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?

The greatest prime factor of k is 11.

Step-by-step explanation:

∵  LCM( 2, 15) = 30,

Thus, the even multiple of 15 must be multiple of 30,

That is, the even multiple of 15 between 295 and 615 are

300, 330, 360,..........600

Which is an AP,

Having first term, a = 300,

Common difference, d = 30,

If n be the number terms,

Last term =

Hence, the sum of the all even multiple of 15 from 295 to 615,

According to the question,

⇒ k = 11 × 450 = 11 × 5 × 5 × 3 × 3 × 2

Hence, the greatest prime factor of k is 11.

the answer is "add 2y and y"

step-by-step explanation:

you need to collect the like terms first.