Difference between revisions of "2018 AMC 12B Problems/Problem 22"
MRENTHUSIASM (talk | contribs) m (→Solution 1) |
MRENTHUSIASM (talk | contribs) (Not done yet. Will return to Sol 2.) |
||
Line 5: | Line 5: | ||
<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> | ||
− | == Solution 1 == | + | == 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. | ||
Line 12: | Line 12: | ||
~pieater314159 ~MRENTHUSIASM | ~pieater314159 ~MRENTHUSIASM | ||
− | == Solution 2 == | + | == 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 \\ | ||
+ | 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}</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> | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
==See Also== | ==See Also== |
Revision as of 16:02, 26 October 2021
Contents
Problem
Consider polynomials of degree at most , each of whose coefficients is an element of . How many such polynomials satisfy ?
Solution 1 (Stars and Bars)
Suppose that This problem is equivalent to counting the ordered quadruples where all of and are integers from through such that Let and Note that both of and are integers from through Moreover, the ordered quadruples and the ordered quadruples have one-to-one correspondence.
We rewrite the given equation as or By the stars and bars argument, there are ordered quadruples
~pieater314159 ~MRENTHUSIASM
Solution 2 (Casework)
Suppose that This problem is equivalent to counting the ordered quadruples where all of and are integers from through such that which rearranges to Note that is an integer from through ........................ So, ......................... We construct the following table:
~BJHHar ~MRENTHUSIASM
Solution 3 (Answer Choices)
Suppose our polynomial is equal to Then we are given that Then the polynomials , also have when So the number of solutions must be divisible by 4. So the answer must be
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.