AMC10 2023 A

AMC10 2023 A · Q12

AMC10 2023 A · Q12. It mainly tests Basic counting (rules of product/sum), Divisibility & factors.

How many three-digit positive integers $N$ satisfy the following properties?

有多少个三位正整数$N$满足以下性质？

(A) 13 13

(B) 14 14

(D) 16 16

(E) 17 17

Answer

Correct choice: (B)

正确答案：（B）

Solution

Multiples of $5$ will always end in $0$ or $5$, and since the numbers have to be a three-digit numbers, it cannot start with 0 (otherwise it would be a two-digit number), narrowing our choices to 3-digit numbers starting with $5$. Since the numbers must be divisible by 7, all possibilities have to be in the range from $7 \cdot 72$ to $7 \cdot 85$ inclusive(504 to 595). (Add 1 to include 72) $85 - 72 + 1 = 14$. $\boxed{\textbf{(B) } 14}$. You can also take 497 away from each of the numbers(removing the hundreds digit and adding three to each of the numbers), resulting in the numbers {7, 14, 21..., 84, 91, 98}. Dividing each of them by 7, you get the numbers {1, 2, 3..., 12, 13, 14}. Therefore, the answer is $\boxed{\textbf{(B) 14}}$

5的倍数总是以$0$或$5$结尾，且由于是三位数，不能以0开头（否则是两位数），因此缩小范围为以$5$开头的三位数。由于必须能被7整除，所有可能数都在$7 \cdot 72$到$7 \cdot 85$之间（即504到595）。（加1包含72） $85 - 72 + 1 = 14$。$\boxed{\textbf{(B) } 14}$。也可以从每个数中减去497（去掉百位并每个数加3），得到{7, 14, 21..., 84, 91, 98}。除以7得到{1, 2, 3..., 12, 13, 14}。因此答案为$\boxed{\textbf{(B) 14}}$

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