AMC12 2000 A

AMC12 2000 A · Q4

AMC12 2000 A · Q4. It mainly tests Sequences in number theory (remainders patterns).

The Fibonacci sequence $1,1,2,3,5,8,13,21,\ldots$ starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?

斐波那契数列 $1,1,2,3,5,8,13,21,\ldots$ 以两个 1 开头，其后每一项是前两项之和。十个数字中，哪个数字最后出现在斐波那契数列中某一项的个位数上？

(A) 0 0

(B) 4 4

(D) 7 7

(E) 9 9

Answer

Correct choice: (C)

正确答案：（C）

Solution

Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in $\bmod{10}$: $1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$ The last digit to appear in the units position of a number in the Fibonacci sequence is $6 \Longrightarrow \boxed{\mathrm{C}}$.

注意，除个位数之外的任何数字都不会影响答案。为了加快计算，我们只需考察斐波那契数列在 $\bmod{10}$ 下的情况： $1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$ 最后一个出现在斐波那契数列某一项个位数上的数字是 $6 \Longrightarrow \boxed{\mathrm{C}}$。

Topics

NT.013 Sequences in number theory (remainders patterns)