• Free Practice Test & Review
How would you score if you took the GMAT

Available with Beat the GMAT members only code

• 1 Hour Free
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

• 5 Day FREE Trial
Study Smarter, Not Harder

Available with Beat the GMAT members only code

• Award-winning private GMAT tutoring
Register now and save up to $200 Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

Available with Beat the GMAT members only code

• Free Veritas GMAT Class
Experience Lesson 1 Live Free

Available with Beat the GMAT members only code

• Magoosh
Study with Magoosh GMAT prep

Available with Beat the GMAT members only code

• Free Trial & Practice Exam
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

If n is an integer greater than 6 + divisibility question

This topic has 6 expert replies and 23 member replies
Goto page
• 1,
• 2
cramya Legendary Member
Joined
28 Aug 2008
Posted:
2469 messages
Followed by:
11 members
331

If n is an integer greater than 6 + divisibility question

Sun Sep 14, 2008 5:16 pm
If n is an integer greater than 6, which of the following must be divisible by 3?
A. n (n+1) (n-4)
B. n (n+2) (n-1)
C. n (n+3) (n-5)
D. n (n+4) (n-2)
E. n (n+5) (n-6)

Answer is given as d Can someone pl explain?

If i take n = 7 then d will not work.

Last edited by cramya on Mon Nov 24, 2008 4:21 am; edited 1 time in total

alescau Junior | Next Rank: 30 Posts
Joined
14 Sep 2008
Posted:
17 messages
4
Sun Sep 14, 2008 5:55 pm
ok, in general every third number in sequence is divisible by three. by taking n, n+1, n+2 you are sure that one of them is divisible by 3. you can add or subtract 3 from either of them without changing divisibility.
n is the same as n-6, n-3, n+3, n+6...
n+1 is the same as n-5, n-2, n+4, n+7...
n+2 is the same as n-4, n-1, n+5, n+8...

now you must pick an answer that has one expression from each of these groups

__A__

gmataug08 Master | Next Rank: 500 Posts
Joined
17 May 2008
Posted:
105 messages
6
Sun Sep 14, 2008 6:07 pm
IMO A

4meonly Legendary Member
Joined
16 Aug 2008
Posted:
891 messages
Followed by:
1 members
27
Target GMAT Score:
750)
GMAT Score:
660(
Tue Sep 16, 2008 7:52 am
Obviously A

n (n+1) (n-4)
if you add to (n-4) 3 you'll get
(n-1) n (n+1)
3 consequtive integers. thier product is always divisible by 3

you can imagine it on number line

Fab Senior | Next Rank: 100 Posts
Joined
15 Jul 2008
Posted:
88 messages
Followed by:
1 members
4
Test Date:
October
Target GMAT Score:
700
Sun Nov 23, 2008 3:59 pm

Thanks.

4meonly Legendary Member
Joined
16 Aug 2008
Posted:
891 messages
Followed by:
1 members
27
Target GMAT Score:
750)
GMAT Score:
660(
Mon Nov 24, 2008 12:27 am
Fab wrote:

Thanks.
U can draw a number line for all given answers

(n-4)-(n-3)-(n-2)-(n-1)-(n)-(n+1)-------->

let n-4 = 1, than n=5, n=6, 1*5*6 is divisible by 3
any number for n-4 will yield a product od (n-4)*(n)*(n+1) that is divisible by 3

mals24 Legendary Member
Joined
22 Jul 2008
Posted:
683 messages
Followed by:
2 members
73
Mon Nov 24, 2008 3:54 am
@ Fab

You can also plug in 7 and 8 as 'n' in each of the options. Only option A will work.

Cramya where did you hunt this question down from???

cramya Legendary Member
Joined
28 Aug 2008
Posted:
2469 messages
Followed by:
11 members
331
Mon Nov 24, 2008 4:22 am
Mals24,
It was sometime back and I dont tremember the source. I agree with all posters above that it should be A) and not D)

May be its a sets question and sometime the answers given cannot be trusted. I will try to loook for the source though if there is any way I could.

Regards,
Cramya

GMAT/MBA Expert

lunarpower GMAT Instructor
Joined
03 Mar 2008
Posted:
3380 messages
Followed by:
1477 members
2256
GMAT Score:
800
Mon Nov 24, 2008 4:51 am
first of all, you must understand this fact: every nth integer is divisible by n.
in other words, every other integer is divisible by 2; every third integer is divisible by 3; etc.
make sure you understand this basic principle first; if you don't, then essentially none of the following will make any sense to you.

for problems like this one, in which you're considering the divisibility of unknown numbers, it can help to do the following:
(1) set up a number line, with ..., n - 1, n, n + 1, n + 2, ... placed along it
(2) write out the different CASES for which ones are divisible by 3 (or divisible by whatever number you're talking about; in this problem it just happens to be 3)

here's what i mean.

