10. Frank's teacher asks him to write down five integers such that the median is one more than the mean, and the mode is one greater than the median. Frank is also told that the median is 10 .
What is the smallest possible integer that he could include in his list?
A 3
B 4
C 5
D 6
E 7
Solution
B The median of the five numbers is 10 , hence the mean is 9 and the mode is 11 . For the mode to be 11 , Frank's list must include more 11s than any other integer. It cannot include three or more $11 \mathrm{~s}$, since then the median would be 11 . So two of the integers are 11 , and the other three integers are all different. For the median to be 10,10 must be one of the integers, and the list must include two integers that are less than 10 and two that are greater than 10 . Hence the five integers are $a, b, 10,11,11$ where $a, b$ are two different integers both less than 10. Say $a<b$. Since the mean is 9 , we have $a+b+10+11+11=5 \times 9=45$ and therefore $a+b=13$. The largest value that $b$ can take is 9 , and hence the smallest possible value for $a$ is 4 .