Question 819681
Let a = the 100's digit
Let b = the 10's
Let c = the units
then
100a+10b+c = "the number"
:
the sum of the digits of a number of three digits is 18.
a + b + c = 18
:
 if the number is divided by the sum of the ten's and hundred's digits.
 the quotient is 38 and the remainder is 1.
{{{((100a+10b+c-1))/((a+b))}}} = 38
Multiply by (a+b)
100a + 10b + c - 1 = 38(a+b)
100a + 10b + c - 1 = 38a + 38b
100a - 38a + 10b - 38b + c = 1
62a - 28b + c = 1
:
if 99 is added to the number, the digits will be revised.
100a+10b+c+99 = 100c+10b+a
100a - a + 10b - 10b + c -  100c = -99
99a  - 99c = -99
Divide by 99
a - c = -1
or
a + 1 = c
In the 1st equation, replace c with (a+1)
a + b + (a+1) = 18
2a + b = 17
In the 2nd equation, replace c with (a+1)
62a - 28b + a + 1 = 1
63a - 28b = 1 - 1
63a - 28b = 0
Use elimination here multiply by 2a + b = 17 by 28, add to the above 
56a + 28b = 476
63a - 28b = 0
---------------adding eliminates b find a
119a = 476
a = 476/119
a = 4
then
c = 4 + 1
c = 5
Find b
4 + b + 5 = 18
b = 18 - 9
b = 9
:
495 is the number
:
See if the works out when you add 99, does it reverse the digits?
495 + 99 = 594