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

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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 1 (Quadratic Function)

Note that each equation is of the form $f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7,$ for some $k\in\{1,2,3\}.$

When we expand $f(k)$ and combine like terms, we obtain a quadratic function of $k:$ $f(k)=ak^2+bk+c,$ where $a,b,$ and $c$ are linear combinations of $x_1,x_2,x_3,x_4,x_5,x_6,$ and $x_7.$

We wish to find $f(4).$ Two solutions follow from here:

Solution 1.1 (Generalized)

We are given that $\begin{alignat*}{10} f(1)&=a+b+c&&=1, \\ f(2)&=4a+2b+c&&=12, \\ f(3)&=9a+3b+c&&=123. \end{alignat*}$ We eliminate $c$ by subtracting the first equation from the second, then subtracting the second equation from the third: $\begin{align*} 3a+b&=11, \\ 5a+b&=111. \end{align*}$ By either substitution or elimination, we get $a=50$ and $b=-139.$ Substituting these back produces $c=90.$

Finally, the answer is $f(4)=16a+4b+c=\boxed{334}.$

~Azjps (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 1.2 (Specified)

~MRENTHUSIASM

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

$\begin{align} x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&=1\\ 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&=12\\ 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&=123\\ 16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&=\end{align}$

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

Solution 4

Notice subtracting the first equation from the second yields $3x_1 + 5x_2 + ... + 15x_7 = 11$ . Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get $2x_1 + 2x_2 + ... +2x_7 = 100$ . Adding this twice to the first obtained equation gives difference of the desired equation and 3rd equation, which is 211. Adding to the 3rd equation, we get $\boxed{334}.$

Solution 5 (Very Cheap: Not Recommended)

We let $(x_4,x_5,x_6,x_7)=(0,0,0,0)$ . Thus, we have

$\begin{cases} x_1+4x_2+9x_3=1\\ 4x_1+9x_2+16x_3=12\\ 9x_1+16x_2+25x_3=123\\ \end{cases}$

Grinding this out, we have $(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)$ which gives $\boxed{334}$ as our final answer.

-Pleaseletmewin

Video Solution

https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx

@@ Line 28: / Line 28: @@
 By either substitution or elimination, we get <math>a=50</math> and <math>b=-139.</math> Substituting these back produces <math>c=90.</math>
-Finally, the answer is <cmath>f(4)=16a-139b+90c=\boxed{334}.</cmath>
+Finally, the answer is <cmath>f(4)=16a+4b+c=\boxed{334}.</cmath>
 ~Azjps (Fundamental Logic)

1989 AIME (Problems • Answer Key • Resources)
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