AMC10 2016 B

AMC10 2016 B · Q15

AMC10 2016 B · Q15. It mainly tests Logic puzzles, Parity (odd/even).

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?

把数字 1、2、3、4、5、6、7、8、9 全部填入一个 $3 \times 3$ 的方格阵列中，每个小方格填一个数，并且要求：如果两个数相邻（差为 1），那么它们所在的方格必须共享一条边。四个角上的数字之和为 18。问：中心格里的数字是多少？

(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）：如第一幅图所示，将方格按棋盘格（黑白相间）的方式涂色。由于相邻的方格中必须放置连续的数字，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，所以涂色的方格包含五个奇数。因此，五个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，这种情况确实可以发生。

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