Difference between revisions of "1989 AIME Problems/Problem 8"
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− | Notice subtracting 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 subtracting 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> |
===Video Solution=== | ===Video Solution=== |
Revision as of 22:28, 29 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
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 subtracting 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.