If integer k is equal to the sum of all even multiples of 15

BTGmoderatorDC 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

OA C

Source: Manhattan GMAT

The first even multiples of 15 between 295 and 615 is 300 and the last one is 600.

Thus, the terms are 300, 330, 360, ... 600.

Total number of terms = [(600 – 300)/30] + 1 = 11 terms

Sum of terms = Number of terms*Average of the first and the last term = 11*[(300 + 600)/2] = 11*450

Taking prime factors of 11*450, we have 11*450 = 11*9*5*10 = 11*3^2*5^2*2

The greatest prime is 11.

The correct answer: C

Hope this helps!

-Jay
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BTGmoderatorDC 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 + 600

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 + 20
Since 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