AMC12 2016 B

AMC12 2016 B · Q12

AMC12 2016 B · Q12. It mainly tests Casework, Logic puzzles.

All the numbers $1,2,3,4,5,6,7,8,9$ are written in a $3 \times 3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$. What number is in the center?

把所有数字 $1,2,3,4,5,6,7,8,9$ 写在一个 $3 \times 3$ 的方格阵列中，每个方格里写一个数，并且满足：如果两个数字是相邻的连续整数，那么它们所在的方格必须有一条公共边相邻。四个角上的数字之和为 $18$。问：中心格里是什么数字？

(A) 5 5

(B) 6 6

(D) 8 8

(E) 9 9

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Shade the squares in a checkerboard pattern as shown in the first figure. Because consecutive numbers must be in adjacent squares, the shaded squares will contain either five odd numbers or five even numbers. Because there are only four even numbers available, the shaded squares contain the five odd numbers. Thus the sum of the numbers in all five shaded squares is $1+3+5+7+9=25$. Because all but the center add up to $18=25-7$, the center number must be 7. The situation described is actually possible, as the second figure demonstrates.

答案（C）：按第一幅图所示以棋盘格方式给方格涂色。由于连续的数字必须位于相邻的方格中，被涂色的方格将包含五个奇数或五个偶数。因为可用的偶数只有四个，所以被涂色的方格包含五个奇数。因此，五个被涂色方格中的数字之和为 $1+3+5+7+9=25$。因为除中心外其余数字之和为 $18=25-7$，所以中心数字必须是 7。第二幅图表明所描述的情况确实是可能的。

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