2013 AMC 12B 第 14 题

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

小提示：

用前两项

a_1, a_2

表示，第七项等于

5a_1 + 8a_2

。

In terms of the first two terms

a_1, a_2,

the seventh term equals

5a_1 + 8a_2

大提示：

令两个第七项相等，得到

5(b_1 - a_1) = 8(a_2 - b_2)

；由于

5

与

8

互质，必须有

b_1 - a_1

是

8

的倍数。

Setting the two seventh terms equal gives

5(b_1 - a_1) = 8(a_2 - b_2);

coprimality of

5

and

8

forces

b_1 - a_1

to be a multiple of

8

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

b_1 - a_1

是

8

的倍数，且

a_2 - b_2

是

5

的倍数。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

b_1 - a_1

to be a multiple of

8

and

a_2 - b_2

to be a multiple of

5.

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

小提示：

大提示：

解答：

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