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

Revision as of 01:08, 24 June 2021

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

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)

~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 8: / Line 8: @@
 == Solution 1 ==
-Note that each equation is of the form <cmath>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,</cmath> where <math>k\in\{1,2,3\}.</math>
+Note that each equation is of the form <cmath>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,</cmath> for some <math>k\in\{1,2,3\}.</math>
+When we expand <math>f(k)</math> and combine like terms, we obtain a quadratic function of <math>k:</math> <cmath>f(k)=ak^2+bk+c,</cmath> where <math>a,b,</math> and <math>c</math> are linear combinations of <math>x_1,x_2,x_3,x_4,x_5,x_6,</math> and <math>x_7.</math>
+We wish to find <math>f(4).</math> Two solutions follow from here:
+===Solution 1.1 (Generalized)===
 ~Azjps (Fundamental Logic)
 ~MRENTHUSIASM (Reconstruction)
+===Solution 1.2 (Specified)===
+~MRENTHUSIASM
 == Solution 2==
@@ Line 23: / Line 34: @@
 Now consider the polynomial given by <math> f(z) := \sum_{i=1}^7 (z+i)^2x_i </math> (we are only treating the <math>x_i</math> as coefficients).
-Notice that <math>f</math> is in fact a quadratic. We are given <math>f(0), \ f(1), \ f(2)</math> as <math>c_1, \ c_2, \ c_3</math> and are asked to find <math>c_4</math>. 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 <math>c_4=334</math>.
+Notice that <math>f</math> is in fact a quadratic. We are given <math>f(0), \ f(1), \ f(2)</math> as <math>c_1, \ c_2, \ c_3</math> and are asked to find <math>c_4</math>. 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 <math>c_4=334</math>.
 Alternatively, applying finite differences, one obtains <math>c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334</math>.

1989 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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