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# If n is a positive integer greater than 6, what is the remai

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If n is a positive integer greater than 6, what is the remai  [#permalink]

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Updated on: 19 May 2014, 03:25
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If n is a positive integer greater than 6, what is the remainder when n is divided by 6?

(1) n^2 – 1 is not divisible by 3.

(2) n^2 – 1 is even.

Originally posted by russ9 on 18 May 2014, 12:58.
Last edited by Bunuel on 19 May 2014, 03:25, edited 1 time in total.
Edited the question.
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Re: If n is a positive integer greater than 6, what is the remai  [#permalink]

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19 May 2014, 03:34
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If n is a positive integer greater than 6, what is the remainder when n is divided by 6?

(1) n^2 – 1 is not divisible by 3 --> (n-1)(n+1) is not divisible by 3 --> neither n-1 not n+1 is divisible by 3. Since from any 3 consecutive integers one is divisible by 3, then n must be divisible by 3. If n is even too (12, 18, ...) then the remainder is 0 but if n is odd (9, 15, ...) then the reminder is 3. Not sufficient.

(2) n^2 – 1 is even --> n^2 = even + 1 = odd --> n=odd. Not sufficient.

(1)+(2) n is odd multiple of 3: 9, 15, 21, ... The remainder when n is divided by 6 is 3. Sufficient.

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Re: If n is a positive integer greater than 6, what?  [#permalink]

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Updated on: 19 May 2014, 15:12
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russ9 wrote:
If N is a positive integer greater than 6, what is the remainder when N is divided by 6?

(1) N^2 – 1 is not divisible by 3.

(2) N^2 – 1 is even.

N is a positive integer > 6, what is the remainder when N is divided by 6?

(1) N^2 – 1 is not divisible by 3.

--> N^2 - 1 = (N-1)*(N+1) , and this is not divisible by three.

Notice that (N-1)*N*(N+1) is the product of three consecutive positive integers. Such a product will always contain exactly one multiple of 3. If this multiple of three is not (N-1) or (N+1), then the multiple of 3 must be N.

If N = 3*3=9, the remainder when divided by 6 is 3.
If N = 3*4=12, the remainder when divided by 6 is 0.
--> Not sufficient.

(Note: The product of K consecutive positive integers is always divisible by K!)

(2) N^2 – 1 is even

--> N^2 = even + 1 = odd --> N is odd.

If N = 7, the remainder when divided by 6 is 1.
If N = 9, the remainder when divided by 6 is 3.
--> Not sufficient.

(1) and (2)
N is an odd multiple of 3 greater than 6.
--> N = 3*(2k-1) = 6k - 3 for k >1.

Thus, since N is three less than a multiple of six, the remainder is always 3. Sufficient.

Originally posted by Reinfrank2011 on 18 May 2014, 22:40.
Last edited by Reinfrank2011 on 19 May 2014, 15:12, edited 1 time in total.
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Re: If n is a positive integer greater than 6, what?  [#permalink]

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18 May 2014, 20:01
How it can be C, because the n value where n^2-1 is not divisible by 6 is 9; and also if we consider n^2-1 is even we can get a lot numbers 48, 120, 80 etc.
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Re: If n is a positive integer greater than 6, what?  [#permalink]

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18 May 2014, 20:53
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krishna789 wrote:
How it can be C, because the n value where n^2-1 is not divisible by 6 is 9; and also if we consider n^2-1 is even we can get a lot numbers 48, 120, 80 etc.

Q:If n is a positive integer greater than 6, what is the remainder when n is divided by 6?

(1) n2 – 1 is not divisible by 3.

(2) n2 – 1 is even.

Ans is C and here is how

St 1 tells us that n^2-1 is not divisible by 3 and we know n>6 so possible values of n =9,12,15,18......
Now if n =12,18,24... then n/6 gives remainder as 0 but if n=9,15,21 then remainder is 3

So from St1 we see that remainder will follow a pattern of 0,1,0,1....

Not sufficient

from St 2 we we have n^2-1=even or n^2= Even +1 =odd, so we see that n is odd.
Now if n=7 then n/6 gives remainder 1
but if n=9 then remainder is 3
if n=11 then remainder is 5

So St2 is not sufficient alone

Combining we see that n is odd and n>6 so possible value of n satisfying the above conditions are n=9,15,21,27.....so on
Remainder will be 3

Ans is C
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Re: If n is a positive integer greater than 6, what is the remai  [#permalink]

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19 May 2014, 17:49
Bunuel wrote:
If n is a positive integer greater than 6, what is the remainder when n is divided by 6?

