It is currently 21 Oct 2017, 02:03

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If n is a positive integer

Author Message
VP
Joined: 18 May 2008
Posts: 1258

Kudos [?]: 527 [0], given: 0

If n is a positive integer [#permalink]

### Show Tags

11 Nov 2008, 04:08
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Pls help
Attachments

integers.JPG [ 57.48 KiB | Viewed 857 times ]

Kudos [?]: 527 [0], given: 0

SVP
Joined: 17 Jun 2008
Posts: 1534

Kudos [?]: 279 [0], given: 0

### Show Tags

11 Nov 2008, 04:13
It will be C.

Try with numbers.

From stmt1:
n = 1,3,5,7,9,11,.....
Hence, remainder of (n-1)(n+1) will be different for different values of n....insufficient.

From stmt2:
n = 1,2,4,5,7,8,10,..... and again the desired remainder will be different for different values of n.

Combining stmt1 and stmt2:

n = 1, 5, 7, 11, ......and remainder will always be 0 for any value of n.....sufficient.

Kudos [?]: 279 [0], given: 0

Manager
Joined: 30 Sep 2008
Posts: 111

Kudos [?]: 20 [0], given: 0

### Show Tags

11 Nov 2008, 04:28
scthakur wrote:
It will be C.

Try with numbers.

From stmt1:
n = 1,3,5,7,9,11,.....
Hence, remainder of (n-1)(n+1) will be different for different values of n....insufficient.

From stmt2:
n = 1,2,4,5,7,8,10,..... and again the desired remainder will be different for different values of n.

Combining stmt1 and stmt2:

n = 1, 5, 7, 11, ......and remainder will always be 0 for any value of n...
..sufficient.

I don't think you can simply calculate some first prime number to get the conclusion

Among 3 consecutive integer n - 1, n, n + 1
n is not factor of 3 => n - 1 or n + 1 is divisible by 3
=> (n-1)(n+1) divisible by 3 (1)

n is not factor of 2 => n - 1 or n + 1 is divisible by 2
between n - 1 and n + 1, if n - 1 is divisible by 2, n + 1 is divisible by 4, if n - 1 is divisible by 4, n + 1 is divisible by 2

=> (n-1)(n+1) is divisible by 8. (2)

from (1) & (2), (n-1)(n+1) is divisible by 12, r = 0

Kudos [?]: 20 [0], given: 0

SVP
Joined: 17 Jun 2008
Posts: 1534

Kudos [?]: 279 [0], given: 0

### Show Tags

11 Nov 2008, 05:16
lylya4 wrote:
I don't think you can simply calculate some first prime number to get the conclusion

Among 3 consecutive integer n - 1, n, n + 1
n is not factor of 3 => n - 1 or n + 1 is divisible by 3
=> (n-1)(n+1) divisible by 3 (1)

n is not factor of 2 => n - 1 or n + 1 is divisible by 2
between n - 1 and n + 1, if n - 1 is divisible by 2, n + 1 is divisible by 4, if n - 1 is divisible by 4, n + 1 is divisible by 2

=> (n-1)(n+1) is divisible by 8. (2)

from (1) & (2), (n-1)(n+1) is divisible by 12, r = 0

I agree with you. However, during actual test, if I can find a pattern with couple of numbers, I will probably go with that approach in case I do not come up with an immediate algebraic approach.

Kudos [?]: 279 [0], given: 0

Manager
Joined: 08 Aug 2008
Posts: 229

Kudos [?]: 48 [0], given: 0

### Show Tags

11 Nov 2008, 08:20
well in fact, from (1) and (2) we can concur that n is prime with the exception of 2 and 3.
i.e n=1,5,7....
(if n=1 and product is 0, remainder is still 0)

Kudos [?]: 48 [0], given: 0

SVP
Joined: 05 Jul 2006
Posts: 1750

Kudos [?]: 430 [0], given: 49

### Show Tags

11 Nov 2008, 08:30
from 1

(n-1),n,(n+1) are 3 consecs where (n-1) and (n+1) are even.

when

(n-1)(n+1) are devided by 2^3 * 3 the remainder could be anything.......insuff

from2

if 3 is not a factor of n then sure either (n-1) or (n+1) is. however.........insuff

both

in the 3 cosnecs one of them is even multiple of 3 , second is even and one of the 2 evens has one more factor of 2 in it, as such

the 2 evens are in the form = 2*2*2*3*x, where x is any +ve intiger

thus remainder is 0..........suff

Kudos [?]: 430 [0], given: 49

Intern
Joined: 17 Oct 2008
Posts: 16

Kudos [?]: 4 [0], given: 0

### Show Tags

11 Nov 2008, 08:38
I think any of the above process is correct. Based on whether one is good at forming calculation with numbers or at forming quick algebric equations, one can take any of the approach.

Kudos [?]: 4 [0], given: 0

Re: GMATPREP integers   [#permalink] 11 Nov 2008, 08:38
Display posts from previous: Sort by