I feel trial and error, with intuition is the best way to solve this in the exam.
However, a formal way I've seen a while ago in an aptitude book:
Let the number be $N$
$N = 11mod(13)$
$N = 9mod(17)$
You can deduce that the given number cannot go beyond the LCM of $13$ and $17 = 221$. Else the remainders would start repeating, and you wouldn't get the lowest possible solution.
Now examine the 2 congruencies individually. Start with the remainder in each case, and increase by 13 or 17. The problem is to find the first common element.
analysing $N =11mod(13)$, $N$ could be $11,24,37,50,63,76,89,102,115,128,141,154,...$
analysing $N = 9mod(17)$, $N$ could be $9,26,43,60,77,94,111,128, 145, 162,...$
So you could conclude that 128 is the smallest such number.
Clearly, our dirty shortcuts are better and more satisfying.