Question

两个非负整数的不降序列有不同的首项。每个序列都满足从第三项开始，每项都是前两项之和，并且两个序列的第七项都是 $N$ 。 $N$ 的最小可能值是多少？

Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N.$ What is the smallest possible value of $N?$

$55$

$89$

$104$

$144$

$273$

答案：C

知识点：斐波那契数列丢番图方程最大公约数

难度评级：1750

解答：

以 $a_1, a_2$ 开始的序列第七项为 $5a_1 + 8a_2$ 。对两个序列， $5a_1 + 8a_2 = 5b_1 + 8b_2$ ，所以 $5(b_1 - a_1) = 8(a_2 - b_2)$ 。因为 $\gcd(5, 8) = 1$ ，需要 $8 \mid b_1 - a_1$ 且 $5 \mid a_2 - b_2$ 。设 $a_1 \lt b_1$ ，并利用不降条件，有 $a_1 \le b_1 - 8 \le b_2 - 8 \le a_2 - 13$ 。取 $a_1 = 0$ 、 $b_1 = b_2 = 8$ 、 $a_2 = 13$ ，得到 $N = 5\cdot 0 + 8\cdot 13 = 104$ 。正确答案是 C。

A sequence starting $a_1, a_2$ has seventh term $5a_1 + 8a_2.$ For the two sequences, $5a_1 + 8a_2 = 5b_1 + 8b_2,$ so $5(b_1 - a_1) = 8(a_2 - b_2).$ Since $\gcd(5, 8) = 1,$ we need $8 \mid b_1 - a_1$ and $5 \mid a_2 - b_2.$ Taking $a_1 \lt b_1$ with nondecreasing terms gives $a_1 \le b_1 - 8 \le b_2 - 8 \le a_2 - 13.$ Choosing $a_1 = 0,$ $b_1 = b_2 = 8,$ $a_2 = 13$ yields $N = 5\cdot 0 + 8\cdot 13 = 104.$ Thus, the correct answer is C.

2013 AMC 12B 第 14 题

答案：C

难度评级：1750

解答：

其他年份的第 14 题