【Question 6】
Using the digits $0, 1, 2, 3, 4, 5$, many positive integers can be formed. Arranging them in ascending order gives $1, 2, 3, 4, 5, 10$, 11, 12, 13, $\cdots$. What is the position of 2015 in this sequence?
【Analysis and Solution】Base and Position Value.

Numbers composed of the digits $0, 1, 2, 3, 4, 5$ can be considered as numbers in base six.
$$
(2015)_{6}=\left(2 \times 6^{3}+0 \times 6^{2}+1 \times 6^{1}+5 \times 6^{0}\right)_{10}=(2 \times 216+0 \times 36+1 \times 6+5 \times 1)_{10}=(443)_{10} ;
$$

Therefore, 2015 is the 443rd number in this sequence.