Difference between revisions of "1989 AIME Problems/Problem 8"

Revision as of 21:13, 26 February 2007

Problem

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that

$x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7=1^{}_{}$ $4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7=12^{}_{}$ $9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7=123^{}_{}$

Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7^{}$ .

Solution

Let us try to derive a way to find the last expression in terms of the three given expressions. The coefficients of $x_i$ in the first equation can be denoted as $y^2$ , making its coefficients in the second equation as $(y+1)^{2}$ and the third as $(y+2)^2$ . We need to find a way to sum them up to make $(y+3)^2$ .

Thus, we can write that $ay^2 + b(y+1)^2 + c(y+2)^2 = (y + 3)^2$ . FOILing out all of the terms, we get $ay^2 + by^2 + cy^2 + 2by + 4cy + b + 4c = y^2 + 6y + 9$ . We can set up the three equation system:

$a + b + c = 1$

$2b + 4c = 6$

$b + 4c = 9$

Subtracting the second and third equations yields that $e = -3$ , so $f = 3$ and $d = 1$ . Thus, we have to add $d \cdot 1 + e \cdot 12 + f \cdot 123 = 1 - 36 + 369 = 334$ .

@@ Line 4: / Line 4: @@
   <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>
 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>.
+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:
+<center>
+<math>a + b + c = 1</math>
+<math>2b + 4c = 6</math>
+<math>b + 4c = 9</math>
+</center>
+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 ==
-* [[1989 AIME Problems/Problem 9|Next Problem]]
+{{AIME box|year=1989|num-b=7|num-a=9}}
-* [[1989 AIME Problems/Problem 7|Previous Problem]]
-* [[1989 AIME Problems]]

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Difference between revisions of "1989 AIME Problems/Problem 8"

Revision as of 21:13, 26 February 2007

Problem

Solution

See also