HMMT 二月 2004 · 冲刺赛 · 第 5 题
HMMT February 2004 — Guts Round — Problem 5
题目详情
英文原题
- [6] Augustin has six 1 × 2 × π bricks. He stacks them, one on top of another, to forma tower six bricks high. Each brick can be in any orientation so long as it rests flat ontop of the next brick below it (or on the floor). How many distinct heights of towerscan he make?
√
√
解析
英文解析
- Augustin has six 1 × 2 × π bricks. He stacks them, one on top of another, to form atower six bricks high. Each brick can be in any orientation so long as it rests flat ontop of the next brick below it (or on the floor). How many distinct heights of towerscan he make?
Solution: 28
If there are k bricks which are placed so that they contribute either 1 or 2 height,
then the height of these k bricks can be any integer from k to 2 k . Furthermore,
towers with different values of k cannot have the same height. Thus, for each k thereare k + 1 possible tower heights, and since k is any integer from 0 to 6, there are
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 possible heights.
√
√