(10 From a material site beside a road, electric poles are to be transported along a straight road in the same direction to be planted at the roadside 500 m away. At the 500 m point, one pole is planted, and then one pole is planted every 50 m along the roadside. Knowing that the transport vehicle can carry a maximum of 3 poles at a time, to complete the task of transporting and planting 20 poles, and returning to the material site, the minimum total distance traveled by the transport vehicle is $\qquad$ $\mathrm{m}$.
1014000 Prompt: First, assume that 21 utility poles need to be transported and planted, with 3 poles each time. Let the round-trip distance for the $k$-th time be $a_{k}(k=1,2, \cdots, 7)$, then $a_{1}=2 \times 600=1200$, and
$$
a_{k+1}=a_{k}+2 \times 150(k=1,2, \cdots, 6) .
$$

Therefore, $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 1200 and a common difference of 300.
Thus, $S_{7}=7 \times 1200+\frac{7 \times 6}{2} \times 300=14700(\mathrm{~m})$.
However, in reality, only 20 poles are transported, so there must be one time when 2 poles are transported, and the other 6 times, 3 poles each.
If the time when 2 poles are transported is the $k$-th time $(k=1,2, \cdots, 7)$, then $a_{1}, a_{2}, \cdots, a_{k-1}$ remain unchanged, and $a_{k}, a_{k+1}, \cdots, a_{7}$ each decrease by $100 \mathrm{~m}$, so
$$
S_{7}(k)=S_{7}-100(8-k)=13800+100 k .
$$

Obviously, when $k=1$, that is, the first time 2 poles are transported, and the other 6 times 3 poles each, the total distance is the smallest, with the minimum value being $14000 \mathrm{~m}$.