Difference between revisions of "1989 AIME Problems/Problem 8"
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<center><math>4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7=12^{}_{}</math></center> | <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^{}_{}</math></center> | <center><math>9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7=123^{}_{}</math></center> | ||
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Find the value of <math>16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7^{}</math>. | Find the value of <math>16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7^{}</math>. | ||
== Solution == | == Solution == | ||
− | { | + | Let us try to derive a way to find the last expression in terms of the three given expressions. The coefficients of <math>x_i</math> in the first equation can be denoted as <math>y^2</math>, making its coefficients in the second equation as <math>(y+1)^{2}</math> and the third as <math>(y+2)^2</math>. We need to find a way to sum them up to make <math>(y+3)^2</math>. |
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+ | Thus, we can write that <math>ay^2 + b(y+1)^2 + c(y+2)^2 = (y + 3)^2</math>. [[FOIL]]ing out all of the terms, we get <math>ay^2 + by^2 + cy^2 + 2by + 4cy + b + 4c = y^2 + 6y + 9</math>. We can set up the three equation system: | ||
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+ | <center> | ||
+ | <math>a + b + c = 1</math> | ||
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+ | <math>2b + 4c = 6</math> | ||
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+ | <math>b + 4c = 9</math> | ||
+ | </center> | ||
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+ | Subtracting the second and third equations yields that <math>e = -3</math>, so <math>f = 3</math> and <math>d = 1</math>. Thus, we have to add <math>d \cdot 1 + e \cdot 12 + f \cdot 123 = 1 - 36 + 369 = 334</math>. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1989|num-b=7|num-a=9}} | |
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Revision as of 21:13, 26 February 2007
Problem
Assume that are real numbers such that
Find the value of .
Solution
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 .
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 |