In order to leave a remainder of 10, the
division must look like this:

     Q
 x)100
    90
    10

So x must be an integer greater than the remainder 10.
Also in order to leave a 10 remainder, the integer 
quotient times the divisor, x, must equal to 90.  
Therefore we have a case of this for every way 
we can break 90 into two factors x and Q, with x 
being greater than 10.  There are 5 such ways:

x*Q = 15*6, 18*5, 30*3, 45*2, 90*1   

The sum of the smallest and largest values of x is
15+90 or 105.

     6
15)100
    90
    10

     5
18)100
    90
    10

     3
30)100
    90
    10

     2
45)100
    90
    10

     1
90)100
    90
    10

Edwin