Difference between revisions of "1989 AIME Problems/Problem 8"
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== Problem == | == Problem == | ||
Assume that <math>x_1,x_2,\ldots,x_7</math> are real numbers such that | Assume that <math>x_1,x_2,\ldots,x_7</math> are real numbers such that | ||
− | <center><math>x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7=1 | + | <center><math>x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7=1</math></center> |
− | <center><math>4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7=12 | + | <center><math>4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7=12</math></center> |
− | <center><math>9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7=123 | + | <center><math>9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7=123</math></center> |
− | Find the value of <math>16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7 | + | Find the value of <math>16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7</math>. |
__TOC__ | __TOC__ |
Revision as of 21:09, 17 July 2008
Problem
Assume that are real numbers such that
Find the value of .
Solution
Solution 1
Let us try to derive a way to find the last expression in terms of the three given expressions. The coefficients of in the first equation can be denoted as , making its coefficients in the second equation as and the third as . We need to find a way to sum them up to make .
Thus, we can write that . FOILing out all of the terms, we get . We can set up the three equation system:
Subtracting the second and third equations yields that , so and . Thus, we have to add .
Solution 2
Notice that we may rewrite the equations in the more compact form:
and
, where and and is what we're trying to find.
Now undergo a paradigm shift: consider the polynomial in (we are only treating the as coefficients). Notice that the degree of must be ; it is 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 .
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 |