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Difference between revisions of "2018 AMC 12B Problems/Problem 22"

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<math>\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286 </math>
 
<math>\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286 </math>
  
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== Solution ==
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== Solution 1 (Stars and Bars) ==
 +
Suppose that <math>P(x)=ax^3+bx^2+cx+d.</math> This problem is equivalent to counting the ordered quadruples <math>(a,b,c,d),</math> where all of <math>a,b,c,</math> and <math>d</math> are integers from <math>0</math> through <math>9</math> such that <cmath>P(-1)=-a+b-c+d=-9.</cmath> Let <math>a'=9-a</math> and <math>c'=9-c.</math> Note that both of <math>a'</math> and <math>c'</math> are integers from <math>0</math> through <math>9.</math> Moreover, the ordered quadruples <math>(a,b,c,d)</math> and the ordered quadruples <math>(a',b,c',d)</math> have one-to-one correspondence.
  
−
Suppose our polynomial is equal to
+
We rewrite the given equation as <math>(9-a)+b+(9-c)+d=9,</math> or <cmath>a'+b+c'+d=9.</cmath> By the stars and bars argument, there are <math>\binom{9+4-1}{4-1}=\boxed{\textbf{(D) } 220}</math> ordered quadruples <math>(a',b,c',d).</math>
−
<cmath>ax^3+bx^2+cx+d</cmath>Then we are given that
 
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<cmath>-9=b+d-a-c.</cmath>If we let <math>-a=a'-9, -c=c'-9</math> then we have
 
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<cmath>9=a'+c'+b+d.</cmath> This way all four variables are within 0 and 9. The number of solutions to this equation is simply <math>\binom{12}{3}=220</math> by stars and bars, so our answer is <math>\boxed{\textbf{D}.}</math>
 
  
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== Solution 2 ==
+
~pieater314159 ~MRENTHUSIASM
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Suppose our polynomial is equal to
+
 
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<cmath>ax^3+bx^2+cx+d</cmath>Then we are given that
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== Solution 2 (Casework) ==
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<cmath>9=b+d-a-c.</cmath>Then the polynomials <cmath>cx^3+bx^2+ax+d</cmath>, <cmath>ax^3+dx^2+cx+b,</cmath> <cmath>cx^3+dx^2+ax+b</cmath>also have <cmath>b+d-a-c=-9</cmath> when <cmath>x=-1.</cmath> So the number of solutions must be divisible by 4. So the answer must be <math>\boxed{\textbf{D}.}</math>
+
Suppose that <math>P(x)=ax^3+bx^2+cx+d.</math> This problem is equivalent to counting the ordered quadruples <math>(a,b,c,d),</math> where all of <math>a,b,c,</math> and <math>d</math> are integers from <math>0</math> through <math>9</math> such that <math>P(-1)=-a+b-c+d=-9,</math> which rearranges to <cmath>b+d+9=a+c.</cmath>
 +
Note that <math>b+d+9</math> is an integer from <math>9</math> through <math>27,</math> and <math>a+c</math> is an integer from <math>0</math> through <math>18.</math> Therefore, both of <math>b+d+9</math> and <math>a+c</math> are integers from <math>9</math> through <math>18.</math> We construct the following table:
 +
<cmath>\begin{array}{c|c|c|c||c}
 +
& & & & \\ [-2.5ex]
 +
\boldsymbol{b+d} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(b,d)} & \boldsymbol{a+c} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(a,c)} & \boldsymbol{\#}\textbf{ of Ordered Quadruples }\boldsymbol{(a,b,c,d)} \\ [0.5ex]
 +
\hline
 +
& & & & \\ [-2ex]
 +
0 & 1 & 9 & 10 & 1\cdot10=10 \\
 +
1 & 2 & 10 & 9 & \phantom{0}2\cdot9=18 \\
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2 & 3 & 11 & 8 & \phantom{0}3\cdot8=24 \\
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3 & 4 & 12 & 7 & \phantom{0}4\cdot7=28 \\
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4 & 5 & 13 & 6 & \phantom{0}5\cdot6=30 \\
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5 & 6 & 14 & 5 & \phantom{0}6\cdot5=30 \\
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6 & 7 & 15 & 4 & \phantom{0}7\cdot4=28 \\
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7 & 8 & 16 & 3 & \phantom{0}8\cdot3=24 \\
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8 & 9 & 17 & 2 & \phantom{0}9\cdot2=18 \\
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9 & 10 & 18 & 1 & 10\cdot1=10
 +
\end{array}</cmath>
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We sum up the counts in the last column to get the answer <math>2\cdot(10+18+24+28+30)=\boxed{\textbf{(D) } 220}.</math>
 +
 
 +
~BJHHar ~MRENTHUSIASM
  
 
==See Also==
 
==See Also==

Latest revision as of 19:29, 14 November 2022

Problem

Consider polynomials $P(x)$ of degree at most $3$, each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. How many such polynomials satisfy $P(-1) = -9$?

$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286$

Solution 1 (Stars and Bars)

Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that \[P(-1)=-a+b-c+d=-9.\] Let $a'=9-a$ and $c'=9-c.$ Note that both of $a'$ and $c'$ are integers from $0$ through $9.$ Moreover, the ordered quadruples $(a,b,c,d)$ and the ordered quadruples $(a',b,c',d)$ have one-to-one correspondence.

We rewrite the given equation as $(9-a)+b+(9-c)+d=9,$ or \[a'+b+c'+d=9.\] By the stars and bars argument, there are $\binom{9+4-1}{4-1}=\boxed{\textbf{(D) } 220}$ ordered quadruples $(a',b,c',d).$

~pieater314159 ~MRENTHUSIASM

Solution 2 (Casework)

Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that $P(-1)=-a+b-c+d=-9,$ which rearranges to \[b+d+9=a+c.\] Note that $b+d+9$ is an integer from $9$ through $27,$ and $a+c$ is an integer from $0$ through $18.$ Therefore, both of $b+d+9$ and $a+c$ are integers from $9$ through $18.$ We construct the following table: \[\begin{array}{c|c|c|c||c} & & & & \\ [-2.5ex] \boldsymbol{b+d} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(b,d)} & \boldsymbol{a+c} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(a,c)} & \boldsymbol{\#}\textbf{ of Ordered Quadruples }\boldsymbol{(a,b,c,d)} \\ [0.5ex] \hline & & & & \\ [-2ex] 0 & 1 & 9 & 10 & 1\cdot10=10 \\ 1 & 2 & 10 & 9 & \phantom{0}2\cdot9=18 \\ 2 & 3 & 11 & 8 & \phantom{0}3\cdot8=24 \\ 3 & 4 & 12 & 7 & \phantom{0}4\cdot7=28 \\ 4 & 5 & 13 & 6 & \phantom{0}5\cdot6=30 \\ 5 & 6 & 14 & 5 & \phantom{0}6\cdot5=30 \\ 6 & 7 & 15 & 4 & \phantom{0}7\cdot4=28 \\ 7 & 8 & 16 & 3 & \phantom{0}8\cdot3=24 \\ 8 & 9 & 17 & 2 & \phantom{0}9\cdot2=18 \\ 9 & 10 & 18 & 1 & 10\cdot1=10 \end{array}\] We sum up the counts in the last column to get the answer $2\cdot(10+18+24+28+30)=\boxed{\textbf{(D) } 220}.$

~BJHHar ~MRENTHUSIASM

See Also

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AMC 12 Problems and Solutions

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