Solution:

( Trong dãy số 1, 1, 2, 3, 5, 8, 13, 21, …. Mỗi số hạng kể từ số thứ 3 bằng tổng của 2 số kề phía trước nó. Hỏi có bao nhiêu giữa 2014 số đầu tiên chia hết cho 4 ? )

The remainder when each term in the sequence divided by 4 is listed as 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, …. We discover the recurring period is 6 and one of the remainder is 0, it means one term in every recurring period in the sequence is divisible by 4.

Since 2014 = 335 x 6 + 4, it means that in the first 2014 terms in the sequence, there are 335 complete period with an incomplete period of four remaining terms. Therefore, there are 335 terms in the sequence that are divisible by 4