Difference between revisions of "2018 AMC 12B Problems/Problem 22"
MRENTHUSIASM (talk | contribs) (Not done yet. Will return to Sol 2.) |
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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== Solution 2 (Casework) == | == 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> | 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> . | + | 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}{ | + | <cmath>\begin{array}{c|c|c|c||c} |
− | + | & & & & \\ [-2.5ex] | |
− | \textbf{ | + | \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 | \hline | ||
− | + | & & & & \\ [-2ex] | |
− | 1 & 1 | + | 0 & 1 & 9 & 10 & 1\cdot10=10 \\ |
− | 2 & | + | 1 & 2 & 10 & 9 & \phantom{0}2\cdot9=18 \\ |
− | 3 & | + | 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> | \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 | ~BJHHar ~MRENTHUSIASM | ||
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==See Also== | ==See Also== |
Latest revision as of 02:20, 28 October 2021
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 |
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