6. There is a sequence of numbers arranged in a row, the 1st number is 1, the 2nd number is also 1, starting from the 3rd number, each number is the sum of the previous two numbers. Then the remainder when the 2025th number is divided by 4 is $(\quad)$.
(A) 0
(B) 1
(C) 2
(D) 3
6. C.

The remainders when this sequence of numbers is divided by 4 are $1,1,2,3,1,0,1,1,2$, $3,1,0, \cdots$, with a period of 6. Also, $2025=337 \times 6+3$, so the 2025th number is congruent to the 3rd number modulo 4, with a remainder of 2.