Difference between revisions of "1989 AIME Problems/Problem 8"
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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>. | ||
− | + | == Solution 1 (Quadratic Function) == | |
+ | Note that each given 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 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> | ||
− | + | 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> | ||
− | + | Finally, the answer is <cmath>f(4)=16a+4b+c=\boxed{334}.</cmath> | |
− | + | ~Azjps (Fundamental Logic) | |
− | + | ~MRENTHUSIASM (Reconstruction) | |
− | < | + | == Solution 2 (Linear Combination) == |
− | <cmath>\begin{align*} | + | 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> 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> | ||
− | + | ~Duohead (Fundamental Logic) | |
− | === | + | ~MRENTHUSIASM (Reconstruction) |
− | + | ||
+ | == Solution 3 (Finite Differences by Arithmetic) == | ||
+ | Note that the second differences of all quadratic sequences must be constant (but nonzero). One example is the following sequence of perfect squares: | ||
+ | |||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(20cm); | ||
+ | |||
+ | for (real i=1; i<=10; ++i) { | ||
+ | label("\boldmath{$"+string(i^2)+"$}",(i-1,0)); | ||
+ | } | ||
+ | |||
+ | for (real i=1; i<=9; ++i) { | ||
+ | label("$"+string(1+2*i)+"$",(i-0.5,-0.75)); | ||
+ | } | ||
+ | |||
+ | for (real i=1; i<=8; ++i) { | ||
+ | label("$2$",(i,-1.5)); | ||
+ | } | ||
+ | |||
+ | for (real i=1; i<=9; ++i) { | ||
+ | draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); | ||
+ | } | ||
+ | |||
+ | for (real i=1; i<=8; ++i) { | ||
+ | draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); | ||
+ | } | ||
+ | |||
+ | for (real i=1; i<=9; ++i) { | ||
+ | draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); | ||
+ | } | ||
+ | |||
+ | for (real i=1; i<=8; ++i) { | ||
+ | draw((0.1+i,-1.35)--(0.4+i,-0.9),red); | ||
+ | } | ||
+ | |||
+ | label("\textbf{First Differences}",(-0.75,-0.75),align=W); | ||
+ | label("\textbf{Second Differences}",(-0.75,-1.5),align=W); | ||
+ | </asy> | ||
+ | |||
+ | Label equations <math>(1),(2),(3),</math> and <math>(4)</math> as Solution 2 does. Since the coefficients of <math>x_1,x_2,x_3,x_4,x_5,x_6,x_7,</math> or <math>(1,4,9,16),(4,9,16,25),(9,16,25,36),(16,25,36,49),(25,36,49,64),(36,49,64,81),(49,64,81,100),</math> respectively, all form quadratic sequences with second differences <math>2,</math> we conclude that the second differences of equations <math>(1),(2),(3),(4)</math> must be constant. | ||
+ | |||
+ | It follows that the second differences of <math>(1,12,123,S)</math> must be constant, as shown below: | ||
− | < | + | <asy> |
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(10cm); | ||
− | + | label("\boldmath{$1$}",(0,0)); | |
+ | label("\boldmath{$12$}",(1,0)); | ||
+ | label("\boldmath{$123$}",(2,0)); | ||
+ | label("\boldmath{$S$}",(3,0)); | ||
− | + | label("$11$",(0.5,-0.75)); | |
+ | label("$111$",(1.5,-0.75)); | ||
+ | label("$d_1$",(2.5,-0.75)); | ||
− | + | label("$100$",(1,-1.5)); | |
+ | label("$d_2$",(2,-1.5)); | ||
+ | for (real i=1; i<=3; ++i) { | ||
+ | draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); | ||
+ | } | ||
− | + | for (real i=1; i<=2; ++i) { | |
+ | draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); | ||
+ | } | ||
− | == | + | for (real i=1; i<=3; ++i) { |
+ | draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); | ||
+ | } | ||
− | + | for (real i=1; i<=2; ++i) { | |
+ | draw((0.1+i,-1.35)--(0.4+i,-0.9),red); | ||
+ | } | ||
− | + | label("\textbf{First Differences}",(-0.75,-0.75),align=W); | |
+ | label("\textbf{Second Differences}",(-0.75,-1.5),align=W); | ||
+ | </asy> | ||
− | <cmath> | + | Finally, we have <math>d_2=100,</math> from which |
− | \begin{align} | + | <cmath>\begin{align*} |
− | + | S&=123+d_1 \\ | |
− | + | &=123+(111+d_2) \\ | |
− | + | &=\boxed{334}. | |
− | + | \end{align*}</cmath> | |
+ | ~MRENTHUSIASM | ||
− | + | == Solution 4 (Finite Differences by Algebra) == | |
+ | 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. | ||
− | + | Now consider the polynomial given by <math> f(z) = \sum_{i=1}^7 (i+z)^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)=c_1, f(1)=c_2, f(2)=c_3</math> and are asked to find <math>f(3)=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=\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) =334.</cmath> | |
− | |||
− | ==Solution 5(Very Cheap | + | ==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 |
− | 4x_1+9x_2+16x_3=12\\ | + | <cmath>\begin{align*} |
− | 9x_1+16x_2+25x_3=123\\ \end{ | + | x_1+4x_2+9x_3&=1,\\ |
+ | 4x_1+9x_2+16x_3&=12,\\ | ||
+ | 9x_1+16x_2+25x_3&=123.\\ | ||
+ | \end{align*}</cmath> | ||
+ | Grinding this out, we have <math>(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)</math> which gives <math>\boxed{334}</math> as our final answer. | ||
− | + | ~Pleaseletmewin | |
− | + | ==Video Solution== | |
https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx | https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx | ||
Latest revision as of 03:08, 6 September 2021
Contents
Problem
Assume that are real numbers such that Find the value of .
Solution 1 (Quadratic Function)
Note that each given equation is of the form for some
When we expand and combine like terms, we obtain a quadratic function of where and are linear combinations of and
We are given that and we wish to find
We eliminate by subtracting the first equation from the second, then subtracting the second equation from the third: By either substitution or elimination, we get and Substituting these back produces
Finally, the answer is
~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2 (Linear Combination)
For simplicity purposes, we number the given equations and in that order. Let Subtracting from subtracting from and subtracting from we obtain the following equations, respectively: Subtracting from and subtracting from we obtain the following equations, respectively: Finally, applying the Transitive Property to and gives from which
~Duohead (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 3 (Finite Differences by Arithmetic)
Note that the second differences of all quadratic sequences must be constant (but nonzero). One example is the following sequence of perfect squares:
Label equations and as Solution 2 does. Since the coefficients of or respectively, all form quadratic sequences with second differences we conclude that the second differences of equations must be constant.
It follows that the second differences of must be constant, as shown below:
Finally, we have from which ~MRENTHUSIASM
Solution 4 (Finite Differences by Algebra)
Notice that we may rewrite the equations in the more compact form as: where and is what we are trying to find.
Now consider the polynomial given by (we are only treating the as coefficients).
Notice that is in fact a quadratic. We are given and are asked to find . 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 .
Alternatively, applying finite differences, one obtains
Solution 5 (Very Cheap: Not Recommended)
We let . Thus, we have Grinding this out, we have which gives 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.