AMC12 2013 B

AMC12 2013 B · Q14

AMC12 2013 B · Q14. It mainly tests Sequences & recursion (algebra), Diophantine equations (integer solutions).

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 $N$?

两个非递减的非负整数序列具有不同的首项。每个序列从第三项开始，每项是前两项之和，且每个序列的第七项均为$N$。$N$的最小可能值为多少？

(A) 55 55

(B) 89 89

(D) 144 144

(E) 273 273

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let the two sequences be $(a_n)$ and $(b_n)$, and assume without loss of generality that $a_1 < b_1$. The definitions of the sequences imply that $a_7 = 5a_1 + 8a_2 = 5b_1 + 8b_2$, so $5(b_1 - a_1) = 8(a_2 - b_2)$. Because $5$ and $8$ are relatively prime, $8$ divides $b_1 - a_1$ and $5$ divides $a_2 - b_2$. It follows that $a_1 \le b_1 - 8 \le b_2 - 8 \le a_2 - 13$. The minimum value of $N$ results from choosing $a_1 = 0$, $b_1 = b_2 = 8$, and $a_2 = 13$, in which case $N = 104$.

答案（C）：设两个数列为$(a_n)$和$(b_n)$，不失一般性，假设$a_1 < b_1$。数列的定义推出 $a_7 = 5a_1 + 8a_2 = 5b_1 + 8b_2$，因此$5(b_1 - a_1) = 8(a_2 - b_2)$。由于$5$和$8$互素，所以$8$整除$b_1 - a_1$，且$5$整除$a_2 - b_2$。于是有$a_1 \le b_1 - 8 \le b_2 - 8 \le a_2 - 13$。当取$a_1 = 0$，$b_1 = b_2 = 8$，$a_2 = 13$时，$N$取得最小值，此时$N = 104$。

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