AMC 8 2020 Problem 20
AMC8 2020 年第 20 题
奇偶性★★★☆☆
A scientist walking through a forest recorded as integers the heights of 5 trees standing in a row. She observed that each tree was either 2 times as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
一个穿过森林的科学家将一排 5 棵树的高度都记录了下来，这 5 棵树的高度均为整数。她观察到，每棵树的高度要么是右边树高的 2 倍，要么是它的一半。不幸的是，由于雨水落在她的笔记本上，导致部分数据丢失。她的笔记如下图所示，空白部分表示丢失的数据。根据她的观察，这个科学家最终能够重建丢失的数据。那么这 5 棵树的平均高度是多少米？
- A.22.2
- B.24.2
- C.33.2
- D.35.2
- E.37.2
正确答案 · Correct Answer
B
解析 · Solution
Solution 1
We will show that 22, 11, 22, 44, and 22 meters are the heights of the trees from left to right. We are given that all tree heights are integers, so since Tree 2 has height 11 meters, we can deduce that Trees 1 and 3 both have a height of 22 meters. There are now three possible cases for the heights of Trees 4 and 5 (in order for them to be integers), namely heights of 11 and 22, 44 and 88, or 44 and 22. Checking each of these, in the first case the average is 17.6 meters, which doesn't end in.2 as the problem requires. With 44 and 88, the average is 37.4 meters, which again does not end in.2, but with 44 and 22 the average is 24.2 meters, which does. Consequently, the answer is 24.2.
Alternatively, notice the average height ends with 0.2; therefore, the sum of all five heights must end with 1 or 6 (since ). We already know Tree 2 is 11 meters tall, and both Tree 1 and Tree 3 must be 22 meters tall. Tree 4 can't be 11 meters for the sum to still end with 1, so Tree 4 is 44 meters tall. Now Tree 5 can be 22 or 88; finding the average for both cases, Tree 5 must be 22 for the average to end with 0.2, giving an average height of 24.2.
As before, we shall show that the heights of the trees are 22, 11, 22, 44, and 22 meters. Let be the sum of the heights, so that the average height will be meters. We note that , so in order for to end in.2, must be one more than a multiple of 5. Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both 22 meters. So far we have Tree 1 = 22, Tree 2 = 11, Tree 3 = 22, with Tree 4 and Tree 5 unknown and the average ending in.2 meters. If Tree 4 has a height of 11, then Tree 5 would need to have height 22, but in that case would equal 88, which is not 1 more than a multiple of 5. So we instead take Tree 4 to have height 44. Then the sum of the heights of the first 4 trees is 22+11+22+44 = 99, so using a height of 22 for Tree 5 gives S=121, which is 1 more than a multiple of 5 (whereas 88 gives S = 187, which is not). Thus the average height of the trees is meters.
Since we know that the tree heights have to be integers, then it is immediate that Tree 1 and 3 have a height of 22. Now using the information given by the last column (that the average of the heights of the trees ends in.2 ), we can tell that the sum of all the heights of the trees ends in either 1 or 6, because those are the only numbers from 0 to 9 that are congruent to 1 after taking modulo 5. The two multiples of eleven (eleven because all of the tree heights have to be a multiple of eleven if they are integers) that come to mind are 66 and 121. Since the sum of the heights of Tree 1, 2, and 3 is already 55, we know that 66 is impossible to obtain. Then, we can decide with relative confidence that the answer should be . P.S.: We can check our solution by trying to obtain 121. Surely enough, we have Tree 1 = 22, Tree 2 = 11, Tree 3 = 22, Tree 4 = 44, and Tree 5 = 22. These all add up to get 121. Since her average ends with 0.2, the sum of her tree heights must be . If Tree 1's height is , Tree 3's is , Tree 4's is , and Tree 5's is , then we and are both 22, since all of the tree heights are integers. Now we have . Simplifying, we get . The only possible combination of and that abides the condition are 44 and 22, respectively. Adding these up, we get 121, and is 24.2. -kn07
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