Difference between revisions of "1989 AIME Problems/Problem 8"
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===Solution 4=== | ===Solution 4=== | ||
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> | ||
+ | |||
+ | ==Solution 5(Very Cheap)(Not advised)== | ||
+ | We let <math>(x_4,x_5,x_6,x_7)=(0,0,0,0)</math>. Thus, we have <cmath>\begin{align} x_1+4x_2+9x_3&=1\\ | ||
+ | 4x_1+9x_2+16x_3&=12\\ | ||
+ | 9x_1+16x_2+25x_3=123\\ \end{align}</cmath> | ||
+ | |||
+ | Grinding this out, we have <math>(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)</math> which gives <math>\boxed{334}</math> as our final answer. | ||
+ | -Pleaseletmewin | ||
===Video Solution=== | ===Video Solution=== |
Revision as of 20:00, 8 August 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
Solution 5(Very Cheap)(Not advised)
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.