2017 AMC 10B 第 16 题

先试着解答 2017 AMC 10B 第 16 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2017 AMC 10B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

16.

小于或等于 20172017 的正整数中,十进制表示含有数字 00 的有多少个?
How many of the base-ten numerals for the positive integers less than or equal to 20172017 contain the digit 0?0?
469469
471471
475475
478478
481481
答案:A
知识点:数字补集计数
小提示:
数补集:不含数字 00 的数。
Count the complement: numbers with no digit 00
大提示:
按一位数、两位数、三位数和到 20172017 为止的四位数分开处理。
Split by one-, two-, three-, and four-digit numbers up to 20172017
解答:
小于 100100 的数中,只有 1010 的倍数才含有数字 00,这样的数有 99 个。
对 100100 到 999999(含端点)的数用补集计数。这个范围共有 900900 个数。没有 00 的数有 9⋅9⋅9=7299\cdot 9\cdot 9=729 个,因为每一位都有 99 种选择使它不是 00。因此这个范围内共有 171171 个。
对 10001000 到 19991999(含端点)的数再次用补集计数。这个范围共有 10001000 个数。没有 00 的数有 1⋅9⋅9⋅9=7291\cdot 9\cdot 9\cdot 9=729 个,因为后 33 位各有 99 种选择使它不是 00,首位必须是 11。因此这个范围内共有 271271 个。
从 20002000 到 20172017(含端点)共有 1818 个数,每个数从左数第二位都是 00。
总数为 9+171+271+18=469。9+171+271+18=469\text{。}
所以正确答案是 A。
For numbers less than 100,100, we only have a 00 if it is a multiple of 10,10, of which there are 9.9.
For numbers between 100100 and 999999 inclusive, we will use complementary counting. There are 900900 total numbers in this range. Also, there are 9⋅9⋅9=7299\cdot 9\cdot 9=729 numbers in this range with no 00 since there are 99 ways to choose each digit to not be 0.0. Thus, the total in this range is 171.171.
For numbers between 10001000 and 19991999 inclusive, we will use complementary counting again. There are 10001000 total numbers in this range. Also, there are 1⋅9⋅9⋅9=7291\cdot 9\cdot 9\cdot 9=729 numbers in this range with no 00 since there are 99 ways to choose each of the last 33 digits to not be 00 and the first digit must be 1.1. Thus, the total in this range is 271.271.
There are 1818 numbers between 20002000 and 20172017 inclusive, each with a 00 in the second digit from the left.
This makes the total 9+171+271+18=469.9+171+271+18=469.
Thus, the correct answer is A .
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