6. The page numbers of a book are from 1 to $n$. When these page numbers were added up, one page number was mistakenly added one extra time. As a result, the incorrect sum obtained was 1986. What is the page number that was added one extra time?
6. Let $k$ be the page number that was added one extra time, then $0<k<n+1$, so $1+2+\cdots+n+k$ is between $1+2+\cdots+n$ and $1+2+\cdots+n+(n+1)$, thus
$$
\frac{n(n+1)}{2}<1988<\frac{(n+1)(n+2)}{2},
$$

which means $n(n+1)<3972<(n+1)(n+2)$. By trial (since $n$ is slightly greater than 60), we get
$$
\begin{array}{l}
n=62, \text{ so } k=1986-\frac{(62)(63)}{2} \\
=1986-1953=33 .
\end{array}
$$