1997 AMC 12 第 16 题

Question

数组

\begin{bmatrix} 4&9&2\\ 8&1&6\\ 3&5&7 \end{bmatrix}

的三个行和与三个列和都相同。至少要改变多少个元素，才能使这六个和互不相同？

The three row sums and the three column sums of the array

\begin{bmatrix} 4&9&2\\ 8&1&6\\ 3&5&7 \end{bmatrix}

are the same. What is the least number of entries that must be altered to make all six sums different from one another?

1

2

3

4

5

答案：D

知识点：arrays invariants construction

难度评级：1860

小提示：

假设只改变三个元素，考察其中没有元素被改变的行和列

With only three altered entries, examine the rows and columns containing no alteration

大提示：

为了给出上界，尝试改变四个元素，使它们对各行和各列造成的变化都不同

For an upper bound, try changing four entries so that their row and column effects are all distinct

解答：

如果至多改变三个元素，那么要么六条行列中至少有两条没有改变，要么某个被改变的元素是它所在行与所在列中唯一被改变的元素。前一种情况下，两条未改变的行或列的和仍然相等；后一种情况下，该元素使其所在行与列的和发生同样的变化，所以这两个和仍然相等。改变四个元素即可：分别将

4

、

1

、

2

、

6

改为

5

、

3

、

7

、

9

。新的三个行和为

21

、

20

、

15

，三个列和为

16

、

17

、

23

。因此最小值为

4

，正确答案是 D。

With at most three alterations, either two of the six lines are unchanged, or some altered entry is the only alteration in both its row and its column. In the first case those two sums remain equal; in the second, that entry changes its row sum and column sum by the same amount, so those two sums remain equal. Four suffice: replace

4,

1,

2,

6

by

5,

3,

7,

9,

respectively. The resulting row sums are

21,

20,

15

and column sums are

16,

17,

23.

Hence the minimum is

4,

and D is correct.

1997 AMC 12 第 16 题

答案：D

难度评级：1860

小提示：

大提示：

解答：

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