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

(Solution 3: Reformatted)
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We are given that
 
We are given that
 
<cmath>\begin{alignat*}{10}
 
<cmath>\begin{alignat*}{10}
f(1)&=a+b+c&&=1, \\
+
f(1)&=\phantom{42}a+b+c&&=1, \\
 
f(2)&=4a+2b+c&&=12, \\
 
f(2)&=4a+2b+c&&=12, \\
 
f(3)&=9a+3b+c&&=123,
 
f(3)&=9a+3b+c&&=123,
Line 34: Line 34:
  
 
== Solution 2 (Linear Combination) ==
 
== Solution 2 (Linear Combination) ==
For simplicity purposes, we number the given equations <math>(1),(2),</math> and <math>(3),</math> in that order.
+
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>
 
+
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> we have <cmath>2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7=100. \hspace{41.25mm}(8)</cmath>
 +
Finally, note that <math>(7)=(6)+(8),</math> or <math>S-123=111+100.</math> The answer is <math>S=\boxed{334}.</math>
  
 
~Duohead (Fundamental Logic)
 
~Duohead (Fundamental Logic)

Revision as of 04:04, 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 $(6),$ we have \[2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7=100. \hspace{41.25mm}(8)\] Finally, note that $(7)=(6)+(8),$ or $S-123=111+100.$ The answer is $S=\boxed{334}.$

~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
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