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: 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 $\boxed{11}$

Answer: E 11