8. There are $n$ balls in a box, and the balls are numbered $1,2,3, \ldots, n$ respectively. One of the balls is removed from the box, and it turns out that the sum of the numbers on the remaining balls in the box is 5048. If the number on the ball removed from the box is $m$, find the value of $m$.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
8. Ans: B
Let $m$ be the number of the removed ball. Then we have $1 \leq m \leq n$, and
$$
\begin{array}{l}
\Longrightarrow \quad 1+2+3+\cdots+n-m=5048 \\
\Longrightarrow \frac{n(n+1)}{2}-m=5048 \\
\Longrightarrow 1 \leq \frac{n(n+1)}{2}-5048 \leq n \\
\Longrightarrow \frac{(n-1) n}{2} \leq 5048 \text { and } \frac{n(n+1)}{2} \geq 5049 .
\end{array}
$$

Observe that $\frac{99 \times 100}{2}=4950, \frac{100 \times 101}{2}=5050$ and $\frac{101 \times 102}{2}=5151$. It follows that $n=100$. Hence $m=\frac{100 \times 101^{2}}{2}-5048=2$.