Difference between revisions of "1989 AIME Problems/Problem 8"
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~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution 3 == | + | == Solution 3 (Generalized) == |
Notice that we may rewrite the equations in the more compact form as: | Notice that we may rewrite the equations in the more compact form as: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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Alternatively, applying finite differences, one obtains <cmath>c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =\boxed{334}.</cmath> | Alternatively, applying finite differences, one obtains <cmath>c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =\boxed{334}.</cmath> | ||
− | ==Solution 4 | + | ==Solution 4 (Very Cheap: Not Recommended)== |
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We let <math>(x_4,x_5,x_6,x_7)=(0,0,0,0)</math>. Thus, we have | We let <math>(x_4,x_5,x_6,x_7)=(0,0,0,0)</math>. Thus, we have | ||
Revision as of 11:45, 24 June 2021
Contents
Problem
Assume that are real numbers such that
Find the value of
.
Solution 1 (Quadratic Function)
Note that each equation is of the form for some
When we expand and combine like terms, we obtain a quadratic function of
where
and
are linear combinations of
and
We are given that
and we wish to find
We eliminate by subtracting the first equation from the second, then subtracting the second equation from the third:
By either substitution or elimination, we get
and
Substituting these back produces
Finally, the answer is
~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2 (Linear Combination)
For simplicity purposes, we number the given equations and
in that order. Let
Subtracting
from
subtracting
from
and subtracting
from
we obtain the following equations, respectively:
Subtracting
from
and subtracting
from
we obtain the following equations, respectively:
Finally, applying the Transitive Property to
and
gives
from which
~Duohead (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 3 (Generalized)
Notice that we may rewrite the equations in the more compact form as:
where
and
is what we are trying to find.
Now consider the polynomial given by (we are only treating the
as coefficients).
Notice that is in fact a quadratic. We are given
as
and are asked to find
. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find
.
Alternatively, applying finite differences, one obtains
Solution 4 (Very Cheap: Not Recommended)
We let . Thus, we have
Grinding this out, we have which gives
as our final answer.
-Pleaseletmewin
Video Solution
https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.