Difference between revisions of "1989 AIME Problems/Problem 8"
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===Solution 4=== | ===Solution 4=== | ||
Notice subracting the first equation from the second yields <math>3x_1 + 5x_2 + ... + 15x_7 = 11</math> Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get <math>2x_1 + 2x_2 + ... +2x_7 = 100</math>. Adding this twice to the first obtained equation gives difference of the desired equation and 3rd equation, which is 211. Adding to the 3rd equation, we get <math>\boxed{334}</math> | Notice subracting the first equation from the second yields <math>3x_1 + 5x_2 + ... + 15x_7 = 11</math> Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get <math>2x_1 + 2x_2 + ... +2x_7 = 100</math>. Adding this twice to the first obtained equation gives difference of the desired equation and 3rd equation, which is 211. Adding to the 3rd equation, we get <math>\boxed{334}</math> | ||
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===Video Solution=== | ===Video Solution=== | ||
https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx | https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx | ||
+ | == See also == | ||
+ | {{AIME box|year=1989|num-b=7|num-a=9}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:07, 28 June 2020
Problem
Assume that are real numbers such that
Find the value of .
Contents
Solution
Solution 1
Notice that because we are given a system of equations with
unknowns, the values
are not fixed; indeed one can take any four of the variables and assign them arbitrary values, which will in turn fix the last three.
Given this, we suspect there is a way to derive the last expression as a linear combination of the three given expressions. Let the coefficent of in the first equation be
; then its coefficients in the second equation is
and the third as
. We need to find a way to sum these to make
[this is in fact a specific approach generalized by the next solution below].
Thus, we hope to find constants satisfying
. FOILing out all of the terms, we get
![$[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.$](http://latex.artofproblemsolving.com/d/e/0/de0a9256131a0d09b604f6d5b019ae307eff232d.png)
Comparing coefficents gives us the three equation system:
Subtracting the second and third equations yields that , so
and
. It follows that the desired expression is
.
Solution 2
Notice that we may rewrite the equations in the more compact form as:
and
where and
is what we're 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 3
Notice that
I'll number the equations for convenience
Let the coefficient of in
be
. Then the coefficient of
in
is
etc.
Therefore,
So
Solution 4
Notice subracting the first equation from the second yields Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get
. Adding this twice to the first obtained equation gives difference of the desired equation and 3rd equation, which is 211. Adding to the 3rd equation, we get
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.