5. Arrange all positive divisors of 8128 in ascending order as $a_{1}, a_{2}, \cdots, a_{n}$, then $\sum_{k=1}^{n} k a_{k}=$ $\qquad$ .
5.211335 .

Given $8128=2^{6}\left(2^{7}-1\right)$, we have
$$
\begin{array}{l}
a_{n}=\left\{\begin{array}{ll}
2^{n-1}, & 1 \leqslant n \leqslant 7 ; \\
2^{n-8}\left(2^{7}-1\right), & 8 \leqslant n \leqslant 14 .
\end{array}\right. \\
\text { Then } \sum_{k=1}^{n} k a_{k}=\sum_{k=1}^{14} k \cdot 2^{k-1}-\sum_{k=1}^{7}(k+7) 2^{k-1} . \\
\text { Also } \sum_{k=1}^{14} k \cdot 2^{k-1}=\sum_{k=1}^{14}\left(k \cdot 2^{k}-k \cdot 2^{k-1}\right) \\
=14 \times 2^{14}+\sum_{k=1}^{13}\left[k \cdot 2^{k}-(k+1) 2^{k}\right]-1 \\
=14 \times 2^{14}-\sum_{k=0}^{13} 2^{k}=14 \times 2^{14}-\left(2^{14}-1\right) \\
=13 \times 2^{14}+1, \\
\sum_{k=1}^{7}(k+7) 2^{k-1}=13 \times 2^{7}-6, \\
\text { Therefore } \sum_{k=1}^{14} k a_{k}=\left(13 \times 2^{14}+1\right)-\left(13 \times 2^{7}-6\right) \\
=13 \times 2^{7}\left(2^{7}-1\right)+7=211335 .
\end{array}
$$