Then you will either --

have to keep trying two-digit numbers from the smallest
two-digit number, 10, all the way through the largest one, 99, and add 18 to
every one of them to see if the answer is the number with the digits reversed.  

Or else --

you can do a little algebra. 

Let t= the tens digit and u= the units or ones digit.

Then (10t+u) is the two digit number.

and (10u+t) is its reverse.

So we increase (10t+u) by 18 and we get the reverse, (10u+t)

To say that in algebra,

(10t+u) + 18 = (10u+t)

    10t+u+18 = 10u+t

    9t-9u+18 = 0

Divide through by 9

   t - u + 2 = 0

Add u to both sides:

        t + 2 = u

That formula is saying "You have to add 2 to the ten's digit 
to get the units digit".  So we have to choose the tens digit 
from 1 through 7, because you can't add 2 to 8 or 9 and get 
a single digit, because you have to then choosethe units digit
as 2 more than the tens digit.  So we just have these 7 numbers:

13, 24, 35, 46, 57, 68, 79

Let's test them all:

13+18=31 That reverses when you add 18.
24+18=42  "        "     "   "   "   " 
35+18=53  "        "     "   "   "   "      
46+18=64  "        "     "   "   "   "
57+18=75  "        "     "   "   "   "        
68+18=86  "        "     "   "   "   "
79+18=97  "        "     "   "   "   "

So they all reverse the number when you add 18.  
That was better than having to add 18 to them all
the way from 10 to 99. 

But you really even don't have to do all the above 
every time.  You can just look at the equation 
t + 2 = u  and see there are only 7 solutions in 
positive integers, because t can't run over 7, so
t can only be an integer from 1 through 7.

Edwin