yangsta8 wrote:
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
Multiples of 15: 15, 30, 45, 60, 75, 90, 105, etc
EVEN multiples of 15: 30, 60, 90, 120, ....
So k = 300 + 330 + 360 + ... + 570 + 600Since each number is a multiple of 30, let's rewrite each value as the product of 30 and
some integer:
300 = 30(
10)
330 = 30(
11)
360 = 30(
12)
390 = 30(
13)
.
.
.
570 = 30(
19)
600 = 30(
20)
So k = 30(
10 + 11 + 12 + ... + 19 + 20)
------------------------------------------------------
Let's examine this sum:
10 + 11 + 12 + ... + 19 + 20Since 20 - 10 + 1 = 11, we know there are
11 numbers to add together.
Since these red numbers are equally spaced (consecutive integers), their sum = (# of values)(average of first and last values)
= [
11][(10+20)/2]
= [
11][15]
=
(11)(15)-------------------------------------------------
So, k = 30(
10 + 11 + 12 + ... + 19 + 20)
= 30
(11)(15)= (2)(3)(5)
(11)(3)(5)We can see that 11 is the greatest prime factor of k
Answer:C
Cheers,
Brent
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