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

(Reformatted the solutions. I will reconstruct Solutions 1 and 3.)
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== Problem ==
 
== Problem ==
 
Assume that <math>x_1,x_2,\ldots,x_7</math> are real numbers such that
 
Assume that <math>x_1,x_2,\ldots,x_7</math> are real numbers such that
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<cmath>\begin{align*}x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&=1\\
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<cmath>\begin{align*}x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&=1,\\
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4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&=12\\
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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*}</cmath>
 
9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&=123.\end{align*}</cmath>
 
   
 
   
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== Solution 1 ==
 
== Solution 1 ==
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Notice that because we are given a system of <math>3</math> equations with <math>7</math> unknowns, the values <math>(x_1, x_2, \ldots, x_7)</math> 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.
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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>
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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 <math>x_i</math> in the first equation be <math>y_i^2</math>; then its coefficients in the second equation is <math>(y_i+1)^{2}</math> and the third as <math>(y_i+2)^2</math>. We need to find a way to sum these to make <math>(y_i+3)^2</math> [this is in fact a specific approach generalized by the next solution below].
 
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Thus, we hope to find constants <math>a,b,c</math> satisfying <math>ay_i^2 + b(y_i+1)^2 + c(y_i+2)^2 = (y_i + 3)^2</math>. [[FOIL]]ing out all of the terms, we get
 
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<center><math>[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.</math></center>
 
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Comparing coefficents gives us the three equation system:
 
  
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<center>
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~Azjps (Fundamental Logic)
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<cmath>\begin{align*}a + b + c &= 1 \\ 2b + 4c &= 6 \\ b + 4c &= 9 \end{align*}</cmath>
 
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</center>
 
  
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Subtracting the second and third equations yields that <math>b = -3</math>, so <math>c = 3</math> and <math>a = 1</math>. It follows that the desired expression is <math>a \cdot (1) + b \cdot (12) + c \cdot (123)  = 1 - 36 + 369 = \boxed{334}</math>.
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~MRENTHUSIASM (Reconstruction)
  
 
== Solution 2==
 
== Solution 2==
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-Pleaseletmewin
 
-Pleaseletmewin
  
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===Video Solution===
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==Video Solution==
 
https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx
 
https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx
  

Revision as of 00:45, 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 \[f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7,\] where $k\in\{1,2,3\}.$

~Azjps (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

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

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

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