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

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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>If we let <math>b=-9-b', d=-9-d'</math> then we have
+
<cmath>-9=b+d-a-c.</cmath>If we let <math>-a=a'-9, -c=c'-9</math> then we have
−
<cmath>-9=a+c+b'+d'.</cmath>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>
+
<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>
  
 
== Solution 2 ==
 
== Solution 2 ==
 
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>
 +
 
 +
 
 +
== Solution 3 ==
 +
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:
 +
<cmath>{0,1,2,3,4,5,6,7,8,9}</cmath>
 +
<cmath>{9,10,11,12,13,14,15,16,17,18}</cmath>
 +
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).
 +
 
 +
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>.
 +
 
 +
~BJHHar
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2018|ab=B|num-b=21|num-a=23}}
 
{{AMC12 box|year=2018|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
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[[Category:Intermediate Combinatorics Problems]]
 +
[[Category:Intermediate Algebra Problems]]

Revision as of 00:42, 2 August 2020

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

Suppose our polynomial is equal to \[ax^3+bx^2+cx+d\]Then we are given that \[-9=b+d-a-c.\]If we let $-a=a'-9, -c=c'-9$ then we have \[9=a'+c'+b+d.\] This way all four variables are within 0 and 9. The number of solutions to this equation is simply $\binom{12}{3}=220$ by stars and bars, so our answer is $\boxed{\textbf{D}.}$

Solution 2

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 $\boxed{\textbf{D}.}$


Solution 3

As before, $-9=-A+B-C+D$. This is $9=(A+B)-(C+D)$. Rephrased, how many two sums of integers from 0-9 have a difference of 9. Make a chart of pairs between these two sets: \[{0,1,2,3,4,5,6,7,8,9}\] \[{9,10,11,12,13,14,15,16,17,18}\] 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).

The answer then is the number of ways to write each component of each pair. This is $(1*10)+(2*9)+(3*8) ...$ or, since it's symmetrical between sum of 4 and 5, $[2\sum\limits_{i=1}^{5} i(11-i)]$. Use summation rules to finally get $\boxed{\textbf{D)}220}$.

~BJHHar

See Also

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