13. List the natural numbers that can be divided by both 5 and 7 in ascending order, starting from 105, and take the first 2013 numbers. The remainder when the sum of these 2013 numbers is divided by 12 is $\qquad$ _.
【Analysis】It is to find the sum of the first 2013 terms of an arithmetic sequence with the first term being 105 and the common difference being $[5,7]=35$. The 2013th term is $105+(2013-1) \times 35=35 \times 3+35 \times 2012=35 \times 2015$. The sum of the first 2013 terms is $(105+35 \times 2015) \times 2013 \div 2=35 \times 1009 \times 2013$. $35 \times 1009 \times 2013 \equiv 11 \times 1 \times 9 \equiv 3(\bmod 12)$, this sum modulo 12 is 3.