AIME 1989 · 第 8 题

Solution 1 (Quadratic Function)

Note that each given 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:

$$\begin{aligned} 3a+b&=11, \\ 5a+b&=111. \end{aligned}$$ 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 ~MRENTHUSIASM

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:

$$\begin{aligned} 3x_1 + 5x_2 + 7x_3 + 9x_4 + 11x_5 + 13x_6 + 15x_7 &=11, \hspace{20mm}&(5) \\ 5x_1 + 7x_2 + 9x_3 + 11x_4 + 13x_5 + 15x_6 + 17x_7 &=111, &(6) \\ 7x_1 + 9x_2 + 11x_3 + 13x_4 + 15x_5 + 17x_6 + 19x_7 &=S-123. &(7) \\ \end{aligned}$$ Subtracting $(5)$ from $(6)$ and subtracting $(6)$ from $(7),$ we obtain the following equations, respectively:

$$\begin{aligned} 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=100, &(8) \\ 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=S-234. \hspace{20mm}&(9) \end{aligned}$$ Finally, applying the Transitive Property to $(8)$ and $(9)$ gives $S-234=100,$ from which $S=\boxed{334}.$

~Duohead ~MRENTHUSIASM

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 $(1),(2),(3),$ and $(4)$ as Solution 2 does. Since the coefficients of $x_1,x_2,x_3,x_4,x_5,x_6,x_7,$ or $(1,4,9,16),(4,9,16,25),(9,16,25,36),(16,25,36,49),(25,36,49,64),(36,49,64,81),(49,64,81,100),$ respectively, all form quadratic sequences with second differences $2,$ we conclude that the second differences of equations $(1),(2),(3),(4)$ must be constant.

It follows that the second differences of $(1,12,123,S)$ must be constant, as shown below:

Finally, we have $d_2=100,$ from which

$$\begin{aligned} S&=123+d_1 \\ &=123+(111+d_2) \\ &=\boxed{334}. \end{aligned}$$ ~MRENTHUSIASM

Solution 4 (Finite Differences by Algebra)

Notice that we may rewrite the equations in the more compact form as:

$$\begin{aligned} \sum_{i=1}^{7}i^2x_i&=c_1, \\ \sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\ \sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\ \sum_{i=1}^{7}(i+3)^2x_i&=c_4, \end{aligned}$$ 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 $f(z) = \sum_{i=1}^7 (i+z)^2x_i$ (we are only treating the $x_i$ as coefficients).

Notice that $f$ is in fact a quadratic. We are given $f(0)=c_1, f(1)=c_2, f(2)=c_3$ and are asked to find $f(3)=c_4$. 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 $c_4=\boxed{334}$.

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 5 (Assumption)

The idea is to multiply the first, second and third equations by $a,b,$ and $c,$ respectively.

We can only consider the coefficients of $x_1,x_2,$ and $x_3:$

$$\begin{aligned} a+4b+9c&=16, \\ 4a+9b+16c&=25, \\ 9a+16b+25c&=36. \end{aligned}$$ Subtracting $(1)$ from $(2),$ we get

$$3a+5b+7c=9. \hspace{15mm}(4)$$ Subtracting $3\cdot(4)$ from $(3),$ we get

$$b+4c=9. \hspace{25.5mm}(5)$$ Subtracting $(1)$ from $4\cdot(5),$ we get

$$7c-a=20. \hspace{23mm}(6)$$ From $(5)$ and $(6),$ we have $(a,b,c)=(7c-20,9-4c,c).$ Substituting this into $(2)$ gives $(a,b,c)=(1,-3,3).$

Therefore, the answer is $1\cdot1+12\cdot(-3) + 123\cdot3 = \boxed{334}.$

Solution 6 (Assumption)

We let $(x_4,x_5,x_6,x_7)=(0,0,0,0)$. Thus, we have

$$\begin{aligned} x_1+4x_2+9x_3&=1,\\ 4x_1+9x_2+16x_3&=12,\\ 9x_1+16x_2+25x_3&=123.\\ \end{aligned}$$ 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

Solution 7 (Similar to Solutions 3 and 4)

Let $s_n = n^2$ be the sequence of perfect squares. By either expanding or via finite differences, one can prove the miraculous recursion

$$s_n = 3s_{n-1} - 3s_{n-2} + s_{n-3}.$$ Hence, the answer is simply

$$3 \cdot 123 - 3 \cdot 12 + 1 = \boxed{334}.$$ I first saw this in a Mathologer video: https://www.youtube.com/watch?v=4AuV93LOPcE

~Ritwin

Note: You can prove this recursion through considering the coefficients for each $x_i$ as $a^2, (a+1)^2, (a+2)^2$ for the equations respectively.

All we are left to do is find $(a+3)^2$ in terms of the above $3$ quadratics. Because we have an $a^2$ as one, we don't have to worry about the $a^2$ term, we can just manipulate it by adding or subtracting any number of the first quadratic.

We are left to find a combination of the second and third quadratics that have the linear term and constant of $(a+3)^2$. With little experimentation we find that $3(a+2)^2 - 3(a+1)^2 = 6a+9$. All we have to add back is one $a^2$.

Thus, the recursion holds. That for any term $x_i$ the coefficients would be recursible by $a^2$ = $3(a-1)^2 - 3(a-2)^2 + (a-3)$(Where we have set $a$ as $a-3$)

~Avi_2009

Solution 8

We number our equations with $(n)$ denoting equation $n$

$$\begin{aligned} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1 &(1)\\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12 &(2) \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123 &(3) \end{aligned}$$ See that we want to find the value of

$$7x_{1}+9x_{2}+11x_{3}+13x_{4}+15x_{5}+17x_{6}+19x_{7}$$ and add that to $(3)$. Keep this in mind.

Subtracting $(1)$ from $(2)$ yields

$$3x_{1}+5x_{2}+7x_{3}+9x_{4}+11x_{5}+13x_{6}+15x_{7} = 11 \ \ (4)$$ and subtracting $(2)$ from $(3)$ yields

$$5x_{1}+7x_{2}+9x_{3}+11x_{4}+13x_{5}+15x_{6}+17x_{7} = 111 \ \ \ (5)$$ subtracting $(4)$ from $(5)$ yields

$$2x_{1}+2x_{2}+2x_{3}+2x_{4}+2x_{5}+2x_{6}+2x_{7} = 100 \ \ \ (6)$$ Adding this back to $(5)$ yields

$$7x_{1}+9x_{2}+11x_{3}+13x_{4}+15x_{5}+17x_{6}+19x_{7}=211 \ \ \ (7)$$ Now, simply adding $(7)$ to $(3)$ yields

$$16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7 = \boxed{334}$$ ~shockfront99(omg this is my first AIME solve)

Video Solution

https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx