Even multiples of 15 are numbers of the form 15(2x) where x is a whole number. Alternatively, you can think of these as multiples of 30, since we have (15*2)x = 30x. The sum in question must be of the form 30(10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 +18 + 19 + 20). Since our first multiple is 300 and our last multiple is 600.
The way to find the sum of the quantity in parenthesis is simple. The formula tells us that we take the average (arithmetic mean) of the numbers and multiply by the number of numbers. There are 11 numbers. The average is very easy to compute in this case since there are an odd number of them and they are consecutive, its the middle number, 15. So our product becomes 30(11*15). We do not want to multiply at this point. We want to prime decompose these numbers:
30 = 2 * 3 * 5 11 = 11 15 = 3 * 5
So 30(11*15) = 2 * 3 * 5 * 11 * 3 * 5 = 2 * (3^3) * (5^2) * 11. We have broken the product up in to powers of its four prime factors 2, 3, 5, and 11, the largest of which is 11.
Users browsing this forum: No registered users and 1 guest
You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum
GMAT(TM) and GMAT CAT (TM) are registered trademarks of the Graduate Management Admission Council(TM). The Graduate Management Admission Council(TM) does not endorse, nor is affiliated in any way with the owner or any content of this site.
