For each positive integer $n$, let $f(n)$ denote the last digit of $1+$ $2+\cdots+n$ (for example, $f(1)=1, f(2)=3$, $f(3)=6)$. Then
$$
f(1)+f(2)+\cdots+f(2008)
$$

has the tens digit ( ).
(A) 1
(B) 2
(C) 3
(D) 4
5.C.

First, list 1 calculates the $f(n)$ for several positive integers:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & & \\
\hline$f(n)$ & 1 & 3 & 6 & 0 & 5 & 1 & 8 & 6 & 5 & 5 & & \\
\hline$n$ & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & & \\
\hline$f(n)$ & 6 & 8 & 1 & 5 & 0 & 6 & 3 & 1 & 0 & 0 & & \\
\hline$n$ & 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 & 29 & 30 & & \\
\hline$f(n)$ & 1 & 3 & 6 & 0 & 5 & 1 & 8 & 6 & 5 & 5 & & \\
\hline$n$ & 31 & 32 & 33 & 34 & 35 & 36 & 37 & 38 & 39 & 40 & 41 & $\cdots$ \\
\hline$f(n)$ & 6 & 8 & 1 & 5 & 0 & 6 & 3 & 1 & 0 & 0 & 1 & $\cdots$ \\
\hline
\end{tabular}

From Table 1, it is observed that every 20 numbers, the corresponding $f(n)$ values are the same.
$$
\begin{array}{l}
\text { Since } f(1)+f(2)+\cdots+f(20)=70, \\
2008=20 \times 100+8,
\end{array}
$$

Therefore,
$$
\begin{array}{l}
f(1)+f(2)+\cdots+f(2008) \\
=70 \times 100+f(2001)+f(2002)+\cdots+f(2008) \\
=7000+30=7030 .
\end{array}
$$