215. The sequence $\left(x_{n}\right)$ is defined by the recurrence relations

$$
x_{1}=1, x_{2}=2, x_{n}=\left|x_{n-1}-x_{n-2}\right|(n>2) .
$$

Find $x_{1994}$.
$\triangle$ First, let's compute the first few terms of the sequence: $x_{1}=1, x_{2}=2, x_{3}=|2-1|=1, x_{4}=|1-2|=1, x_{5}=|1-1|=0, x_{6}=$ $=|0-1|=1, x_{7}=1, x_{8}=0, x_{9}=1, x_{10}=1, x_{11}=0$.

Starting from the third term, the terms of this sequence are either 1 or 0. Only the terms with indices

$$
5,8,11, \ldots, 5+3 k=2+(3+3 k)=2+3(k+1)
$$

are equal to 0, i.e., the terms whose indices, when divided by 3, give a remainder of 2. (The latter can be strictly proven by mathematical induction).

Let's check if the remainder of the division of the number 1994 by 3 is 2:

$$
1994=3 \cdot 664+2
$$

Therefore, $x_{1994}=0$.

Answer: 0.