AIME 2000 I · 第 10 题

Solution

Let the sum of all of the terms in the sequence be $\mathbb{S}$. Then for each integer $k$, $x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k$. Summing this up for all $k$ from $1, 2, \ldots, 100$,

$$\begin{aligned}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &= 1 + 2 + \cdots + 100\\ 100\mathbb{S} - 2\mathbb{S} &= \frac{100 \cdot 101}{2} = 5050\\ \mathbb{S}&=\frac{2525}{49}\end{aligned}$$ Now, substituting for $x_{50}$, we get $2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}$, and the answer is $75+98=\boxed{173}$.

Solution 2

Consider $x_k$ and $x_{k+1}$. Let $S$ be the sum of the rest 98 terms. Then $x_k+k=S+x_{k+1}$ and $x_{k+1}+(k+1)=S+x_k.$ Eliminating $S$ we have $x_{k+1}-x_k=-\dfrac{1}{2}.$ So the sequence is arithmetic with common difference $-\dfrac{1}{2}.$

In terms of $x_{50},$ the sequence is $x_{50}+\dfrac{49}{2}, x_{50}+\dfrac{48}{2},\cdots,x_{50}+\dfrac{1}{2}, x_{50}, x_{50}-\dfrac{1}{2}, \cdots, x_{50}-\dfrac{49}{2}, x_{50}-\dfrac{50}{2}.$ Therefore, $x_{50}+50=99x_{50}-\dfrac{50}{2}$.

Solving, we get $x_{50}=\dfrac{75}{98}.$ The answer is $75+98=\boxed{173}.$

- JZ

- edited by erinb28lms

Solution 3 (Sum of equations)

Like Solution 1, let the sum of all of the terms in this sequence be $\mathbb{S}$. By definition:

$$x_1 + 1 = x_2+x_3+x_4+...+x_{100}$$ $$x_2 + 2 = x_1+x_3+x_4+...+x_{100}$$ $$x_3 + 3 = x_1+x_2+x_4+...+x_{100}$$ $$...$$ $$x_{99} + 99 = x_1+x_2+x_3+...+x_{98}+x_{100}$$ $$x_{100} + 100 = x_1+x_2+x_3+...+x_{98}+x_{99}$$ .

Adding up all of these equations yields:

$$\mathbb{S} + TR(100) = 99\mathbb{S}$$ Here $TR(100)$ represents the $100$th triangular number, which is $5050$. Solving for $\mathbb{S}$ yields:

$$\mathbb{S} = \frac{2525}{49}$$ .

$\mathbb{S}$ can also be written as $x_{50} + (x_{50} + 50)$. Solving for $x_{50}$,

$$2x_{50} = \frac{2525-2450}{49}$$ $$2x_{50} = \frac{75}{49}$$ $$x_{50} = \frac{75}{98}$$ The requested sum is therefore $75+98 = \boxed{173}$.

~mathwizard123123

Solution 4 (Clever Substitution)

Due to the problem's unique setup, we can actually represent every $x_k$ in terms of its sum, call it $S$. For any $x_k$,

$$x_k = \frac{S}{2} - \frac{k}{2}$$ We can show that this works because the sum of the other $99$ terms is now equal to $\frac{S}{2} + \frac{k}{2}$, and the difference is indeed $k$. The motivation to find that we need an $\frac{S}{2}$ term lies in that the difference between $x_k$ and the $99$ other terms is an integer, and not an expression in terms of $S$, and allowing $x_k$ to be in terms of $\frac{S}{2}$ is the only way to achieve this.

Now, finding the numerical value of $S$ solves the problem. Summing all terms from $x_1$ to $x_{100}$, we find

$$S = \sum_{k=1}^{100} x_k = 100\frac{S}{2} - \sum_{k=1}^{100} \frac{k}{2} = 50S - \frac{100\cdot\frac{1 + 100}{2}}{2} = 50S - 2525$$ We obtain $S = \frac{2525}{49}$, and we now only need to find $x_{50}$, which is represented as $\frac{S}{2} - \frac{50}{2}$. Substituting,

$$x_{50} = \frac{2525}{49\cdot2} - 25 = \frac{25\cdot101 - 25\cdot98}{98} = \frac{75}{98}$$ Our desired answer is thus $75 + 98 = \boxed{173}$

~faliure167

Video solution

https://www.youtube.com/watch?v=TdvxgrSZTQw