k = 15x20 + ... 15x40 = 15 x ( 20 + 22 + 24 .... + 40)
Using AP sum = 11/2 ( 2x20 + 10*2) = 11/2 * (60) = 330
So, k = 15 * 330 = 5 x 3 x 11 x 3 x 5 x 2
So, [spoiler]{11}[/spoiler]
Number Systems
This topic has expert replies
Source: Beat The GMAT — Problem Solving |
-
theCodeToGMAT
- Legendary Member
- Posts: 1556
- Joined: Tue Aug 14, 2012 11:18 pm
- Thanked: 448 times
- Followed by:34 members
- GMAT Score:650
GMAT/MBA Expert
-
Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Hmmmmm, not sure about "even multiples." 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. Is this correct?? Of course, if this is the case, why not just say "multiples of 30"?sukhman 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
What's the source of this question?
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
GMAT/MBA Expert
-
Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Assuming "even multiples of 15" is the same as "multiples of 30," let's examine some terms to add up.sukhman 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
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
Brent Hanneson - Creator of GMATPrepNow.com
-
sukhman
- Master | Next Rank: 500 Posts
- Posts: 202
- Joined: Sun Sep 08, 2013 11:51 am
- Thanked: 3 times
- Followed by:2 members
Its a document called 700-800 Quant probably by IVY Gmat in which questions are clubbed topic wise,I dont know how many of them actually are 700 + but seem to be good source of difficult questions,Next I also have 2 year old question set of GMATClub (the old non-adaptive test )downloaded from Amazon.
