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

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m (Solution 2 (Linear Combination))
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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
<cmath>\begin{align*}x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&=1\\
+
<cmath>\begin{align*}
4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&=12\\
+
x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\
9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&=123.\end{align*}</cmath>
+
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>
 
Find the value of <math>16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7</math>.
 
Find the value of <math>16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7</math>.
  
__TOC__
+
== Solution 1 (Quadratic Function) ==
 +
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>
  
== Solution ==
+
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>
=== Solution 1===
 
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.  
 
  
 +
We are given that
 +
<cmath>\begin{alignat*}{10}
 +
f(1)&=\phantom{42}a+b+c&&=1, \\
 +
f(2)&=4a+2b+c&&=12, \\
 +
f(3)&=9a+3b+c&&=123,
 +
\end{alignat*}</cmath>
 +
and we wish to find <math>f(4).</math>
  
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].
+
We eliminate <math>c</math> by subtracting the first equation from the second, then subtracting the second equation from the third:
 +
<cmath>\begin{align*}
 +
3a+b&=11, \\
 +
5a+b&=111.
 +
\end{align*}</cmath>
 +
By either substitution or elimination, we get <math>a=50</math> and <math>b=-139.</math> Substituting these back produces <math>c=90.</math>
  
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
+
Finally, the answer is <cmath>f(4)=16a+4b+c=\boxed{334}.</cmath>
  
<center><math>[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.</math></center>
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~Azjps (Fundamental Logic)
  
Comparing coefficents gives us the three equation system:
+
~MRENTHUSIASM (Reconstruction)
  
<center>
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== Solution 2 (Linear Combination) ==
<cmath>\begin{align*}a + b + c &= 1 \\ 2b + 4c &= 6 \\ b + 4c &= 9 \end{align*}</cmath>
+
For simplicity purposes, we number the given equations <math>(1),(2),</math> and <math>(3),</math> in that order. Let <cmath>16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7=S. \hspace{29.5mm}(4)</cmath>
</center>  
+
Subtracting <math>(1)</math> from <math>(2),</math> subtracting <math>(2)</math> from <math>(3),</math> and subtracting <math>(3)</math> from <math>(4),</math> we obtain the following equations, respectively:
 +
<cmath>\begin{align*}
 +
3x_1 + 5x_2 + 7x_3 +  9x_4 + 11x_5 + 13x_6 + 15x_7 &=11, \hspace{20mm}&(5) \\
 +
5x_1 + 7x_2 +  9x_3 + 11x_4 + 13x_5 + 15x_6 + 17x_7 &=111, &(6) \\
 +
7x_1 + 9x_2 + 11x_3 + 13x_4 + 15x_5 + 17x_6 + 19x_7 &=S-123. &(7) \\
 +
\end{align*}</cmath>
 +
Subtracting <math>(5)</math> from <math>(6)</math> and subtracting <math>(6)</math> from <math>(7),</math> we obtain the following equations, respectively:
 +
<cmath>\begin{align*}
 +
2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=100, &(8) \\
 +
2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=S-234. \hspace{20mm}&(9)
 +
\end{align*}</cmath>
 +
Finally, applying the Transitive Property to <math>(8)</math> and <math>(9)</math> gives <math>S-234=100,</math> from which <math>S=\boxed{334}.</math>
  
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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~Duohead (Fundamental Logic)
  
=== Solution 2===
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~MRENTHUSIASM (Reconstruction)
 +
 
 +
== Solution 3 ==
 
Notice that we may rewrite the equations in the more compact form as:
 
Notice that we may rewrite the equations in the more compact form as:
 +
<cmath>\begin{align*}
 +
\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, \\
 +
\sum_{i=1}^{7}(i+3)^2x_i&=c_4,
 +
\end{align*}</cmath>
 +
where <math>c_1=1, c_2=12, c_3=123,</math> and <math>c_4</math> is what we are trying to find.
  
