let
x------> the first integer
y------> the second integer
z-----> the third integer
x < y < z < 25
we know that
Median=13
Median=(x+y+z)/3
13=(x+y+z)/3
x+y+z=39------> equation 1
range=10
The range is the difference between the largest number within the set and the smallest number in the set
so
range=z-x
z-x=10------> z=x+10-----> equation 2
to find the solution we will assume different values of x (See the attached table)
terms
x < y < z < 25
z=x+10
y=39-z-x
The solution can be
7,15,17
8,13,18
9,11,19
