MGMAT CAT 1: PS Problem

josh80 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?

-5
-7
-11
-13
-17

Ans. C

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NOTE: I doubt that the GMAT would use the term "even multiples."
Yes, this term MAY BE intuitively apparent, but I believe the GMAT test-makers would provide additional text to avoid any ambiguity. Presumably even multiples of 15 are 30, 60, 90, etc.
In other words, we're looking for multiples of 30

So, k = 300 + 330 + 360 + ... + 570 + 600

Let's examine some terms in this series. . . .

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)

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Now, 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)

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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

Hi Mathsbuddy,

You made a minor error in your calculation that is significant in these types of questions:

The multiples of 30 from 300 to 600, INCLUSIVE is 11, not 10

300
330
360
390
420

450
480
510
540
570

600

This is what's called a "fence post" problem. The math requires a slight adjustment:

(600-300)/30 + 1 = 11 terms

The "+1" account for the term "300", which was subtracted out during the first part of the calculation, but needs to be factored back in because it is INCLUSIVE.

GMAT assassins aren't born, they're made,
Rich