(1) n^2 – 1 is not divisible by 3 --> (n-1)(n+1) is not divisible by 3 --> neither n-1 not n+1 is divisible by 3. Since from any 3 consecutive integers one is divisible by 3, then n must be divisible by 3. If n is even too (12, 18, ...) then the remainder is 0 but if n is odd (9, 15, ...) then the reminder is 3. Not sufficient.

(2) n^2 – 1 is even --> n^2 = even + 1 = odd --> n=odd. Not sufficient.

(1)+(2) n is odd multiple of 3: 9, 15, 21, ... The remainder when n is divided by 6 is 3. Sufficient.

Thanks for the explanation. I can definitely follow what you have outlined above but I wouldn't have been able to derive it on my own.

Two questions:
1) Is it better to plug in numbers for these type of problems or work through the theory?
2) How do I get better at these specific types of problems? Is it just practice and if so, can you suggest similar problems please?

Thanks!
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Re: If n is a positive integer greater than 6, what is the remai  [#permalink]

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20 May 2014, 01:38
russ9 wrote:
Bunuel wrote:
If n is a positive integer greater than 6, what is the remainder when n is divided by 6?

(1) n^2 – 1 is not divisible by 3 --> (n-1)(n+1) is not divisible by 3 --> neither n-1 not n+1 is divisible by 3. Since from any 3 consecutive integers one is divisible by 3, then n must be divisible by 3. If n is even too (12, 18, ...) then the remainder is 0 but if n is odd (9, 15, ...) then the reminder is 3. Not sufficient.

(2) n^2 – 1 is even --> n^2 = even + 1 = odd --> n=odd. Not sufficient.

(1)+(2) n is odd multiple of 3: 9, 15, 21, ... The remainder when n is divided by 6 is 3. Sufficient.

Thanks for the explanation. I can definitely follow what you have outlined above but I wouldn't have been able to derive it on my own.

Two questions:
1) Is it better to plug in numbers for these type of problems or work through the theory?
2) How do I get better at these specific types of problems? Is it just practice and if so, can you suggest similar problems please?

Thanks!

1. Questions on remainders are good for plug-in, but not all of them. It depends on a question and your skills/preferences what approach to choose.

2. By studying theory and practicing:
Theory on remainders problems: remainders-144665.html

Units digits, exponents, remainders problems: new-units-digits-exponents-remainders-problems-168569.html

All DS remainders problems to practice: search.php?search_id=tag&tag_id=198
All PS remainders problems to practice: search.php?search_id=tag&tag_id=199

Hope this helps.
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Re: If n is a positive integer greater than 6, what is the remai  [#permalink]

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14 Apr 2017, 20:11
Here is my take!

Statement 1: n^2 - 1 is not divisible by 3 --> (n-1)(n+1) is not divisible by 3 ---> n is divisible by 3 --> insuff

Statement 2: n^2 - 1 is odd --> (n-1)(n+1) is odd ---> n -1 is even ---> n is odd ---> insuff

Statement 1+2: n is divisible by 3 ---> n = 3a

n is even ---> n = 2b +1

So, n = 6c +3 ---> remainder when n is divided by 6 is 3

Hence, C!
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Re: If n is a positive integer greater than 6, what is the remai  [#permalink]

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22 May 2017, 20:44
russ9 wrote:
If n is a positive integer greater than 6, what is the remainder when n is divided by 6?

(1) n^2 – 1 is not divisible by 3.

(2) n^2 – 1 is even.

Goal: We need to know the remainder when n is divided by 6.

Statement 1: (N^2-1)/3 =int. So (n+1)(n-1)/3 is not an integer. Thus, n must be a multiple of three because every three consecutive integers has at least one multiple of three within it. Not sufficient because all it tells us is that n is a multiple of 3, so its remainder when divided by 6 has multiple potential values.

Statement 2: N^2 - 1 is even. Hence (n+1)*(n-1) is even, so n must be odd. Not sufficient because some odd numbers have different remainders when divided by 6.

Combined: We know that N is an integer greater than 6 that is odd and a multiple of 3. Start testing cases: n=9 (r=3); n=15 (r=3); and so on. Sufficient.
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Re: If n is a positive integer greater than 6, what is the remai  [#permalink]

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03 Aug 2019, 12:48
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