Difference between revisions of "1989 AIME Problems/Problem 8"
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=100, &(8) \ | 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=100, &(8) \ | ||
− | 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=S-234 \hspace{ | + | 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=S-234. \hspace{20mm}&(9) |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Finally, applying the Transitive Property to <math>(8)</math> and <math>(9)</math> gives <math>S-234=100,</math> from which <math>S=\boxed{334}.</math> | Finally, applying the Transitive Property to <math>(8)</math> and <math>(9)</math> gives <math>S-234=100,</math> from which <math>S=\boxed{334}.</math> |
Revision as of 10:00, 24 June 2021
Contents
[hide]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
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
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 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.