Difference between revisions of "1989 AIME Problems/Problem 8"
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I'll number the equations for convenience | I'll number the equations for convenience | ||
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<math>(1) x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&=1</math> | <math>(1) x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&=1</math> | ||
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<math>(2) 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&=12</math> | <math>(2) 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&=12</math> | ||
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<math>(3) 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&=123</math> | <math>(3) 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&=123</math> | ||
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<math>(4) 16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7</math> | <math>(4) 16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7</math> | ||
Revision as of 07:40, 5 August 2011
Problem
Assume that are real numbers such that
Find the value of .
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
$(1) x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&=1$ (Error compiling LaTeX. Unknown error_msg)
$(2) 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&=12$ (Error compiling LaTeX. Unknown error_msg)
$(3) 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&=123$ (Error compiling LaTeX. Unknown error_msg)
Let the coefficient of in (1) be n. Then the coeffienct of
in (2) is n+1 etc.
Therefore,
So
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 |