$10 \cdot 34$ Write a sequence of consecutive positive integers starting with 1 on the blackboard, and erase one of the numbers. The arithmetic mean of the remaining numbers is $35 \frac{7}{17}$. What is the number that was erased?
(A) 6.
(B) 7.
(C) 8.
(D) 9.
(E) Cannot be determined.
(33rd American High School Mathematics Examination, 1982)
[Solution] Let $n$ be the largest integer written on the blackboard. If 1 is the number that is erased, then the maximum arithmetic mean is
$$
\frac{2+3+\cdots+n}{n-1}=\frac{\frac{(n+1) n}{2}-1}{n-1}=\frac{n+2}{2};
$$

If $n$ is the number that is erased, then the minimum arithmetic mean is
$$
\frac{1+2+3+\cdots+n-1}{n-1}=\frac{n(n-1)}{2(n-1)}=\frac{n}{2}.
$$

Thus, we have
$$
\begin{array}{l}
\frac{n}{2} \leqslant 35 \frac{7}{17} \leqslant \frac{n+2}{2}, \\
n \leqslant 70 \frac{14}{17} \leqslant n+2,
\end{array}
$$

Therefore, $68 \frac{14}{17} \leqslant n \leqslant 70 \frac{14}{17}$. Hence, $n=69$ or 70.
Since $35 \frac{7}{17}$ is the arithmetic mean of $n-1$ integers, $\left(35 \frac{7}{17}\right)(n-1)$
must be an integer. When $n=70$, this is impossible, so $n=69$.
If $x$ is the number that is erased, then
$$
\frac{\frac{69 \cdot 70}{2}-x}{68}=35 \frac{7}{17},
$$

which simplifies to $69 \cdot 35 - x = 35 \frac{7}{17} \cdot 68 = 35 \cdot 68 + 28$,
or $35 - x = 28$, giving $x=7$.
Therefore, the answer is (B).