} \left( 6 \right) = 0$ .

$\textbf{Case 2.3}$ : $b_2 < 0 < b_3$

We have $d_{22} \left( 7 \right) = \binom{3}{2} = 3$ .

Therefore, $D \left( 7 \right) = 24 \cdot 5$ .

$\underline{\textbf{SCENARIO}}$ $m = 9$ .

We have $b_i \in \left\{ - 4 , \cdots , 4 \right\}$ .

$\textbf{Case 1}$

$\textbf{Case 1.1}$ : $b_1 = 0$

The only solution is $\left( b_2 , b_3 , b_4 \right) = \left( 2, 3, 4 \right)$ . Thus, $d_{11} \left( 9 \right) = 1$ .

$\textbf{Case 1.2}$ : $b_2 = 0$

The feasible solutions are $\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)$ , $\left( - 4 , 1 , 3 \right)$ . Thus, $d_{12} \left( 9 \right) = 2$ .

$\textbf{Case 2}$

$\textbf{Case 2.1}$ : $b_1 > 0$

There is no feasible solution. So $d_{24} \left( 9 \right) = 0$ .

$\textbf{Case 2.2}$ : $b_1 < 0 < b_2$

To get $3$ distinct positive integers, we have $b_2 + b_3 + b_4 \geq 1 + 2 + 3 = 6$ . This implies $b_1 = - 6$ . However, this is out of the range of $b_1$ . So $d_{23} \left( 9 \right) = 0$ .

$\textbf{Case 2.3}$ : $b_2 < 0 < b_3$

We have $d_{22} \left( 9 \right) = 8$ .

Therefore, $D \left( 9 \right) = 24 \cdot 14$ .

$\underline{\textbf{SCENARIO}}$ $m = 11$ .

We have $b_i \in \left\{ - 5 , \cdots , 5 \right\}$ .

$\textbf{Case 1}$

$\textbf{Case 1.1}$ : $b_1 = 0$

The only solution is $\left( b_2 , b_3 , b_4 \right) = \left( 2, 4, 5 \right)$ . Thus, $d_{11} \left( 11 \right) = 1$ .

$\textbf{Case 1.2}$ : $b_2 = 0$

The feasible solutions are $\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)$ , $\left( - 4 , 1 , 3 \right)$ , $\left( - 5 , 1 , 4 \right)$ , $\left( - 5 , 2 , 3 \right)$ . Thus, $d_{12} \left( 11 \right) = 4$ .

$\textbf{Case 2}$

$\textbf{Case 2.1}$ : $b_1 > 0$

The only feasible solution is $\left( b_1 , b_2 , b_3 , b_4 \right) = \left( 1 , 2, 3, 5 \right)$ . So $d_{24} \left( 11 \right) = 1$ .

$\textbf{Case 2.2}$ : $b_1 < 0 < b_2$

The only feasible solution is $\left( b_1 , b_2 , b_3 , b_4 \right) = \left( -1 , 3, 4, 5 \right)$ . So $d_{23} \left( 11 \right) = 1$ .

$\textbf{Case 2.3}$ : $b_2 < 0 < b_3$

We have $d_{22} \left( 11 \right) = 16$ .

Therefore, $D \left( 11 \right) = 24 \cdot 30$ .

We know that $q \left( m \right) = D \left( m \right)$ for odd $m \geq 5$ .

Plugging $m = 5, 7, 9, 11$ into this equation, we get $\begin{align*} c_3 5^3 + c_2 5^2 + c_1 5 + c_0 & = 24 \cdot 1 && (1.1) \\ c_3 7^3 + c_2 7^2 + c_1 7 + c_0 & = 24 \cdot 5 && (1.2) \\ c_3 9^3 + c_2 9^2 + c_1 9 + c_0 & = 24 \cdot 14 && (1.3) \\ c_3 11^3 + c_2 11^2 + c_1 11 + c_0 & = 24 \cdot 30 && (1.4) \end{align*}$

