Difference between revisions of "2005 AMC 12A Problems/Problem 24"
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~Williamgolly | ~Williamgolly | ||
+ | ==Quickest Solution== | ||
+ | Obviously, <math>\deg Q = 2</math> and the RHS has roots <math>1, 2, 3.</math> Thus, the only requirement is that <math>Q</math> is a quadratic that maps each of the numbers <math>1, 2, 3</math> to one of <math>{1, 2, 3}.</math> It suffices to count the number of such mappings, as this uniquely determines <math>Q</math>, through polynomial interpolation. Thus there are <math>3^3</math> possibilities. But we have three functions that are constant (<math>Q(1) =Q(2) = Q(3)</math>), and <math>2</math> that are linear (<math>\{1, 2, 3\} \rightarrow \{3, 2, 1\}</math> and <math>\{1, 2, 3\} \rightarrow \{1, 2, 3\}</math>). Hence there are <math>3^3-3-2 = \boxed{22}</math> solutions, so <math>\boxed{B}</math>. | ||
+ | ~Maximilian113 | ||
− | ==Guessing Solution( | + | ==Guessing Solution(desperate),== |
<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> | <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> |
Latest revision as of 18:17, 8 April 2024
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
Quickest Solution
Obviously, and the RHS has roots
Thus, the only requirement is that
is a quadratic that maps each of the numbers
to one of
It suffices to count the number of such mappings, as this uniquely determines
, through polynomial interpolation. Thus there are
possibilities. But we have three functions that are constant (
), and
that are linear (
and
). Hence there are
solutions, so
.
~Maximilian113
Guessing Solution(desperate),
rewrite it as
say Q(x)= 2nd degree polymonial
that means (Q(x)-1) must equal to 2 factors of (R(x) times P(x))
we have 6 factors
We need 2 factors,so it must be 6 choices, choose 2 or
6!/4!=30
none of choices are 30, so lets use the answers
it cannot be E because it is above 30. Now we look for answers that are similar
so we see 22,27,32 which shows that we added or subtracted 5 from 27 in the actual solution. since it can't be greater than 30, the answer should be 22 or 27. Which means choose B or D if you are desperate.
only use if you are desperate
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.