AMC10 2013 B

AMC10 2013 B · Q21

AMC10 2013 B · Q21. It mainly tests Sequences & recursion (algebra), Number theory misc.

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?

有两个非递减的非负整数序列，它们的首项不同。每个序列从第三项开始，每一项都是前两项之和，并且每个序列的第七项都是 N。N 的最小可能值是多少？

(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）：设两个数列分别为 $(a_n)$ 和 $(b_n)$，不失一般性地假设 $a_1<b_1$。由数列的定义可得 $a_7=5a_1+8a_2=5b_1+8b_2$，因此 $5(b_1-a_1)=8(a_2-b_2)$。由于 5 与 8 互素，故 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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