Difference between revisions of "2005 AMC 12A Problems/Problem 24"
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== Problem == | == Problem == | ||
− | Let <math>P(x)=(x-1)(x-2)(x-3)</math>. For how many [[polynomial]]s <math>Q(x)</math> does there exist a polynomial <math>R(x)</math> of degree 3 such that <math>P(Q(x))=P(x) | + | Let <math>P(x)=(x-1)(x-2)(x-3)</math>. For how many [[polynomial]]s <math>Q(x)</math> does there exist a polynomial <math>R(x)</math> of degree 3 such that <math>P(Q(x))=P(x) \cdot R(x)</math>? |
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== Solution == | == Solution == | ||
− | + | We can write the problem as | |
+ | <div style="text-align:center;"> | ||
+ | <math>P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)\cdot R(x)</math>. | ||
+ | </div> | ||
+ | |||
+ | |||
+ | Since <math>\deg P(x) = 3</math> and <math>\deg R(x) = 3</math>, <math>\deg P(x)\cdot R(x) = 6</math>. Thus, <math>\deg P(Q(x)) = 6</math>, so <math>\deg Q(x) = 2</math>. | ||
<div style="text-align:center;"> | <div style="text-align:center;"> | ||
<math> | <math> | ||
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− | However, we have included <math>Q(x)</math> which are not quadratics. Namely, | + | However, we have included <math>Q(x)</math> which are not quadratics: lines. Namely, |
<div style="text-align:center;"> | <div style="text-align:center;"> | ||
<math> | <math> | ||
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</math> | </math> | ||
</div> | </div> | ||
− | Clearly, we could not have included any other constant functions. For any linear function, we have <math>2\cdot Q(2) = Q(1) + Q(3)</math>. | + | Clearly, we could not have included any other constant functions. For any linear function, we have <math>2\cdot Q(2) = Q(1) + Q(3)</math> because <math>Q(2)</math> is y-value of the midpoint of <math>(1, Q(1))</math> and <math>(3, Q(3))</math>. So we have not included any other linear functions. Therefore, the desired answer is <math>27 - 5 = \boxed{\textbf{(B) }22}</math>. |
+ | |||
+ | ==Quicker Solution== | ||
+ | We see that | ||
+ | <cmath>P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)\cdot R(x).</cmath> | ||
+ | Therefore, <math>P(x) | P(Q(x))</math>. Since <math>\deg Q = 2,</math> we must have <math>x-1, x-2, x-3</math> divide <math>P(Q(x))</math>. So, we pair them off with one of <math>Q(x)-1, Q(x)-2,</math> and <math>Q(x)-3</math> to see that there are <math>3!+3 \cdot 2 \cdot \binom{3}{2} = 24</math> without restrictions. (Note that this count was made by pairing off linear factors of <math>P(x)</math> with <math>Q(x)-1, Q(x)-2,</math> and <math>Q(x)-3</math>, and also note that the degree of <math>Q</math> is 2.) However, we have two functions which are constant, which are <math>Q(x) = x</math> and <math>Q(x) = 4-x.</math> So, we subtract <math>2</math> to get a final answer of <math>\boxed{22} \implies \boxed{B}</math>. | ||
+ | |||
+ | ~Williamgolly | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 02:16, 3 March 2021
Contents
Problem
Let . For how many polynomials does there exist a polynomial of degree 3 such that ?
Solution
We can write the problem as
.
Since and , . Thus, , so .
Hence, we conclude , , and must each be , , or . Since a quadratic is uniquely determined by three points, there can be different quadratics after each of the values of , , and are chosen.
However, we have included which are not quadratics: lines. Namely,
Clearly, we could not have included any other constant functions. For any linear function, we have because is y-value of the midpoint of and . So we have not included any other linear functions. Therefore, the desired answer is .
Quicker Solution
We see that Therefore, . Since we must have divide . So, we pair them off with one of and to see that there are without restrictions. (Note that this count was made by pairing off linear factors of with and , and also note that the degree of is 2.) However, we have two functions which are constant, which are and So, we subtract to get a final answer of .
~Williamgolly
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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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