Question

小于或等于 $2017$ 的正整数中，十进制表示含有数字 $0$ 的有多少个？

How many of the base-ten numerals for the positive integers less than or equal to $2017$ contain the digit $0?$

$469$

$471$

$475$

$478$

$481$

答案：A

知识点：数字补集计数

难度评级：1480

解答：

小于 $100$ 的数中，只有 $10$ 的倍数才含有数字 $0$ ，这样的数有 $9$ 个。

对 $100$ 到 $999$ （含端点）的数用补集计数。这个范围共有 $900$ 个数。没有 $0$ 的数有 $9\cdot 9\cdot 9=729$ 个，因为每一位都有 $9$ 种选择使它不是 $0$ 。因此这个范围内共有 $171$ 个。

对 $1000$ 到 $1999$ （含端点）的数再次用补集计数。这个范围共有 $1000$ 个数。没有 $0$ 的数有 $1\cdot 9\cdot 9\cdot 9=729$ 个，因为后 $3$ 位各有 $9$ 种选择使它不是 $0$ ，首位必须是 $1$ 。因此这个范围内共有 $271$ 个。

从 $2000$ 到 $2017$ （含端点）共有 $18$ 个数，每个数从左数第二位都是 $0$ 。

总数为 $9+171+271+18=469.$

所以正确答案是 A。

For numbers less than $100,$ we only have a $0$ if it is a multiple of $10,$ of which there are $9.$

For numbers between $100$ and $999$ inclusive, we will use complementary counting. There are $900$ total numbers in this range. Also, there are $9\cdot 9\cdot 9=729$ numbers in this range with no $0$ since there are $9$ ways to choose each digit to not be $0.$ Thus, the total in this range is $171.$

For numbers between $1000$ and $1999$ inclusive, we will use complementary counting again. There are $1000$ total numbers in this range. Also, there are $1\cdot 9\cdot 9\cdot 9=729$ numbers in this range with no $0$ since there are $9$ ways to choose each of the last $3$ digits to not be $0$ and the first digit must be $1.$ Thus, the total in this range is $271.$

There are $18$ numbers between $2000$ and $2017$ inclusive, each with a $0$ in the second digit from the left.

This makes the total $9+171+271+18=469.$

Thus, the correct answer is A .

2017 AMC 10B 第 16 题

答案：A

难度评级：1480

解答：

其他年份的第 16 题