Difference between revisions of "2020 AMC 8 Problems/Problem 20"
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For the heights of the trees to be integers, we must have the height of Tree 1, Tree 2, Tree 3 as <math>22, 11, 22</math>. Now, there are two possible cases. | For the heights of the trees to be integers, we must have the height of Tree 1, Tree 2, Tree 3 as <math>22, 11, 22</math>. Now, there are two possible cases. | ||
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Case 1: The length of tree 4 is <math>11</math>- | Case 1: The length of tree 4 is <math>11</math>- |
Revision as of 08:11, 18 November 2020
A scientist walking through a forest recorded as integers the heights of trees standing in a row. She observed that each tree was either twice 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?
Solution 1
It is not too hard to construct as the heights of the trees from left to right. The average is
.
Solution 2
For the heights of the trees to be integers, we must have the height of Tree 1, Tree 2, Tree 3 as . Now, there are two possible cases.
Case 1: The length of tree 4 is -
So, we have the first 4 tree lengths as
. For the length of Tree 5 to be an integer, we must have the length of Tree
as
. But, when we take the average of these
integers, it results in
, which doesn't satisfy our conditions.
Case 2: The length of tree 4 is -
So, we have the first
tree lengths as
. Now, using quick modular arithmetic, we see that when the length of Tree 5 is
, the average of the heights of the 5 trees is
. This is where our condition is satisfied.
~ATGY
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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