Smallest number in a set 
A is the set of seven consequtive two digit numbers, none of these being multiple of 10. Reversing the numbers in set A forms numbers in set B. The difference between the sum of elements in set A and those in set B is 63. The smallest number in set A can be :

I tried to write some sets and reverse them and calculate their value, but I am not able to arrive at the answer. 

Let $A = \{10a+b,10a+b+1,10a+b+2,10a+b+3,10a+b+4,10a+b+5,10a+b+6\}$, where $a \in \{1,2,\ldots,9\}$ and since none of them is divisible by $10$, we have that $b\in\{1,2,3\}$. Then $$B =\{10b+a,10b+a+10,10b+a+20,10b+a+30,10b+a+40,10b+a+50,10b+a+60\}$$
Sum of elements in $A$ is $70a+7b+21$ and sum of elements in $B$ is $70b+7a+210$. We are given $$\vert 63(a-b) - 189 \vert = 63 \implies \vert (a-b) - 3 \vert = 1$$
If $b=1$, we get that $\vert a-4\vert = 1 \implies a=3,5$.
If $b=2$, we get that $\vert a-5\vert = 1 \implies a=4,6$.
If $b=3$, we get that $\vert a-6\vert = 1 \implies a=5,7$.
The smallest number in $A$ is obtained by choosing $a=3$ and $b=1$.
Hence, the smallest number in the set $A$ is $31$.