Now, we solve this system of equations. Taking $\frac{(1.2)-(1.1)}{2}$ , $\frac{(1.3)-(1.2)}{2}$ , $\frac{(1.4)-(1.2)}{2}$ , we get $\begin{align*} c_3 109 + c_2 12 + c_1 & = 48 && (2.1) \\ c_3 193 + c_2 16 + c_1 & = 108 && (2.2) \\ c_3 301 + c_2 20 + c_1 & = 192 && (2.3) \end{align*}$

Taking $\frac{(2.2)-(2.1)}{4}$ , $\frac{(2.3)-(2.2)}{4}$ , we get $\begin{align*} c_3 21 + c_2 & = 15 && (3.1) \\ c_3 27 + c_2 & = 21 && (3.2) \end{align*}$

Taking $\frac{(3.2)-(3.1)}{6}$ , we get $c_3 = 1$ .

Plugging $c_3$ into Equation (3.1), we get $c_2 = - 6$ .

Plugging $c_2$ and $c_3$ into Equation (2.1), we get $c_1 = 11$ .

Therefore, the answer is $\boxed{\textbf{(E) }11}$ .

~Steven Chen (www.professorchenedu.com)

Solution 3

As before, note that we have $m(m-1)(m-2)$ numbers we can choose as $a,b,c.$ From here, there is exactly one possible value of $1 \leq d \leq m$ that could make $S=a+b+c+d$ divisible by $m.$ However, there is a $\frac{3}{m}$ chance that this value of $d$ has already been chosen as $a,b$ or $c$ . Thus our polynomial is $m(m-1)(m-2)\left(1-\frac{3}{m}\right)=m(m-1)(m-2)\left(\frac{m-3}{m}\right)=(m-1)(m-2)(m-3)$ . By Vieta's, $c_1 = 2+3+6=\boxed{\textbf{(E) } 11}$ .

~Dhillonr25

Video Solution

https://youtu.be/YExRIOt819Y

~MathProblemSolvingSkills.com

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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 25"

Latest revision as of 00:04, 29 April 2023

Contents

Problem

Solution 1 (Complete Residue System)

Solution 2 (Symmetric Congruent Numbers and Interpolation)

