AMC10 2004 B

AMC10 2004 B · Q21

AMC10 2004 B · Q21. It mainly tests Arithmetic sequences basics, GCD & LCM.

Let 1, 4, \dots and 9, 16, \dots be two arithmetic progressions. The set $S$ is the union of the first 2004 terms of each sequence. How many distinct numbers are in $S$?

让 1, 4, \dots 和 9, 16, \dots 是两个等差数列。集合 $S$ 是每个数列的前 2004 项的并集。$S$ 中有多少个不同的数？

(A) 3722 3722

(B) 3732 3732

(D) 3924 3924

(E) 4007 4007

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A) The smallest number that appears in both sequences is 16. The two sequences have common differences 3 and 7, whose least common multiple is 21, so a number appears in both sequences if and only if it is in the form $$16+21k,$$ where $k$ is a nonnegative integer. Such a number is in the first 2004 terms of both sequences if and only if $$16+21k\le 1+2003(3)=6010.$$ Thus $0\le k\le 285$, so there are 286 duplicate numbers. Therefore the number of distinct numbers is $4008-286=3722$.

（A）两个数列中同时出现的最小数是 16。两个数列的公差分别为 3 和 7，它们的最小公倍数是 21，因此一个数同时出现在两个数列中当且仅当它具有形式 $$16+21k,$$ 其中 $k$ 为非负整数。这样的数同时属于两个数列的前 2004 项当且仅当 $$16+21k\le 1+2003(3)=6010.$$ 因此 $0\le k\le 285$，所以重复的数有 286 个。因此不同的数的个数为 $4008-286=3722$。

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