12. A sequence of positive integers, the first term is 8, the second term is 1, and from the third term onwards, each term is equal to the sum of the two preceding terms. What is the remainder when the 2013th term is divided by 105? $\qquad$
[Answer] 16
[Solution] Since $105=3 \times 5 \times 7$, we first calculate the remainders when divided by $3, 5, 7$.
(1) Remainder when divided by 3: $\underbrace{2,1,0,1,1,2,0,2}_{\text {one cycle }}, 2,1,0,1, \cdots$, since $2013=8 \times 251+5$, the remainder is 1
(2) Remainder when divided by 5: $\underbrace{3,1,4,0,4,4,3,2,0,2,2,4,1,0,1,1,2,3,0,3}_{\text {one cycle }}, 3,1,4,0, \cdots$, since $2013=20 \times 100+13$, the remainder is 1
(3) Remainder when divided by 7: $\underbrace{1,1,2,3,5,1,6,0,6,6,5,4,2,6,1,0}_{\text {one cycle }}, 1,1,2,3, \cdots$, since $2013=16 \times 125+13$, the remainder is 2

In summary, this number leaves a remainder of 1 when divided by 3; a remainder of 1 when divided by 5; and a remainder of 2 when divided by 7. It is easy to see that the smallest such number is 16, so the final remainder is 16