There are 5 one-digit positive integers; there are $5 \times 6=30$ two-digit positive integers; there are $5 \times 6^{2}=180$ three-digit positive integers; there are $5 \times 6^{3}=1080$ four-digit positive integers; among the five-digit positive integers starting with 1, the numbers with thousands place being $0,1,2$ each have $6^{3}=216$; among the five-digit positive integers starting with 1 and having thousands place as 3, the numbers with hundreds place being 0,1 each have $6^{2}=36$; at this point, the total number of positive integers that meet the criteria is $5+30+180+1080+3 \times 216+2 \times 36=2015$, so the 2021st number is 13205.
3. Arrange all positive integers whose digits do not exceed 5 in ascending order. What is the 2021st number?