MGMAT CAT 1: PS Problem

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by Brent@GMATPrepNow » Mon Nov 11, 2013 8:51 am
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
When posting questions, please use the spoiler function to hide the correct answer. This will allow others to attempt the question without seeing the final answer.

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)

------------------------------------------------------

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)

-------------------------------------------------
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

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by ganeshrkamath » Mon Nov 11, 2013 8:55 am
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
300,330,...,600
This series is an arithmetic progression.
The difference d between any two consecutive terms is a constant (=30).

The sum of such series is given by
k = (average of first term and last term) * number of terms
Here, the number of terms = 11
k = (300+600)/2 * 11
k = 450 * 11
k = 9*5*10*11
k = 2 * 3^2 * 5^2 * 11

Clearly, the biggest prime factor of k is 11.

Choose C

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by Mathsbuddy » Tue Nov 12, 2013 9:03 am
Using 300 to 600 inclusive, even multiples of 15 are multiples of 30.
number of multiples = n = (600-300)/30 = 10 such multiples

Sum of 1 to 10 = 55

55 * 30 = 1650 = 2 * 3 * 5^2 * 11 (as a product of prime factors)

Therefore the highest prime factor = 11

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by [email protected] » Tue Nov 12, 2013 2:11 pm
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.

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