Problem:
A list of 9 real numbers consists of 1,2.2,3.2,5.2,6.2, and 7 , as well as x,y, and z with x≤y≤z. The range of the list is 7 , and the mean and the median are both positive integers. How many ordered triples (x,y,z) are possible?
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. infinitely many
Solution:
Because the range is 7, the values of x,y, and z are in the interval [0,8]. Because the median is an integer, it is one of x,y, or z and is either 3,4,5, or 6 . The sum of the list is s=24.8+x+y+z, which is between 24.8+0+0+3=27.8 and 24.8+6+8+8=46.8. Because the mean is an integer, s is an integer multiple of 9 , so s=36 or 45 , and x+y+z=11.2 or 20.2.
- If the median is 3 , then x≤y≤z=3, so x+y+z≤9<11.2. Therefore this case cannot occur.
- If the median is 4 , then x≤y=4≤z, so x+z=7.2, and the list spans one of the intervals [0,7],[1,8], or [x,z]. If x=0, then z=7.2>7, and if z=8, then x=−0.8<0. Therefore these cases cannot occur. Otherwise z−x=7, giving (x,y,z)=(0.1,4,7.1).
- If the median is 5 , then x≤y=5≤z, so x+z=6.2. In order to have a range of 7,x must be 0 , and (x,y,z)=(0,5,6.2).
- If the median is 6 , then x=6≤y≤z, so y+z=14.2. In order to have a range of 7,z must be 8 , and (x,y,z)=(6,6.2,8).
Thus there are (C)3​ possible ordered triples (x,y,z).
The problems on this page are the property of the MAA's American Mathematics Competitions