This puzzle comes from here:
In each of the four sectors of the outer circle, there is a two-digit number which is equal to the sum of the three numbers at the corners of its sector. The numbers in the individual circles can only be 1 to 9 and each number can be used only once. One number has been provided to get you started. Find the remaining four numbers
As usual, what interests me is not the solution but how one arrives at it.
Let's start with the lowest number - 10. We have to provide three digits which when added together, will give the sum 10. But as we know that one of the digits is 3, we have to find two digits which add up to 7. There are several pairs - (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - so at the moment it's not clear which pair is needed. In order to get extra information, let's look at the other digit which is based on 3 - 18. Again, we have to find a pair of digits which will add up to 15 - of which one must come from the first set of pairs. As this digit has to be less than 10, the only digit which it can be is 6: 3 + 6 + 9 = 18. Thus the third digit which is required for 10 is 1.
Once this is known, the rest falls into place automatically: 12 = 3 + 1 + 8, 20 = 3 + 8 + 9.