AMC 10B/12B 2024 (Problem 15/10)
A list of 9 real numbers consists of and , as well as , , and , with . The range of the list is , and the mean and the median are both positive integers. How many ordered triples are possible?
The range is . Use this fact to organize the casework according to the minimum and maximum elements of the list.
Case 1: Minimum = 1, Maximum = 8
Since , we must have .
The mean is a positive integer, so
must be an integer.
Since ,
Thus, the total sum must be either or :
Therefore,
Since , we must have .
The mean is a positive integer, so
must be an integer.
Since ,
Thus, the total sum must be either or :
Therefore,
The list consists of
Since the median is an integer, it must be either or , because all the other numbers that could occupy the middle position are nonintegers.
Case 1.1:
For example,
In this case, and are both at most , so the median is , which is not an integer. ❌
Case 1.2:
Some possible pairs are
The only pair that gives an integer median is
The list in increasing order is
Therefore,
This gives the ordered triple
Since the median is an integer, it must be either or , because all the other numbers that could occupy the middle position are nonintegers.
Case 1.1:
For example,
In this case, and are both at most , so the median is , which is not an integer. ❌
Case 1.2:
Some possible pairs are
The only pair that gives an integer median is
The list in increasing order is
Therefore,
This gives the ordered triple
Case 2: Minimum = 0, Maximum = 7
Since , we have .
The list consists of
The mean must be an integer:
Since ,
Thus, the total sum must be either or :
Therefore,
Since , we have .
The list consists of
The mean must be an integer:
Since ,
Thus, the total sum must be either or :
Therefore,
The list consists of
Since the median must be an integer, it must be either or .
Case 2.1:
For example,
Here, both and are at most , so the median is , which is not an integer. ❌
Case 2.2:
Some possible pairs are
The only pair that gives an integer median is
The list in increasing order is
Therefore,
This gives the ordered triple
Since the median must be an integer, it must be either or .
Case 2.1:
For example,
Here, both and are at most , so the median is , which is not an integer. ❌
Case 2.2:
Some possible pairs are
The only pair that gives an integer median is
The list in increasing order is
Therefore,
This gives the ordered triple
Case 3: Minimum = , Maximum =
In this case,
Since the range is ,
The list consists of
For the median to be an integer, it must be . Consequently,
The mean must also be an integer:
Equivalently,
Since , we have . Since the mean is an integer, the total sum must be a multiple of .
Therefore,
In this case,
Since the range is ,
The list consists of
For the median to be an integer, it must be . Consequently,
The mean must also be an integer:
Equivalently,
Since , we have . Since the mean is an integer, the total sum must be a multiple of .
Therefore,
From Hint 6,
Since , the value is impossible. ❌
Thus,
Since or , we must have (because would give ). ❌
Therefore,
Since ,
Hence,
Since , the value is impossible. ❌
Thus,
Since or , we must have (because would give ). ❌
Therefore,
Since ,
Hence,
Combining the results of Hints 3, 5, and 7, the possible ordered triples are
Therefore, there are exactly
possible ordered triples.
Therefore, there are exactly
possible ordered triples.
C
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