3. Let $a_{n}$ denote the last digit of the number $n^{4}$. Then $a_{1}+a_{2}+\cdots+a_{2008}=$ $\qquad$
3.6632 .

Obviously, the unit digit of $(10+n)^{4}$ is the same as that of $n^{4}$, that is,
$$
1^{4}, 2^{4}, 3^{4}, 4^{4}, 5^{4}, 6^{4}, 7^{4}, 8^{4}, 9^{4}, 10^{4}
$$

the unit digits are
$$
\begin{array}{l}
1,6,1,6,5,6,1,6,1,0 \text {. } \\
\text { Then } a_{1}+a_{2}+\cdots+a_{10} \\
=1+6+1+6+5+6+1+6+1+0=33 \text {. } \\
\text { Therefore, } a_{1}+a_{2}+\cdots+a_{2008} \\
=200 \times 33+1+6+1+6+5+6+1+6 \\
=6632 \text {. }
\end{array}
$$