Question

四个两位数的和是 $221$ 。这八个数字中没有 $0$ ，且互不相同。下列哪一个数字没有出现在这八个数字中？

The sum of four two-digit numbers is $221.$ None of the eight digits is $0$ and no two of them are the same. Which of the following is not included among the eight digits?

$1$

$2$

$3$

$4$

$5$

答案：D

知识点：数字位值分类讨论

难度评级：1660

解答：

八个数字来自 $1$ 到 $9$ ，全部非零数字的和为 $45$ ，所以这八个数字的总和在 $45 - 9 = 36$ 与 $45 - 1 = 44$ 之间。

设个位数字和为 $U$ ，十位数字和为 $T$ ，则 $10T + U = 221$ ，所以 $U$ 的个位为 $1$ 。又 $1+2+3+4 = 10$ $\le U \le$ $6+7+8+9 = 30$ ，故 $U = 11$ 或 $U = 21$ 。

若 $U = 11$ 则 $10T = 210$ ，所以 $T = 21$ ，八个数字总和为 $32$ ，小于 $36$ ，不可能。因此 $U = 21$ ， $T = 20$ ，总和为 $41$ 。

缺失数字为 $45 - 41 = 4$ 。例如 $13 + 25 + 86 + 97 = 221$ 。

所以正确答案是 D。

The eight digits are distinct and chosen from $1$ through $9,$ whose total is $45.$ So the eight used digits sum to between $45 - 9 = 36$ and $45 - 1 = 44.$

Let the four units digits sum to $U$ and the four tens digits sum to $T.$ Then $10T + U = 221,$ so $U$ ends in $1.$ Since $1+2+3+4 = 10$ $\le U \le$ $6+7+8+9 = 30,$ we have $U = 11$ or $U = 21.$

If $U = 11,$ then $10T = 210,$ so $T = 21$ and the eight digits sum to $32,$ which is below $36.$ So $U = 21,$ giving $T = 20$ and total $41.$

The missing digit is $45 - 41 = 4.$ For example, $13 + 25 + 86 + 97 = 221.$

Thus, the correct answer is D.

2005 AMC 12B 第 15 题

答案：D

难度评级：1660

解答：

其他年份的第 15 题