Difference between revisions of "2018 AMC 12B Problems/Problem 22"
MRENTHUSIASM (talk | contribs) (→Solution 3 (Answer Choices): Hmm, this solution is not very convincing that the answer needs to be divisible by 4. What if a=c and b!=d? If we swap (a,c) and (b,d), we only get two solutions.) |
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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> | 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> | ||
| − | ~pieater314159 | + | ~pieater314159 |
== Solution 2 (Casework) == | == Solution 2 (Casework) == | ||
Revision as of 19:18, 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 ![\[P(-1)=-a+b-c+d=-9.\]](http://latex.artofproblemsolving.com/c/7/f/c7f49b5a6d16402c6df0bb8f3d4ef443afb6e6a7.png)
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 ![\[a'+b+c'+d=9.\]](http://latex.artofproblemsolving.com/e/8/0/e80a8be115b429c588370434b0a5c09bb6d6194c.png)
By the stars and bars argument, there are 
ordered quadruples 
~pieater314159
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 ![\[b+d+9=a+c.\]](http://latex.artofproblemsolving.com/f/c/5/fc5bc84eeac495c6d334862b85e8814da91fe252.png)
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:
![\[\begin{array}{c|c|c|c||c} & & & & \\ [-2.5ex] \boldsymbol{b+d} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(b,d)} & \boldsymbol{a+c} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(a,c)} & \boldsymbol{\#}\textbf{ of Ordered Quadruples }\boldsymbol{(a,b,c,d)} \\ [0.5ex] \hline & & & & \\ [-2ex] 0 & 1 & 9 & 10 & 1\cdot10=10 \\ 1 & 2 & 10 & 9 & \phantom{0}2\cdot9=18 \\ 2 & 3 & 11 & 8 & \phantom{0}3\cdot8=24 \\ 3 & 4 & 12 & 7 & \phantom{0}4\cdot7=28 \\ 4 & 5 & 13 & 6 & \phantom{0}5\cdot6=30 \\ 5 & 6 & 14 & 5 & \phantom{0}6\cdot5=30 \\ 6 & 7 & 15 & 4 & \phantom{0}7\cdot4=28 \\ 7 & 8 & 16 & 3 & \phantom{0}8\cdot3=24 \\ 8 & 9 & 17 & 2 & \phantom{0}9\cdot2=18 \\ 9 & 10 & 18 & 1 & 10\cdot1=10 \end{array}\]](http://latex.artofproblemsolving.com/f/a/d/fad98d5ef1f983b454a15b20860917a7638da9b8.png)
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. 
