Difference between revisions of "1989 AIME Problems/Problem 8"
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From <math>(5)</math> and <math>(6),</math> we have <math>(a,b,c)=(7c-20,9-4c,c).</math> Substituting this into <math>(2)</math> gives <math>(a,b,c)=(1,-3,3).</math> | From <math>(5)</math> and <math>(6),</math> we have <math>(a,b,c)=(7c-20,9-4c,c).</math> Substituting this into <math>(2)</math> gives <math>(a,b,c)=(1,-3,3).</math> | ||
− | Therefore, the answer is <math>1+12\cdot(-3) + 123 | + | Therefore, the answer is <math>1\cdot1+12\cdot(-3) + 123\cdot3 = \boxed{334}.</math> |
==Solution 6 (Assumption)== | ==Solution 6 (Assumption)== |
Revision as of 01:33, 30 June 2022
Contents
Problem
Assume that are real numbers such that Find the value of .
Solution 1 (Quadratic Function)
Note that each given 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 ~MRENTHUSIASM
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 ~MRENTHUSIASM
Solution 3 (Finite Differences by Arithmetic)
Note that the second differences of all quadratic sequences must be constant (but nonzero). One example is the following sequence of perfect squares:
Label equations and as Solution 2 does. Since the coefficients of or respectively, all form quadratic sequences with second differences we conclude that the second differences of equations must be constant.
It follows that the second differences of must be constant, as shown below:
Finally, we have from which ~MRENTHUSIASM
Solution 4 (Finite Differences by Algebra)
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 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 5 (Assumption)
The idea is to multiply the first, second and third equations by and respectively.
We can only consider the coefficients of and Subtracting from we get Subtracting from we get Subtracting from we get From and we have Substituting this into gives
Therefore, the answer is
Solution 6 (Assumption)
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.