A number is said to be in a generalised form, if it is expressed as the sum of the products of its digits with their respective place values as ab = a x 10 + b and abc = a x 100 + b x 10 + c.
73 = 10 x 7 + 3    [ \(\because\)ab =  10a + b ]

The given number is 616.
Sum of digits = 6 +1+6 = 13
Now, 13 \(\div\) 9 = 1and remainder = 4
which is not divisible by 9.
Hence, 616 is not divisible by 9.

1 or 5 or 9

0 or 5

No

\(\begin{matrix} \quad \quad 2\quad A\quad B \\ +\quad A\quad B\quad 1 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad B\quad 1\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have two letters A and B whose values are to be found.
Studying the addition in the one's column, we have B + 1 which gives 8, therefore B must be 7.
Then, the puzzle becomes
\(\begin{matrix} \quad \quad 2\quad A\quad 7 \\ +\quad A\quad 7\quad 1 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad 7\quad 1\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Studying the addition in ten's digit column. We have A + 7 which gives 1, i.e. a number whose unit's digit is 1.
So, A must be 4.
Then, the puzzle is solved as shown below:
\(\begin{matrix} \quad \quad 2\quad 4\quad 7 \\ +\quad 4\quad 7\quad 1 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad 7\quad 1\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 4 and B = 7.

It is given that, 48101B095 is divisible by 33 = 3 x11.
Hence, it will be divisible by 3 and 11.
If 48101B095 is divisible by 3. Sum of digits
= 4 + 8 + 1 + 0 + 1 + B + 0 + 9 + 5 = 28 + B
(28 + B) is a multiple of 3.
If 481018095 is divisible by 11,
Sum of digits at odd places = 5 + 0 + 1 + 1+ 4 = 11
Sum of digits at even places = 9 + B + 0 + 8
= 17 + B
Difference = 17 + B -11 = 6 + B       (6 + 8) is a multiple of 11.
Least value of 6 + B = 11
B=11-6=5
Since, (28 + B) is a multiple of 3.
Least possible value of 28 +B = 30
8 = 30 - 28 = 2
Which is impossible because we have obtained
B = 5.
Now, we take next value, 28 + B = 33
B = 33 -28 =5
Hence, B = 5 is the solution.

39

148101 B095 is divisible by 33. So, it is divisible by 3 also because 33 is a multiple of 3.
Then, the sum of all digits
=1 + 4 + 8 + 1 + 0 + 1 + B + 0 + 9 + 5= 29 + B
Now, it is a multiple of 3
So, 29 + B is multiple of 3.
\(\Rightarrow\)29 + B = 0, 3, 9 ... , 33, 36, ... , 39, 42
\(\therefore\)B = 1 or 4 or 7
So, B = 4 and required number is 1481014095.