1991 AMC 8 第 5 题

先试着解答 1991 AMC 8 第 5 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1991 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

5.

一个“多米诺骨牌”由两个小正方形组成:
。下面哪一个“棋盘”不能被若干个不重叠的完整多米诺骨牌恰好完全覆盖?
A “domino” is made up of two small squares:
. Which of the “checkerboards” illustrated below CANNOT be covered exactly and completely by a whole number of non-overlapping dominoes?
3×43 \times 4
3×53 \times 5
4×44 \times 4
4×54 \times 5
6×36 \times 3
答案:B
知识点:铺砖奇偶性
小提示:
每个多米诺骨牌恰好覆盖两个小正方形,所以先看每个棋盘有多少个小正方形
Each domino covers exactly two small squares, so focus on how many small squares each board has.
大提示:
棋盘要被完全覆盖,总格数必须是偶数;用两个维度相乘求每个棋盘的格数
A board can be fully covered only if its total number of squares is even; find the number of squares in each board by multiplying its two dimensions.
解答:
每个多米诺骨牌覆盖 22 个小正方形,所以能被不重叠多米诺完全覆盖的棋盘必须有偶数个小正方形。
各棋盘的格数为:3×4=123 \times 4 = 12,3×5=153 \times 5 = 15,4×4=164 \times 4 = 16,4×5=204 \times 5 = 20,6×3=186 \times 3 = 18。只有 3×5=153 \times 5 = 15 是奇数,所以不能被完全覆盖。(每个格数为偶数的棋盘都有一条边长是偶数,很容易用多米诺骨牌铺满。)
所以正确答案是 B。
Every domino covers exactly 22 squares, so any board that is completely covered by non-overlapping dominoes must contain an even number of small squares.
Counting squares: 3×4=12,3 \times 4 = 12, 3×5=15,3 \times 5 = 15, 4×4=16,4 \times 4 = 16, 4×5=20,4 \times 5 = 20, and 6×3=18.6 \times 3 = 18. Only 3×5=153 \times 5 = 15 is odd, so that board cannot be covered. (Each of the even boards has a side of even length and is easily tiled with dominoes.)
Thus, the correct answer is B .
LIVE by Po-Shen Loh
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