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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 25"
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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>
  
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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>
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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]
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\hline
 
\hline
 
& \\ [-2ex]
 
& \\ [-2ex]
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(a'_1, a'_2, a'_3, a'_4) & S+4(0) \\ [0.5ex]
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(a'_1, a'_2, a'_3, a'_4) & S'+4(0) \\ [0.5ex]
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(a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S+4(1) \\ [0.5ex]
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(a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S'+4(1) \\ [0.5ex]
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(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>

Revision as of 23:03, 28 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

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AMC 12 Problems and Solutions

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