'1-188 Write the numbers $1,2,3, \cdots, 1986,1987$ on the blackboard. At each step, determine some of the numbers written and erase them, replacing them with the remainder of their sum divided by 7. After several steps, two numbers remain on the blackboard, one of which is 987. What is the second remaining number?
［Solution］We notice that after each "erasing" and writing of the "remainder", the sum of all numbers still written on the blackboard remains unchanged modulo 7. Let the last remaining number be $x$, then $x+987$ and the following sum
$$
1+2+\cdots+1986+1987=1987 \cdot 7 \cdot 142
$$

are congruent modulo 7, and both are equal to 0. Since 987 is divisible by 7, $x$ must also be divisible by 7. Because after each "erasing", a "remainder" is always written, the last remaining numbers must include at least one "remainder". However, 987 is not a remainder when divided by 7, so $x$ must be a remainder when divided by 7. Therefore, $0 \leqslant x \leqslant 6$, and thus $x$ can only be 0.