Greatest prime factor

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prernamalhotra: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Wed Apr 02, 2014 2:01 amFollowed by: 1 member; Thanked: 2 times

by prernamalhotra » Fri May 30, 2014 6:17 am

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?

1)5
2)7
3)11
4)13
5)17

Thank you,
Prerna

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Source: Beat The GMAT — Problem Solving | Fri May 30, 2014 6:17 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri May 30, 2014 7:22 am

prernamalhotra 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?

1)5
2)7
3)11
4)13
5)17

Thank you,
Prerna

For any EVENLY SPACED SET:
Number of terms = (biggest-smallest)/interval + 1, where the interval is the distance between successive terms.
Average = median = (biggest + smallest)/2.
Sum = (number of terms)(average).

The even multiples of 15 between 295 and 615 are the MULTIPLES OF 30 between 295 and 615.
Thus:
k = 300 + 330 + 360 + ... + 600.
Here, the interval between successive terms is 30.
Thus:
Number of terms = (600-300)/30 + 1 = 11.
Average = (600+300)/2 = 450.
Sum = 11*450.

Prime-factoring k = (11)(450), we get:
(11)(2*3*3*5*5).
Thus, the greatest prime factor of k is 11.

The correct answer is C.

Last edited by GMATGuruNY on Fri May 30, 2014 8:09 am, edited 1 time in total.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Fri May 30, 2014 7:22 am

prernamalhotra 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?

1)5
2)7
3)11
4)13
5)17

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.
Aside: A nice rule says: the number of integers from x to y inclusive equals y - x + 1

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

Brent Hanneson - Creator of GMATPrepNow.com

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Fri May 30, 2014 8:03 am

k = 300, 330, ...... 600

11/2 ( 2*300 + 10*30)

= 11/2 * (600 + 300) = 450*11

450 = 5x5x2x3x3

So, 11

R A H U L

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon Sep 08, 2014 10:03 pm

oops - double post

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