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

Revision as of 23:42, 1 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

@@ Line 18: / Line 18: @@
-== Solution 3 (BASH)==
+== 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>

2018 AMC 12B (Problems • Answer Key • Resources)
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