BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Prime Factors in a number sequence

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 am
Thanked: 146 times
Followed by:2 members

by parallel_chase » Mon Aug 04, 2008 10:00 am
first find the multiples of 15 between 295 and 615

[300-615] / 15 = 21+1 =22 integers

even integers would be 11

sum of 11 even integers.

use the formula n/2 [2a+(n-1)d]
n=11
a=300
d=30 (because we need to find the sum of even integers)

11/2[600+300]

Therefore largest prime factor would be 11.

Whats the OA?
Top
Legendary Member
Posts: 661
Joined: Tue Jul 08, 2008 12:58 pm
Location: France
Thanked: 48 times

by pepeprepa » Mon Aug 04, 2008 10:27 am
I don't follow your formula Chase, where does it come from?

I did it by calculation, it's ok for the timing, but can be complicated if there are more numbers.
Top
Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 am
Thanked: 146 times
Followed by:2 members

by parallel_chase » Mon Aug 04, 2008 10:49 am
pepeprepa wrote:I don't follow your formula Chase, where does it come from?

I did it by calculation, it's ok for the timing, but can be complicated if there are more numbers.
I used two formulas in my post.

First formula is to find the number of integers.

How many integers are there between 1-1000 that are divisible by 8

first integer =8 (first integer divisible by 8)
last integer = 1000 (last integer multiplied by 8)

[1000-8]/8 = 124 + 1 (we need to add 1 because we have subtracted 8.

Therefore there are 125 integers between 1 and 1000 that are divisible by 8.


Second formula is to find the sum of integers.

What is the sum of all the integers between 1-1000 that are divisible by 8

n = no. of integers i.e. 125 from the above calculations
a= first term i.e. 8 (we find the sum of multiples of 8, therefore the first term would be 8 and not 1)
d= common difference between each term i.e. 8

The formula is n/2[2a+(n-1)d]

125/2[2*8 + (125-1)8]

= 125/2[16+992]

=125/2[1008]

= 125 * 504 = would be the answer.


I used such big numbers to show that both the formulas can work with any possible numbers.

Hope this helps. Let me know your thoughts.
Top
Master | Next Rank: 500 Posts
Posts: 167
Joined: Tue Feb 26, 2008 4:37 am
Thanked: 5 times
Followed by:1 members

by pranavc » Mon Aug 04, 2008 11:45 am
Is C the answer?
Top
Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 am
Thanked: 146 times
Followed by:2 members

by parallel_chase » Mon Aug 04, 2008 11:48 am
Yes C is the answer.
Top
Master | Next Rank: 500 Posts
Posts: 167
Joined: Tue Feb 26, 2008 4:37 am
Thanked: 5 times
Followed by:1 members

by pranavc » Mon Aug 04, 2008 11:52 am
Thanks for the prompt response!!!
Top
Senior | Next Rank: 100 Posts
Posts: 62
Joined: Thu Jul 03, 2008 4:52 am

by [email protected] » Mon Aug 04, 2008 12:09 pm
Thanks for you responses.

OA is C
Top
Legendary Member
Posts: 1578
Joined: Sun Dec 28, 2008 1:49 am
Thanked: 82 times
Followed by:9 members
GMAT Score:720

by maihuna » Sun May 17, 2009 1:27 am
parallel_chase wrote:first find the multiples of 15 between 295 and 615

[300-615] / 15 = 21+1 =22 integers

even integers would be 11

sum of 11 even integers.

use the formula n/2 [2a+(n-1)d]
n=11
a=300
d=30 (because we need to find the sum of even integers)

11/2[600+300]

Therefore largest prime factor would be 11.

Whats the OA?
Hi _|_ chase, the no of items here will not be 22 but 11, its asking for even multiples of 15: they are 300 to 600 i.e. 15*20 to 30*40

sum = 15*20 + 15*21 + 15*22 + ... + 15*40
= 15*2(10+11+12+..+20)
= 15*2*11/2*15
= 2*3*3*5*11

so 11 is largest factor
Charged up again to beat the beast :)
Top
Legendary Member
Posts: 1169
Joined: Sun Jul 06, 2008 2:34 am
Thanked: 25 times
Followed by:1 members

by aj5105 » Sun May 17, 2009 2:33 am
Hey Main Hoon Na!

