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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 1 ==
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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 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.
  
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~pieater314159 ~MRENTHUSIASM
 
~pieater314159 ~MRENTHUSIASM
  
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== Solution 2 ==
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== Solution 2 (Casework) ==
 +
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> ........................ So, ......................... We construct the following table:
 +
<cmath>\begin{array}{c||c|c|c|c|c||c}
 +
& & & & & & \\ [-2.5ex]
 +
\textbf{Row} & \boldsymbol{x_2} & \boldsymbol{x_3} & \boldsymbol{x_4} & \boldsymbol{x_5} & \boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \textbf{Valid?} \\ [0.5ex]
 +
\hline
 +
& & & & & & \\ [-2ex]
 +
1 & 1 & 1 & 2 & 2 & 0 & \checkmark \\
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2 & 1 & 2 & 1 & 2 & 0 & \checkmark \\
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3 & 1 & 2 & 2 & 1 & 2 & \\
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4 & 2 & 1 & 1 & 2 & 1 & \\
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5 & 2 & 1 & 2 & 1 & 0 & \checkmark \\
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6 & 2 & 2 & 1 & 1 & 0 & \checkmark
 +
\end{array}</cmath>
 +
 
 +
~BJHHar ~MRENTHUSIASM
 +
 
 +
== Solution 3 (Answer Choices) ==
 
Suppose our polynomial is equal to
 
Suppose our polynomial is equal to
 
<cmath>ax^3+bx^2+cx+d</cmath>Then we are given that
 
<cmath>ax^3+bx^2+cx+d</cmath>Then we are given that
 
<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>
 
<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>
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== Solution 3 ==
 
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As before, <math>-9=-A+B-C+D</math>. This is <math>9=(A+B)-(C+D)</math>. Rephrased, how many two sums of integers from 0-9 have a difference of 9. Make a chart of pairs between these two sets:
 
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<cmath>{0,1,2,3,4,5,6,7,8,9}</cmath>
 
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<cmath>{9,10,11,12,13,14,15,16,17,18}</cmath>
 
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Observe how there is one way to sum 2 numbers to 0 and two ways to 1, 3 ways to 2, and so on. At 9, there are 10 possible ways. Recall that only integers between 0-9 are valid. Now observe how there is 1 way to to sum to 18 in this fashion (9+9), 2 ways to sum to 17, and so forth again (to optionally prove that this pattern holds, apply stars and bars up to 9 and notice the symmetry).
 
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The answer then is the number of ways to write each component of each pair. This is <math>(1*10)+(2*9)+(3*8) ...</math> or, since it's symmetrical between sum of 4 and 5, <math>[2\sum\limits_{i=1}^{5} i(11-i)]</math>. Use summation rules to finally get <math>\boxed{\textbf{D)}220}</math>.
 
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~BJHHar
 
  
 
==See Also==
 
==See Also==

Revision as of 16:02, 26 October 2021

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 $27,$ ........................ So, ......................... We construct the following table: \[\begin{array}{c||c|c|c|c|c||c} & & & & & & \\ [-2.5ex] \textbf{Row} & \boldsymbol{x_2} & \boldsymbol{x_3} & \boldsymbol{x_4} & \boldsymbol{x_5} & \boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \textbf{Valid?} \\ [0.5ex] \hline & & & & & & \\ [-2ex] 1 & 1 & 1 & 2 & 2 & 0 & \checkmark \\ 2 & 1 & 2 & 1 & 2 & 0 & \checkmark \\ 3 & 1 & 2 & 2 & 1 & 2 & \\ 4 & 2 & 1 & 1 & 2 & 1 & \\ 5 & 2 & 1 & 2 & 1 & 0 & \checkmark \\ 6 & 2 & 2 & 1 & 1 & 0 & \checkmark \end{array}\]

~BJHHar ~MRENTHUSIASM

Solution 3 (Answer Choices)

Suppose our polynomial is equal to \[ax^3+bx^2+cx+d\]Then we are given that \[9=b+d-a-c.\]Then the polynomials \[cx^3+bx^2+ax+d\], \[ax^3+dx^2+cx+b,\] \[cx^3+dx^2+ax+b\]also have \[b+d-a-c=-9\] when \[x=-1.\] So the number of solutions must be divisible by 4. So the answer must be $\boxed{\textbf{D}.}$

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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