A number is divisible by 11 if the difference between the sum of digits at its odd places and that of the digits at the even places is divisible by 11.
(1 + 8) - 9 = 0 which is divisible by 11.
198 is even number and thus, divisible by 2.
The sum of digits 1 + 9 + 8 = 18 which is divisible by 3.
So, 198 is also divisible by 3.
Since, the number is divisible by both 2 and 3, it is also divisible by 6.
A number to be divisible by 5 must end in 0 or 5. Hence, 198 is not divisible by 5.
A number is divisible by 8 if the number formed by the last 3 digits of the number are divisible by 8, hence 198 is not divisible by 8.

Study the addition in the ones column:
Q + 3 = 1, a number whose ones digit is 1.

If Q = 8, then 8 + 3 = 11.
So, the puzzle can be solved as shown below:
$$\begin{array}{c}318\\ +183\\ 501\end{array}$$
Hence, Q = 8

Let the number be abc.

abc = 100a + 10b + c. 
By rearranging the digits, let us say the two other numbers formed are cab and bca.
Expressing these two numbers in the expanded form,

cab = 100c + 10a + b

bca = 100b + 10c + a

abc + cab + bca
= (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)
= 100a + 10a + a + 10b + b + 100b + c + 100c + 10c
= 111a + 111b + 111c
= 111(a + b + c)
= 37 $\times$ 3 $\times$ (a + b + c), since 111 is the product of 37 and 3.
Hence the sum is divisible by 37.

If the ones digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.
For the number 244 in the ones place we find 4 which is divisible by 2. Therefore we can say that the number 244 is divisible by 2.

990 and 9999 are both divisible by 99. Hence, their sum is also divisible by 99.

99 is divisible by 9 and 11. 9 is divisible by 3. Hence, the sum of 990 and 9999 is divisible by 9, 11 as well as 3.

The sum of three A's is such a number that has A in its ones place.

Therefore, the sum of two A's must be a number whose ones digit is 0.

This happens only for A = 0 and A = 5

If A = 0, then the sum is 0 + 0 + 0 = 0, which makes B = 0.
Given that B is a natural number, therefore B cannot be equal to 0. So we don't consider this possibility.
Hence, A = 5$$\begin{array}{c}5\\ +5\\ +5\\ 15\end{array}$$

Therefore, A= 5 and B= 1

Ones digit of 3 $\times$ A is A. So, it must be either A = 0 or A = 5.

Now, if B = 1, then BA $\times$ B3 would at most be equal to 15 $\times$ 13; (Taking A= 5 to find the maximum value of the product) that is, it would at most be equal to 195. But the product here is 57A, which is more than 500. So we cannot have B =1.

If B = 3, then BA $\times$ B3 would be at least 30 $\times$ 33; (Taking A = 0 to find the minimum value of the product) that is 990. But 57A is less than 990. So, B cannot be equal to 3.

Putting these two facts together, we get B = 2.
So, the multiplication is either 20 $\times$ 23, or
25 $\times$ 23.

The first possibility fails, since
20 $\times$ 23 = 460. But, the second one works out correctly, since
25 $\times$ 23 = 575. So the answer is A = 5 and B = 2. 
$$\begin{array}{c}25\\ \times 23\\ 575\end{array}$$

Given: $\frac{3BA}{10}=34$
Since there is no remainder, 3BA is completely divisible by 10.
Now we know that all numbers exactly divisible by 10 will have 0 in the one's place.
$\Rightarrow$ A = 0
Now $\frac{3B0}{10}=34$
So, 3B = 34
$\Rightarrow$ B = 4
Therefore, the number is 340.

Let a be the tens place digit and b be the units place digit, then  the original number is 10 x a + 1 x b.The new number after reversing the digits is 10 x b + 1 x b. Therefore the sum will be:
(10a + b) + (10b + a) = 11a + 11b

                                 = 11 (a + b)
Which is divisible by 11.