Solution 3

Video Solution

See Also

@@ Line 5: / Line 5: @@
 <math>\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11</math>
-==Solution==
+==Solution 1 (Complete Residue System)==
 For a fixed value of <math>m,</math> there is a total of <math>m(m-1)(m-2)(m-3)</math> possible ordered quadruples <math>(a_1, a_2, a_3, a_4).</math>
 Let <math>S=a_1+a_2+a_3+a_4.</math> We claim that exactly <math>\frac1m</math> of these <math>m(m-1)(m-2)(m-3)</math> ordered quadruples satisfy that <math>m</math> divides <math>S:</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 system of residues 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 satisfy that <math>m</math> divides <math>S:</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 has sum <math>0</math> modulo <math>m:</math>
 <cmath>\begin{array}{c|c}
 & \\ [-2.5ex]
@@ Line 18: / Line 18: @@
 \hline
 & \\ [-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]
-(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>
 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>
@@ Line 29: / Line 29: @@
 ~MRENTHUSIASM
-==Solution 2 (if you're running out of time)==
+== Solution 2 (Symmetric Congruent Numbers and Interpolation) ==
-Note that you see numbers with absolute value <math>1</math>, <math>6</math>, and <math>11</math> in the answer choices. What is special about those numbers? Well, you should notice that they are the coefficients of the polynomial <math>(x+1)(x+2)(x+3)</math> when expanded (if you've already memmed this). Then, you can probably guess the polynomial is some form of <math>(x+1)(x+2)(x+3)</math> whether negative or positive. Since <math>c_1</math> is asked, the answer should be reasoned out as <math>1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3 = \boxed{11}.</math> you can gain further confidence in your guess since that is the only answer choice with absolute value <math>11</math>
+Define
+<cmath>
+\[
+b_i =
+\left\{
+\begin{array}{ll}
+a_i & \mbox{ if } 1 \leq a_i \leq \frac{m-1}{2} \\
+a_i - m & \mbox{ if } \frac{m-1}{2} + 1 \leq a_i \leq m - 1 \\
+& \mbox{ if } a_i = m
+\end{array}
+\right..
+\]
+</cmath>
--fidgetboss_4000
+Hence, <math>b_i</math> is a one-to-one and onto function of <math>a_i</math>, and the range of <math>b_i</math> is <math>\left\{- \frac{m-1}{2}  , \cdots , \frac{m-1}{2} \right\}</math>.
+Therefore, to solve this problem, it is equivalent for us to count the number of tuples <math>\left( b_1 , b_2 , b_3 , b_4 \right)</math> that are all distinct and satisfy <math>m | b_1 + b_2 + b_3 + b_4</math>.
+Denote by <math>d \left( m \right)</math> the number of such tuples that are also subject to the constraint <math>b_1 < b_2 < b_3 < b_4</math>.
+Hence, <math>D \left( m \right) = 4! d \left( m \right) = 24 d \left( m \right)</math>.
+We do the following casework analysis to compute <math>d \left( m \right) </math>.
+<math>\textbf{Case 1}</math>: There is one <math>0</math> in <math>\left( b_1 , b_2 , b_3 , b_4 \right)</math>.
+Denote by <math>d_{1i} \left( m \right)</math> the number of tuples with <math>b_i = 0</math>.
+By symmetry, <math>d_{11} \left( m \right)= d_{14} \left( m \right)</math> and <math>d_{12} \left( m \right)= d_{13} \left( m \right)</math>.
+<math>\textbf{Case 2}</math>: There is no <math>0</math> in <math>\left( b_1 , b_2 , b_3 , b_4 \right)</math>.
+Denote by <math>d_{2i} \left( m \right)</math> the number of tuples with <math>i</math> positive entries.
+By symmetry, <math>d_{20} \left( m \right) = d_{24} \left( m \right)</math> and <math>d_{21} \left( m \right) = d_{23} \left( m \right)</math>.
+Therefore,
+<cmath>
+\begin{align*}
+D \left( m \right) & = 24 d \left( m \right) \\
