Difference between revisions of "2018 AMC 12B Problems/Problem 22"
(→Solution) |
MRENTHUSIASM (talk | contribs) |
||
(15 intermediate revisions by 9 users not shown) | |||
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 == | + | == 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. | ||
− | + | 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> | ||
− | == Solution 2 == | + | ~pieater314159 ~MRENTHUSIASM |
− | Suppose | + | |
− | < | + | == 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> 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 \\ | ||
+ | 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}</cmath> | ||
+ | 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== | ||
Line 21: | Line 39: | ||
{{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}} | ||
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 20:29, 14 November 2022
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
and
is an integer from
through
Therefore, both of
and
are integers from
through
We construct the following table:
We sum up the counts in the last column to get the answer
~BJHHar ~MRENTHUSIASM
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.