Problem Solving

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 Prep

Already answered: https://www.beatthegmat.com/if-integer- ... 03645.html

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

OA C

Source: Manhattan Prep

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

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 Prep

The smallest even multiple of 15 between 295 and 615 is 15(20) = 300, and the largest is 15(40) = 600. Therefore, k = 300 + 330 + 360 + ... + 600 = (300 + 600)/2 x 11 = 450 x 11 = 45 x 10 x 11 = 3^2 x 5 x 2 x 5 x 11. So the largest prime factor of k is 11.

Answer: C