in this case, there are 3 different cases, because every third integer is divisible by 3. (in general, for divisibility by n, there will be n different ways in which to draw the number line.)
here are the 3 cases. in each case, the multiples of 3 are in orange, and the non-multiples of 3 are in the default color (black on my view, although i'm not sure whether that's the default for everyone else).

case 1
... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...

case 2
... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...

case 3
... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...

once you've done this, just look at the 3 cases for each of the answer choices. if any of the factors is in orange, then that particular product is a multiple of 3 (because, if there's a 3 anywhere in the factorization, then the whole product is a multiple of 3 - as with any other factor).

for choice (a), all three cases include one orange factor apiece: n is orange in case 1, n+1 is orange in case 3, and n-4 is orange in case 2. therefore, (a) is the correct answer.

for choice (b), case 3 contains no orange factors, so (b) doesn't have to be a multiple of 3.

for choice (c), case 2 contains no orange factors, so (b) doesn't have to be a multiple of 3.

for choice (d), case 2 contains no orange factors, so (b) doesn't have to be a multiple of 3.

for choice (e), case 3 contains no orange factors, so (b) doesn't have to be a multiple of 3.

note that the correct answer is a, not d.

--

note that you could also just plug in a ton of numbers and see what happens with the factorizations: just plug in 7, 8, 9, ... (per the directions) for the choices and see what the products are. don't multiply the products if you do that; just see whether the numbers in the products are divisible by 3.

_________________
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions Ã  Ron en franÃ§ais
Voit esittÃ¤Ã¤ kysymyksiÃ¤ Ron:lle myÃ¶s suomeksi

--

Quand on se sent bien dans un vÃªtement, tout peut arriver. Un bon vÃªtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Free Manhattan Prep online events - The first class of every online Manhattan Prep course is free. Classes start every week.
cramya Legendary Member
Joined
28 Aug 2008
Posted:
2469 messages
Followed by:
11 members
331
Mon Nov 24, 2008 4:57 am
Thanks Ron for suggesting a nice approach and a detailed explanation!

Fab Senior | Next Rank: 100 Posts
Joined
15 Jul 2008
Posted:
88 messages
Followed by:
1 members
4
Test Date:
October
Target GMAT Score:
700
Mon Nov 24, 2008 6:15 am
Quote:
@ Fab

You can also plug in 7 and 8 as 'n' in each of the options. Only option A will work.

If it's 8...C works...

mals24 Legendary Member
Joined
22 Jul 2008
Posted:
683 messages
Followed by:
2 members
73
Mon Nov 24, 2008 6:42 am
Quote:
If it's 8...C works...
But 7 doesn't work. I said plug in 7 AND 8, and not 7 OR 8.

It should work for both of them not either of them.

Fab Senior | Next Rank: 100 Posts
Joined
15 Jul 2008
Posted:
88 messages
Followed by:
1 members
4
Test Date:
October
Target GMAT Score:
700
Mon Nov 24, 2008 6:53 am
Got it.

alina_linlin Newbie | Next Rank: 10 Posts
Joined
13 Feb 2009
Posted:
1 messages
1
Test Date:
April 1, 09
Target GMAT Score:
700
Fri Feb 13, 2009 11:17 pm
I just come out an idea that, whether any 3 intergers whose sum could be divided by 3, can also make a product divided by 3?

take the answer A for instance, n+n+1+n-4=3n-3, can be divided by 3.

for the other 4 answers, no one can meet the above rules.

sureshbala Master | Next Rank: 500 Posts
Joined
04 Feb 2009
Posted:
319 messages
Followed by:
9 members
84
Sat Feb 14, 2009 4:32 am
alina_linlin wrote:
I just come out an idea that, whether any 3 intergers whose sum could be divided by 3, can also make a product divided by 3?

take the answer A for instance, n+n+1+n-4=3n-3, can be divided by 3.

for the other 4 answers, no one can meet the above rules.
Friend this is not true in all the cases....

If a+b+c is divisible by 3, you can't conclude that axbxc will always be divisible by 3.

Here is an example....

7+10+13 is divisible by 3, but 7x10x13 is not divisible by 3.

Similarly the converse is also not true...

i.e If axbxc is divisible by 3, we can conclude that a+b+c will be divisible by 3.

Here is an example....

9x10x13 is divisible by 3, but 9+10+13 is not divisible by 3.

Top First Responders*

1 GMATGuruNY 69 first replies
2 Rich.C@EMPOWERgma... 41 first replies
3 Brent@GMATPrepNow 40 first replies
4 Jay@ManhattanReview 24 first replies
5 Terry@ThePrinceto... 10 first replies
* Only counts replies to topics started in last 30 days
See More Top Beat The GMAT Members

Most Active Experts

1 GMATGuruNY

The Princeton Review Teacher

134 posts
2 Jeff@TargetTestPrep

Target Test Prep

111 posts
3 Rich.C@EMPOWERgma...

EMPOWERgmat

111 posts
4 Scott@TargetTestPrep

Target Test Prep

103 posts
5 Max@Math Revolution

Math Revolution

92 posts
See More Top Beat The GMAT Experts