2020 AMC 12A 第 4 题

先试着解答 2020 AMC 12A 第 4 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 2020 AMC 12A 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。

4.

有多少个 44 位正整数,也就是从 10001000 到 99999999(含端点)的整数,只含偶数数字且能被 55 整除?
How many 44-digit positive integers (that is, integers between 10001000 and 9999,9999, inclusive) having only even digits are divisible by 5?5?
8080
100100
125125
200200
500500
答案:B
知识点:乘法原理整除性数字
小提示:
能被 55 整除的数末位是 00 或 55,但末位还必须是偶数
A number divisible by 55 ends in 00 or 5,5, but the last digit must also be even
大提示:
分别数每一位的选择数:首位不能是 00。
Count the choices for each digit: the leading digit cannot be 00
解答:
要能被 55 整除,末位必须是 00 或 55,又因为所有数字都要是偶数,所以末位只能是 00。
首位是非零偶数:2,4,6,82, 4, 6, 8,共有 44 种选择。中间两位各可以是任意偶数 0,2,4,6,80, 2, 4, 6, 8,各有 55 种选择。
总数为 4⋅5⋅5⋅1=1004 \cdot 5 \cdot 5 \cdot 1 = 100。
因此,正确答案是 B。
To be divisible by 55 the last digit is 00 or 5,5, and to be even it must be 0.0. So the units digit is fixed.
The leading digit is a nonzero even digit: 2,4,6,82, 4, 6, 8 give 44 choices. Each of the two middle digits is any even digit 0,2,4,6,8,0, 2, 4, 6, 8, giving 55 choices each.
The total is 4⋅5⋅5⋅1=100.4 \cdot 5 \cdot 5 \cdot 1 = 100.
Thus, B is the correct answer.
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