If integer k is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of k?

a. 5
b. 7
c. 11
d. 13
e. 17

I have no idea....

Hey Greg,

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.

Take care,
Steve