Watch closely. Parallel Chase was right, he too has mentioned 11 :)
maihuna wrote:
parallel_chase wrote:first find the multiples of 15 between 295 and 615

[300-615] / 15 = 21+1 =22 integers

even integers would be 11

sum of 11 even integers.

use the formula n/2 [2a+(n-1)d]
n=11
a=300
d=30 (because we need to find the sum of even integers)

11/2[600+300]

Therefore largest prime factor would be 11.

Whats the OA?
Hi _|_ chase, the no of items here will not be 22 but 11, its asking for even multiples of 15: they are 300 to 600 i.e. 15*20 to 30*40

sum = 15*20 + 15*21 + 15*22 + ... + 15*40
= 15*2(10+11+12+..+20)
= 15*2*11/2*15
= 2*3*3*5*11

so 11 is largest factor
Top
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Sun Oct 11, 2009 8:37 pm

by firang » Sun May 09, 2010 4:44 pm
There is no way all this will occur to me in the 2 mins under the time pressure in the exam.

just realized this is my first post, although i have been here for a while, very nice and dirty question, should we expect to see a lot of these type of ques on the real gmat? when I say these, I mean such cumbersome calculation types, or is there a shortcut here that we are missing?
Top
Master | Next Rank: 500 Posts
Posts: 385
Joined: Sun Jul 12, 2009 10:16 pm
Thanked: 29 times
Followed by:2 members
GMAT Score:710

by debmalya_dutta » Sun May 09, 2010 5:05 pm
Yep..11 is the largest prime factor.
Need to establish value of k first
k is equal to the sum of all even multiples of 15 between 295 and 615
300 is the first number in the series with a common difference of 30 . The last number in the series is 600 and hence 11 terms
so , K is the sum which is equal to 4950
Largest prime factor of 4950 is 11
Top
User avatar
Legendary Member
Posts: 1022
Joined: Mon Jul 20, 2009 11:49 pm
Location: Gandhinagar
Thanked: 41 times
Followed by:2 members

by shashank.ism » Sun May 09, 2010 11:01 pm
firang wrote:There is no way all this will occur to me in the 2 mins under the time pressure in the exam.

just realized this is my first post, although i have been here for a while, very nice and dirty question, should we expect to see a lot of these type of ques on the real gmat? when I say these, I mean such cumbersome calculation types, or is there a shortcut here that we are missing?
It is not so cumbersome . It is just using the formula for sum of an A.P so obviously this question can be asked and it is just a 1 min question. Though this question is tougher than other question as it includes a bit of calculation but calculation is not very tough....
My Websites:
www.mba.webmaggu.com - India's social Network for MBA Aspirants

www.deal.webmaggu.com -India's online discount, coupon, free stuff informer.

www.dictionary.webmaggu.com - A compact free online dictionary with images.

Nothing is Impossible, even Impossible says I'm possible.
Top
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Tue Jan 22, 2013 8:48 pm

by thinktank08 » Sun Apr 07, 2013 4:10 am
An alternate method to find the sum of all the numbers between two limits can be done by using the formula

n/2 [first term + last term]

Where n= no of terms in the sample space, which can be found out using the formula


***Quote from parallel_chase
How many integers are there between 1-1000 that are divisible by 8

first integer =8 (first integer divisible by 8)
last integer = 1000 (last integer multiplied by 8)

[1000-8]/8 = 124 + 1 (we need to add 1 because we have subtracted 8.

Therefore there are 125 integers between 1 and 1000 that are divisible by 8.

Thanks
TT
Top
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Fri Apr 05, 2013 11:56 pm
Thanked: 3 times

by bharat.bondalapati » Sun Apr 07, 2013 8:50 am
Simple solution,

List the even multiples of 15 between 295 and 615
300
330
360
390
420
450
480
510
540
570
600

11 numbers in Arithmetic progression. Mean is the mid number. Sum is mean*n

450*11 = 2*(3^2)*(5^2)*11

11 is the largest prime number
Best,
Bharat

If you find my post helpful, please click on the 'Thank' icon.
Top
Post new topic Post Reply