【Question 17】
For any positive integer $\mathrm{n}$, let $\mathrm{f}(\mathrm{n})$ denote the last digit of $1+2+3+\cdots \cdots+\mathrm{n}$, such as $\mathrm{f}(1)=1, \mathrm{f}(2)=3, \mathrm{f}(5)=5$, etc. Find the value of $f(2)+f(4)+f(6)+\cdots+f$ (2012).
$$
1+2+\cdots+20=(1+20) \times 20 \div 2=210 \text {; }
$$

That is, $f(20)=0$;

Thus, $f(n)$ has a cycle every 20;
$$
\begin{array}{l}
f(1)=1, f(2)=3, f(3)=6, f(4)=0, f(5)=5, \\
f(6)=1, \quad f(7)=8, f(8)=6, f(9)=5, f(10)=5, \\
f(11)=6, f(12)=8, f(13)=1, f(14)=5, f(15)=0, \\
f(16)=6, \quad f(17)=3, f(18)=1, f(19)=0, f(20)=0 ; \\
2012 \div 20=100 \cdots \cdots \cdot 12 ; \\
f(2)+f(4)+\cdots+f(20)=3+0+1+6+5+8+5+6+1+0=35 ; \\
f(2)+f(4)+\cdots+f(12)=3+0+1+6+5+8=23 ; \\
f(2)+f(4)+\cdots+f(2012)=35 \times 100+23=3523
\end{array}
$$