Question

Positive integers a, b, c, d and e are such that a < b < c < d < e. If the average (arithmetic mean) of the five numbers is 6 and d − b = 3, then what is the greatest possible range of the five numbers?

Answer 1

Answer: Range(max) = e(max)-a(min) = 19-1 =18

Explanation:

The Range is expressed as e-a.

Given that d-b=3 or d=b+3 and Average = 6 thus a+b+c+d+e=30

To maximize range, we need to minimize a and maximize e. Minimum a is 1, so we choose that.

Since sum is fixed, we should choose minimum possible values for b,c,d to maximize e.

b>a, minimum choice is 2

c>b, minimum choice is 3

d=b+3 so d=5

So e(max) = 30 - a(min) -b(min) -c(min) - d(min) = 30-1-2-3-5 = 19

Range(max) = e(max)-a(min) = 19-1 =18