1993 AIME 第 1 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

1.

4000400070007000 之间,有多少个四位数字互不相同的偶数?

How many even integers between 40004000 and 70007000 have four different digits?

答案:728
知识点:基本计数分类讨论数字
难度评级:1920
小提示:

按千位数字是偶数还是奇数分类

Separate the cases according to whether the thousands digit is even or odd

大提示:

选定千位和个位数字后,依次计算两个中间数位的选择数

After choosing the thousands and units digits, count the choices for the two middle digits in order

解答:

千位数字是 445566。若千位数字是 4466,则个位数字有 44 种选择,即从 0022446688 中排除已经用过的千位数字;随后百位和十位数字分别有 88 种和 77 种选择。这两种情况共得到 2487=4482\cdot4\cdot8\cdot7=448 个数。若千位数字是 55,则 55 个偶数都可作为个位数字,共得到 587=2805\cdot8\cdot7=280 个数。因此总数为 448+280=728448+280=728

The thousands digit is 4,4, 5,5, or 6.6. If it is 44 or 6,6, the units digit has 44 choices among 0,0, 2,2, 4,4, 6,6, and 8,8, after which the hundreds and tens digits have 88 and 77 choices. These two cases contribute 2487=448.2\cdot4\cdot8\cdot7=448. If the thousands digit is 5,5, all 55 even units digits are available, contributing 587=280.5\cdot8\cdot7=280. Thus the total is 448+280=728.448+280=728.

完整试卷

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