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

Revision as of 18:16, 4 July 2013

Problem

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that $\begin{align*}x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&=1\\ 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&=12\\ 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&=123\end{align*}.$

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

Solution

Solution 1

Notice that because we are given a system of $3$ equations with $7$ unknowns, the values $(x_1, x_2, \ldots, x_7)$ 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 $x_i$ in the first equation be $y_i^2$ ; then its coefficients in the second equation is $(y_i+1)^{2}$ and the third as $(y_i+2)^2$ . We need to find a way to sum these to make $(y_i+3)^2$ [this is in fact a specific approach generalized by the next solution below].

Thus, we hope to find constants $a,b,c$ satisfying $ay_i^2 + b(y_i+1)^2 + c(y_i+2)^2 = (y_i + 3)^2$ . FOILing out all of the terms, we get

$[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.$

Comparing coefficents gives us the three equation system:

$\begin{align*}a + b + c &= 1 \\ 2b + 4c &= 6 \\ b + 4c &= 9 \end{align*}$

Subtracting the second and third equations yields that $b = -3$ , so $c = 3$ and $a = 1$ . It follows that the desired expression is $a \cdot (1) + b \cdot (12) + c \cdot (123) = 1 - 36 + 369 = \boxed{334}$ .

Solution 2

Notice that we may rewrite the equations in the more compact form as:

$\sum_{i=1}^{7}i^2x_i=c_1,\ \ \sum_{i=1}^{7}(i+1)^2x_i=c_2,\ \ \sum_{i=1}^{7}(i+2)^2x_i=c_3,$ and $\sum_{i=1}^{7}(i+3)^2x_i=c_4,$

where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we're trying to find.

Now consider the polynomial given by $f(z) := \sum_{i=1}^7 (z+i)^2x_i$ (we are only treating the $x_i$ as coefficients).

Notice that $f$ is in fact a quadratic. We are given $f(0), \ f(1), \ f(2)$ as $c_1, \ c_2, \ c_3$ and are asked to find $c_4$ . 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 $c_4=334$ .

Alternatively, applying finite differences, one obtains $c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334$ .

Solution 3

Notice that $3(n+2)^2-3(n+1)^2+n^2=(n+3)^2$

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)

$(4) 16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$

Let the coefficient of $x_i$ in $(1)$ be $n^2$ . Then the coefficient of $x_i$ in $(2)$ is $(n+1)^2$ etc.

Therefore, $3*(3)-3*(2)+(1)=(4)$

So $(4)=3*123-3*12+1=\boxed{334}$

Revision as of 06:43, 5 August 2011 (view source) Duohead (talk \| contribs) (→‎Solution 3) ← Older edit		Revision as of 18:16, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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	[[Category:Intermediate Algebra Problems]]		[[Category:Intermediate Algebra Problems]]
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