Another way to go at it is to think of the number (A+B) as a bunch of stacked blocks, each stack representing a digit. Here is 43 for example:
 _
|_| _
|_||_|
|_||_|
|_||_|

Now, to find all the A's and B's that can form S(A+B) = S(43), for example 12+31 or 22+21, etc, we can represent this as chopping each stack into two parts, for example in the case of 12+31:

 _
|_| _
 _ |_|
|_||_|
|_| _
|_||_|


Now, if we take the general case of having two digits, if the first digit is 1, we can't split the first digit in two since the first digit of A or B can't be 0. However, if the first digit is 2, we can split it in one way (in the middle), and then we can split the second digit depending on what digit it is. If it is 2, we can split it in 3 different ways ( (0,2),(1,1),(2,0) ). If it is 3, we can split it in 4 different ways, etc. So, if the first digit of the two-digit number is 2, we can split the whole number in 1*(1+2+3+4+5+6+7+8+9+10)=1+(10*11/2) different ways. If the first digit is 3, we can split the whole number in 2*(1+2+3+4+5+6+7+8+9+10)=2*(10*11/2) different ways, etc. If we sum everything we get (8*9)/2*(10*11)/2 ways to split a two digit number. For an n digit number continuing the same way we get (8*9)/2*((10*11)/2)^(n-1).

See, kettles number 1 - 5 will drink coffee at the same time (00:13:58) and only in this time kettle number 6 may begin to boil water for himself ..... He will need 2:48 seconds for that. Why 2:48 ? ...... because in this time 200 ml water will be in the kettle with 20 degrees .... but kettle number 6 want 100 degrees in the main kettle .... then we take:
deltaT = 100 - 20 = 80
C and P you know from statement
m = 200
using formula from statement (deltaT = P * littleT / C * m) we will take that
littleT = deltaT * C * m / P = 02:48        --- this is time that we need to boil water.
Result : 00:13:58 + 00:02:48 = 00:16:46