Maths Problems!!!

I prefer Eric Fong's answer more.

First of all, the sum of the 4th and 5th integers (1st is the smallest, the 8th is the largest) must be 16 since the median is 8 while the number of integers are even.

Secondly, the sum of the remaining 6 integers must be 48 because it adds the 4th and the 5th to make it 64 (mean is 8 for 8 integers).

Thirdly, the 6th and the 7th integers must not be smaller than 8, and as small as possible for making the largest 8th integer. So, they can both only be 8. Then, the 4th and the 5th will be 8 as well.

Fourthly, the smaller three integers must be as small as possible for the same purpose, but the difference between the smallest and the largest (range) is 8. So, those three integers must be the same. Using simultaneous equations, we'll have

x - y = 8 --- (1)
x + 3y = 32 --- (2)

where x and y are the largest and the smallest integers respectively.

Solving (1) and (2), we'll have the smallest and the largest integers as 6 and 14 respectively. In other words, the answer is 14.

In the list A, mean, median, mode and the range are all 8.