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If n is a positive integer

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If n is a positive integer [#permalink]

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New post 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
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Re: GMATPREP integers [#permalink]

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New post 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.

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Re: GMATPREP integers [#permalink]

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New post 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

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Re: GMATPREP integers [#permalink]

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New post 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.

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Re: GMATPREP integers [#permalink]

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New post 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....
answer is C.
(if n=1 and product is 0, remainder is still 0)

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Re: GMATPREP integers [#permalink]

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New post 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

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Re: GMATPREP integers [#permalink]

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New post 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.

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Re: GMATPREP integers   [#permalink] 11 Nov 2008, 08:38
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