AMC12 2022 B

AMC12 2022 B · Q6

AMC12 2022 B · Q6. It mainly tests Divisibility & factors, Inequalities with integers (floor/ceiling basics).

Consider the following $100$ sets of $10$ elements each: \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} How many of these sets contain exactly two multiples of $7$?

考虑以下100个每个包含10个元素的集合： \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} 这些集合中有多少个包含恰好两个7的倍数？

(A) 40 40

(B) 42 42

(D) 49 49

(E) 50 50

Answer

Correct choice: (B)

正确答案：（B）

Solution

There are floor(10007)=142 numbers divisible by 7. We split these into 100 sets containing 10 numbers each, giving us 1.42 multiples of 7 per set. After the first set, the numbers come evenly, and we multiply 100 by $1.42 - 1 = \boxed{\textbf{(B)}\ 42}.$

有$\lfloor1000/7\rfloor=142$个被7整除的数。我们将这些分成100个每个包含10个数的集合，每集合平均1.42个7的倍数。第一集合后，数字均匀分布，我们计算100乘以$1.42 - 1 = \boxed{\textbf{(B)}\ 42}$。

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