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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 25"
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==Solution 2 (Educated Guess)==
 
==Solution 2 (Educated Guess)==
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Note that you see numbers with absolute value <math>1,6,</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>  
+
Note that you see numbers with absolute value <math>1,6,</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 memorized 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{\textbf{(E)}\ 11}.</math>  
 +
 
 +
Furthermore, you can gain confidence in your guess since that is the only answer choice with absolute value <math>11.</math>  
  
 
~fidgetboss_4000
 
~fidgetboss_4000

Revision as of 12:21, 24 November 2021

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 satisfy that $m$ divides $S:$ \[\begin{array}{c|c} & \\ [-2.5ex] \textbf{Ordered Quadruple} & \textbf{Sum Modulo }\boldsymbol{m} \\ [0.5ex] \hline & \\ [-2ex] (a'_1, a'_2, a'_3, a'_4) & S+4(0) \\ [0.5ex] (a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S+4(1) \\ [0.5ex] (a'_1+2, a'_2+2, a'_3+2, a'_4+2) & S+4(2) \\ [0.5ex] \cdots & \cdots \\ [0.5ex] (a'_1+m-1, a'_2+m-1, a'_3+m-1, a'_4+m-1) & S+4(m-1) \\ [0.5ex] \end{array}\] 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 (Educated Guess)

Note that you see numbers with absolute value $1,6,$ and $11$ in the answer choices. What is special about those numbers? Well, you should notice that they are the coefficients of the polynomial $(x+1)(x+2)(x+3)$ when expanded (if you've already memorized this). Then, you can probably guess the polynomial is some form of $(x+1)(x+2)(x+3)$ whether negative or positive. Since $c_1$ is asked, the answer should be reasoned out as $1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3 = \boxed{\textbf{(E)}\ 11}.$

Furthermore, you can gain confidence in your guess since that is the only answer choice with absolute value $11.$

~fidgetboss_4000

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

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