AMC10 2022 B

AMC10 2022 B · Q8

AMC10 2022 B · Q8. It mainly tests Rounding & estimation, Divisibility & factors.

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*} 其中有多少个集合恰好包含两个 $7$ 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 $\text{floor}\left(\frac{1000}{7}\right) = 142$ 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 $1.42 - 1 = \boxed{\textbf{(B)}\ 42}.$

$1000$ 中有 $\lfloor\frac{1000}{7}\rfloor = 142$ 个 $7$ 的倍数。将这些分配到 $100$ 个每个含 $10$ 个数的集合中，每集合平均 $1.42$ 个 $7$ 的倍数。第一组后，其余均匀分布，因此有 $100 \times (1.42 - 1) = \boxed{\textbf{(B)}\ 42}$ 个集合恰好有两个。

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