Hi Skeifer,
This question can be solved without any special formulas - you can use 'bunching' to do it.
To start, we need to better-define the numbers we are interested in. All of the multiples of 3 from '14 to 50' really means from 15 to 48 (since those are the smallest and largest multiples of 3, respectively, in this sequence).
If we take those two numbers (again, the smallest and the largest), we get a sum of 15 + 48 = 63
If we take the next smallest and the next largest, we get... 18 + 45 = 63
and then the next smallest and the next largest, we get... 21 + 42 = 63
So we're clearly dealing with a bunch of 63s... the question now is 'how many' (and is there a 'middle', unpaired number in the sequence?)? There are a number of different ways of working through this step (and the easiest would probably be just to list out the pairs - the last of which would be 30 + 33). That's a total of six 63s = 6(63) = 378.
GMAT assassins aren't born, they're made,
Rich
Contact Rich at: [email protected]