2018 AMC 12A Problems/Problem 6
Problem
For positive integers and such that , both the mean and the median of the set are equal to . What is ?
Solution 1
The mean and median are so and . Solving this gives for . (trumpeter)
Solution 2
This is an alternate solution if you don't want to solve using algebra. First, notice that the median is the average of and . Therefore, , so the answer is , which must be odd. This leaves two remaining options: and . Notice that if the answer is , then is odd, while is even if the answer is . Since the average of the set is an integer , the sum of the terms must be even. is odd by definition, so we know that must also be odd, thus with a few simple calculations is odd. Because all other answers have been eliminated, is the only possibility left. Therefore, . ∎ --anna0kear
Solution 3
Since the median is , then , or . Plug this in for values to get . Plug it back in to get , thus .
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Video Solution 1
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See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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All AMC 12 Problems and Solutions |
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