+& = 24 \left( \sum_{i=1}^4 d_{1i} \left( m \right) + \sum_{i=0}^4 d_{2i} \left( m \right) \right) \\
+& = 24 \left( 2 d_{11} \left( m \right) + 2 d_{12} \left( m \right) + 2 d_{24} \left( m \right) + 2 d_{23} \left( m \right) + d_{22} \left( m \right) \right) .
+\end{align*}
+</cmath>
+Now, we compute <math>D \left( m \right)</math> for <math>m = 5 , 7 , 9 , 11</math>.
+<math>\underline{\textbf{SCENARIO}}</math> <math>m = 5</math>.
+We have <math>b_i \in \left\{ - 2 , \cdots , 2 \right\}</math>.
+<math>\textbf{Case 1}</math>
+<math>\textbf{Case 1.1}</math>: <math>b_1 = 0</math>
+We cannot have <math>3</math> distinct positive integers. So <math>d_{11} \left( 5 \right) = 0</math>.
+<math>\textbf{Case 1.2}</math>: <math>b_2 = 0</math>
+Because there are <math>2</math> positive integers, we must have <math>b_3 = 1</math>, <math>b_4 = 2</math>.
+Hence, <math>b_1 = - 3</math>. However, this is out of the range of <math>b_i</math>.
+Thus, <math>d_{12} \left( 5 \right) = 0</math>.
+<math>\textbf{Case 2}</math>
+<math>\textbf{Case 2.1}</math>: <math>b_1 > 0</math>
+We cannot have <math>4</math> distinct positive integers. So <math>d_{24} \left( 5 \right) = 0</math>.
+<math>\textbf{Case 2.2}</math>: <math>b_1 < 0 < b_2</math>
+We cannot have <math>3</math> distinct positive integers. So <math>d_{23} \left( 5 \right) = 0</math>.
+<math>\textbf{Case 2.3}</math>: <math>b_2 < 0 < b_3</math>
+The only solution is <math>\left( b_1 , b_2 , b_3 , b_4 \right) = \left( - 2 , - 1 , 1 , 2 \right)</math>. So <math>d_{22} \left( 5 \right) = 1</math>.
+Therefore, <math>D \left( 5 \right) = 24</math>.
+<math>\underline{\textbf{SCENARIO}}</math> <math>m = 7</math>.
+We have <math>b_i \in \left\{ - 3 , \cdots , 3 \right\}</math>.
+<math>\textbf{Case 1}</math>
+<math>\textbf{Case 1.1}</math>: <math>b_1 = 0</math>
+We have no feasible solution. Thus, <math>d_{11} \left( 7 \right) = 0</math>.
+<math>\textbf{Case 1.2}</math>: <math>b_2 = 0</math>
+The only solution is <math>\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)</math>.
+Thus, <math>d_{12} \left( 7 \right) = 1</math>.
+<math>\textbf{Case 2}</math>
+<math>\textbf{Case 2.1}</math>: <math>b_1 > 0</math>
+We cannot have <math>4</math> distinct positive integers. So <math>d_{24} \left( 7 \right) = 0</math>.
+<math>\textbf{Case 2.2}</math>: <math>b_1 < 0 < b_2</math>
+To get <math>3</math> distinct positive integers, we have <math>\left( b_2 , b_3 , b_4 \right) = \left( 1 , 2 , 3 \right)</math>. This implies <math>b_1 = - 6</math>. However, this is out of the range of <math>b_1</math>. So <math>d_{23} \left( 6 \right) = 0</math>.
+<math>\textbf{Case 2.3}</math>: <math>b_2 < 0 < b_3</math>
+We have <math>d_{22} \left( 7 \right) = \binom{3}{2} = 3</math>.
+Therefore, <math>D \left( 7 \right) = 24 \cdot 5</math>.
+<math>\underline{\textbf{SCENARIO}}</math> <math>m = 9</math>.
+We have <math>b_i \in \left\{ - 4 , \cdots , 4 \right\}</math>.
+<math>\textbf{Case 1}</math>
+<math>\textbf{Case 1.1}</math>: <math>b_1 = 0</math>
+The only solution is <math>\left( b_2 , b_3 , b_4 \right) = \left(  2, 3, 4 \right)</math>.
+Thus, <math>d_{11} \left( 9 \right) = 1</math>.
+<math>\textbf{Case 1.2}</math>: <math>b_2 = 0</math>
+The feasible solutions are <math>\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)</math>, <math>\left( - 4 , 1 , 3 \right)</math>.
+Thus, <math>d_{12} \left( 9 \right) = 2</math>.
+<math>\textbf{Case 2}</math>
+<math>\textbf{Case 2.1}</math>: <math>b_1 > 0</math>
+There is no feasible solution. So <math>d_{24} \left( 9 \right) = 0</math>.
+<math>\textbf{Case 2.2}</math>: <math>b_1 < 0 < b_2</math>
+To get <math>3</math> distinct positive integers, we have <math>b_2 + b_3 + b_4 \geq 1 + 2 + 3 = 6</math>. This implies <math>b_1 = - 6</math>. However, this is out of the range of <math>b_1</math>. So <math>d_{23} \left( 9 \right) = 0</math>.
