Question

从集合 $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ 中选出六个不同数字，填入一个由竖直三格和水平四格组成、且两者共享一个方格的图形中，使竖直列三个数之和为 $23$ ，水平行四个数之和为 $12$ 。所用六个数字之和是

Six different digits from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ are placed in a figure made of a vertical column of three squares and a horizontal row of four squares that overlap in one shared square, so that the sum of the three entries in the vertical column is $23$ and the sum of the four entries in the horizontal row is $12$ . The sum of the six digits used is

$27$

$29$

$31$

$33$

$35$

答案：B

知识点：数字逻辑推理

难度评级：1090

解答：

从 $1$ 到 $9$ 中选三个不同数字和为 $23$ ，只能是 $6, 8, 9$ 。水平行另外三个数字最小为 $1 + 2 + 3 = 6$ ，所以共享方格至多是 $12 - 6 = 6$ 。因此共享数字是 $6$ 。

六个数字是 $6, 8, 9$ 和 $1, 2, 3$ ，总和为 $29$ 。等价地， $23 + 12 - 6 = 29$ 。

所以正确答案是 B。

Three distinct digits from $1$ through $9$ summing to $23$ must be $6, 8, 9$ . The row's other three digits are at least $1 + 2 + 3 = 6$ , so the shared square (belonging to both the column and the row) is at most $12 - 6 = 6$ . Hence the shared digit is $6$ .

The six digits are then $6, 8, 9$ and $1, 2, 3$ , whose sum is $29$ . Equivalently, $23 + 12 - 6 = 29$ .

Thus, the correct answer is B .

1996 AMC 8 第 14 题

答案：B

难度评级：1090

解答：

其他年份的第 14 题