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

Latest revision as of 21:37, 16 September 2025

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 $D(m)=\frac1m\cdot[m(m-1)(m-2)(m-3)]=(m-1)(m-2)(m-3),$ 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

Solution 4

We can solve this by directly counting quadruples and deriving a closed formula for $D(m)$ .

For an odd integer $m \geq 5$ , we need to count ordered quadruples $(a_1,a_2,a_3,a_4)$ of distinct integers where $1 \leq a_i \leq m$ and $m$ divides their sum.

Consider how many total ordered quadruples of distinct integers exist in the range $[1,m]$ : - We choose 4 distinct numbers from $m$ possible values: $\binom{m}{4}$ - We arrange these 4 numbers in order: $4!$

Therefore, total quadruples = $\binom{m}{4} \cdot 4! = \frac{m!}{(m-4)!}$

Now, these quadruples can be partitioned into $m$ groups based on the remainder of their sum when divided by $m$ . By symmetry, these remainders are uniformly distributed modulo $m$ .

Therefore, exactly $\frac{1}{m}$ of all quadruples have a sum divisible by $m$ :

$D(m) = \frac{1}{m} \cdot \frac{m!}{(m-4)!} = \frac{m(m-1)(m-2)(m-3)}{m} = (m-1)(m-2)(m-3)$

Expanding this expression: $(m-1)(m-2)(m-3) = m^3 - 6m^2 + 11m - 6$

Comparing with $q(x) = c_3x^3 + c_2x^2 + c_1x + c_0$ , we identify: $c_3 = 1$ , $c_2 = -6$ , $c_1 = 11$ , $c_0 = -6$

Therefore, $c_1 = 11$ , and the answer is $\boxed{(E) 11}$ .

~ brandonyee

Solution 5 (Principle of Inclusion-Exclusion)

If we do not have to worry about whether or not the numbers are distinct, then there would be $m^3$ possible quadruples. This is because the first $3$ numbers can be anything, and then the last number is uniquely chosen so that the sum is a multiple of $m$ .

Now, we must subtract the quadruples with at least two of the same number. There are $\binom{4}{2}$ ways to choose the positions for the two identical numbers, $m$ ways to choose the number for these two positions, and $m$ ways to choose a number for another position. The last position is uniquely determined as before. Thus, our expression becomes $m^3-\binom{4}{2}\cdot m\cdot m = m^3 - 6m^2$ .

However, we have subtracted the cases in which there are three identical numbers too many times. Each such case was subtracted $\binom{3}{2}=3$ times, and there are $\binom{4}{3}\cdot m=4m$ such cases. Thus, we need to add back $8m$ .

We also subtracted the cases in which there are two pairs of two identical numbers too many times. There are $\frac{\binom{4}{2}m}{2}=3m$ such cases. This is because we first choose the two positions for the first two identical numbers, then multiply by $m$ to determine which number goes in these positions. The last two positions can be determined uniquely because $m$ is odd. We divide by $2$ because we counted each case twice. In our original counting, we subtracted each such case twice, so we add back $3m$ . In total, our expression is now $m^3-6m^2+8m+3m=m^3-6m^2+11m$ .

We can stop here, because we know that the number of times $(5,5,5,5)$ was overcounted is just a constant. Thus, the answer is $\boxed{(E) 11}$ .

~ndv1tt

Video Solution

https://youtu.be/YExRIOt819Y

~MathProblemSolvingSkills.com

@@ Line 24: / Line 24: @@
 (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>
+We conclude that <math>D(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>
 By Vieta's Formulas, we get <math>c_1=1\cdot2+1\cdot3+2\cdot3=\boxed{\textbf{(E)}\ 11}.</math>
@@ Line 280: / Line 280: @@
 ~ brandonyee
+== Solution 5 (Principle of Inclusion-Exclusion) ==
+If we do not have to worry about whether or not the numbers are distinct, then there would be <math>m^3</math> possible quadruples. This is because the first <math>3</math> numbers can be anything, and then the last number is uniquely chosen so that the sum is a multiple of <math>m</math>.
+Now, we must subtract the quadruples with at least two of the same number. There are <math>\binom{4}{2}</math> ways to choose the positions for the two identical numbers, <math>m</math> ways to choose the number for these two positions, and <math>m</math> ways to choose a number for another position. The last position is uniquely determined as before. Thus, our expression becomes <math>m^3-\binom{4}{2}\cdot m\cdot m = m^3 - 6m^2</math>.
+However, we have subtracted the cases in which there are three identical numbers too many times. Each such case was subtracted <math>\binom{3}{2}=3</math> times, and there are <math>\binom{4}{3}\cdot m=4m</math> such cases. Thus, we need to add back <math>8m</math>.
+We also subtracted the cases in which there are two pairs of two identical numbers too many times. There are <math>\frac{\binom{4}{2}m}{2}=3m</math> such cases. This is because we first choose the two positions for the first two identical numbers, then multiply by <math>m</math> to determine which number goes in these positions. The last two positions can be determined uniquely because <math>m</math> is odd. We divide by <math>2</math> because we counted each case twice. In our original counting, we subtracted each such case twice, so we add back <math>3m</math>. In total, our expression is now <math>m^3-6m^2+8m+3m=m^3-6m^2+11m</math>.
+We can stop here, because we know that the number of times <math>(5,5,5,5)</math> was overcounted is just a constant. Thus, the answer is <math>\boxed{(E) 11}</math>.
+~ndv1tt
 ==Video Solution==

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

Latest revision as of 21:37, 16 September 2025

Contents

Problem

Solution 1 (Complete Residue System)

Solution 2 (Symmetric Congruent Numbers and Interpolation)

Solution 3

Solution 4

Solution 5 (Principle of Inclusion-Exclusion)

Video Solution

See Also