<math>\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,</math> and <math>\sum_{i=1}^{7}(i+3)^2x_i=c_4,</math>
+
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).
 
 
where <math>c_1=1, c_2=12, c_3=123,</math> and <math>c_4</math> is what we're trying to find.
 
 
 
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>.
 
 
 
 
 
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>.
 
 
 
=== Solution 3 ===
 
 
 
Notice that <math>3(n+2)^2-3(n+1)^2+n^2=(n+3)^2</math>
 
 
 
I'll number the equations for convenience
 
 
 
<cmath>
 
\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}</cmath>
 
 
 
Let the coefficient of <math>x_i</math> in <math>(1)</math> be <math>n^2</math>. Then the coefficient of <math>x_i</math> in <math>(2)</math> is <math>(n+1)^2</math> etc.
 
  
Therefore, <math>3*(3)-3*(2)+(1)=(4)</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>.
  
So <math>(4)=3*123-3*12+1=\boxed{334}</math>
+
Alternatively, applying finite differences, one obtains <cmath>c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =\boxed{334}.</cmath>
  
===Solution 4===
+
==Solution 4==
 
Notice subtracting the first equation from the second yields <math>3x_1 + 5x_2 + ... + 15x_7 = 11</math>. Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get <math>2x_1 + 2x_2 + ... +2x_7 = 100</math>. 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 <math>\boxed{334}.</math>
 
Notice subtracting the first equation from the second yields <math>3x_1 + 5x_2 + ... + 15x_7 = 11</math>. Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get <math>2x_1 + 2x_2 + ... +2x_7 = 100</math>. 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 <math>\boxed{334}.</math>
  
==Solution 5(Very Cheap)(Not advised)==
+
==Solution 5 (Very Cheap: Not Recommended)==
 
We let <math>(x_4,x_5,x_6,x_7)=(0,0,0,0)</math>. Thus, we have  
 
We let <math>(x_4,x_5,x_6,x_7)=(0,0,0,0)</math>. Thus, we have  
  
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-Pleaseletmewin
 
-Pleaseletmewin
  
===Video Solution===
+
==Video Solution==
 
https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx
 
https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx
  

Revision as of 11:00, 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 (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 are given that \begin{alignat*}{10} f(1)&=\phantom{42}a+b+c&&=1, \\ f(2)&=4a+2b+c&&=12, \\ f(3)&=9a+3b+c&&=123, \end{alignat*} and we wish to find $f(4).$

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 2 (Linear Combination)

For simplicity purposes, we number the given equations $(1),(2),$ and $(3),$ in that order. Let \[16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7=S. \hspace{29.5mm}(4)\] Subtracting $(1)$ from $(2),$ subtracting $(2)$ from $(3),$ and subtracting $(3)$ from $(4),$ we obtain the following equations, respectively: \begin{align*} 3x_1 + 5x_2 +  7x_3 +  9x_4 + 11x_5 + 13x_6 + 15x_7 &=11, \hspace{20mm}&(5) \\ 5x_1 + 7x_2 +  9x_3 + 11x_4 + 13x_5 + 15x_6 + 17x_7 &=111, &(6) \\ 7x_1 + 9x_2 + 11x_3 + 13x_4 + 15x_5 + 17x_6 + 19x_7 &=S-123. &(7) \\ \end{align*} Subtracting $(5)$ from $(6)$ and subtracting $(6)$ from $(7),$ we obtain the following equations, respectively: \begin{align*} 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=100, &(8) \\ 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=S-234. \hspace{20mm}&(9) \end{align*} Finally, applying the Transitive Property to $(8)$ and $(9)$ gives $S-234=100,$ from which $S=\boxed{334}.$

~Duohead (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 3

Notice that we may rewrite the equations in the more compact form as: \begin{align*} \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, \\ \sum_{i=1}^{7}(i+3)^2x_i&=c_4, \end{align*} where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are 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 \[c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =\boxed{334}.\]

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 (ProblemsAnswer KeyResources)
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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