+<math>\textbf{Case 2.3}</math>: <math>b_2 < 0 < b_3</math>
+We have <math>d_{22} \left( 9 \right) = 8</math>.
+Therefore, <math>D \left( 9 \right) = 24 \cdot 14</math>.
+<math>\underline{\textbf{SCENARIO}}</math> <math>m = 11</math>.
+We have <math>b_i \in \left\{ - 5 , \cdots , 5 \right\}</math>.
+<math>\textbf{Case 1}</math>
+<math>\textbf{Case 1.1}</math>: <math>b_1 = 0</math>
+The only solution is <math>\left( b_2 , b_3 , b_4 \right) = \left(  2, 4, 5 \right)</math>.
+Thus, <math>d_{11} \left( 11 \right) = 1</math>.
+<math>\textbf{Case 1.2}</math>: <math>b_2 = 0</math>
+The feasible solutions are <math>\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)</math>, <math>\left( - 4 , 1 , 3 \right)</math>, <math>\left( - 5 , 1 , 4 \right)</math>, <math>\left( - 5 , 2 , 3 \right)</math>.
+Thus, <math>d_{12} \left( 11 \right) = 4</math>.
+<math>\textbf{Case 2}</math>
+<math>\textbf{Case 2.1}</math>: <math>b_1 > 0</math>
+The only feasible solution is <math>\left( b_1 , b_2 , b_3 , b_4 \right) = \left(  1 , 2, 3, 5 \right)</math>. So <math>d_{24} \left( 11 \right) = 1</math>.
+<math>\textbf{Case 2.2}</math>: <math>b_1 < 0 < b_2</math>
+The only feasible solution is <math>\left( b_1 , b_2 , b_3 , b_4 \right) = \left(  -1 , 3, 4, 5 \right)</math>.  So <math>d_{23} \left( 11 \right) = 1</math>.
+<math>\textbf{Case 2.3}</math>: <math>b_2 < 0 < b_3</math>
+We have <math>d_{22} \left( 11 \right) = 16</math>.
+Therefore, <math>D \left( 11 \right) = 24 \cdot 30</math>.
+We know that <math>q \left( m \right) = D \left( m \right)</math> for odd <math>m \geq 5</math>.
+Plugging <math>m = 5, 7, 9, 11</math> into this equation, we get
+<cmath>
+\begin{align*}
+c_3 5^3 + c_2 5^2 + c_1 5 + c_0 & = 24 \cdot 1 && (1.1) \\
+c_3 7^3 + c_2 7^2 + c_1 7 + c_0 & = 24 \cdot 5 && (1.2) \\
+c_3 9^3 + c_2 9^2 + c_1 9 + c_0 & = 24 \cdot 14 && (1.3) \\
+c_3 11^3 + c_2 11^2 + c_1 11 + c_0 & = 24 \cdot 30 && (1.4)
+\end{align*}
+</cmath>
+Now, we solve this system of equations.
+Taking <math>\frac{(1.2)-(1.1)}{2}</math>, <math>\frac{(1.3)-(1.2)}{2}</math>, <math>\frac{(1.4)-(1.2)}{2}</math>, we get
+<cmath>
+\begin{align*}
+c_3 109 + c_2 12 + c_1 & = 48 && (2.1) \\
+c_3 193 + c_2 16 + c_1 & = 108 && (2.2) \\
+c_3 301 + c_2 20 + c_1 & = 192 && (2.3)
+\end{align*}
+</cmath>
+Taking <math>\frac{(2.2)-(2.1)}{4}</math>, <math>\frac{(2.3)-(2.2)}{4}</math>, we get
+<cmath>
+\begin{align*}
+c_3 21 + c_2  & = 15 && (3.1) \\
+c_3 27 + c_2 & = 21 && (3.2)
+\end{align*}
+</cmath>
+Taking <math>\frac{(3.2)-(3.1)}{6}</math>, we get <math>c_3 = 1</math>.
+Plugging <math>c_3</math> into Equation (3.1), we get <math>c_2 = - 6</math>.
+Plugging <math>c_2</math> and <math>c_3</math> into Equation (2.1), we get <math>c_1 = 11</math>.
+Therefore, the answer is <math>\boxed{\textbf{(E) }11}</math>.
+~Steven Chen (www.professorchenedu.com)
+==Solution 3==
+As before, note that we have <math>m(m-1)(m-2)</math> numbers we can choose as <math>a,b,c.</math> From here, there is exactly one possible value of <math>1 \leq d \leq m</math> that could make <math>S=a+b+c+d</math> divisible by <math>m.</math> However, there is a <math>\frac{3}{m}</math> chance that this value of <math>d</math> has already been chosen as <math>a,b</math> or <math>c</math>. Thus our polynomial is <math>m(m-1)(m-2)\left(1-\frac{3}{m}\right)=m(m-1)(m-2)\left(\frac{m-3}{m}\right)=(m-1)(m-2)(m-3)</math>. By Vieta's, <math>c_1 = 2+3+6=\boxed{\textbf{(E) } 11}</math>.
+~Dhillonr25
+==Video Solution==
+https://youtu.be/YExRIOt819Y
+~MathProblemSolvingSkills.com
 ==See Also==
 {{AMC12 box|year=2021 Fall|ab=A|num-b=24|after=Last Problem}}
 {{MAA Notice}}