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