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# If n is a positive integer and r is the remainder when (n-1)

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If n is a positive integer and r is the remainder when (n-1) [#permalink]  31 Aug 2009, 04:03
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Difficulty:

75% (hard)

Question Stats:

53% (02:29) correct 47% (01:51) wrong based on 69 sessions
If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?

(1) n is not divisible by 2
(2) n is not divisible by 3

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-n-is-a-positive-integer-and-r-is-the-remainder-when-n-126393.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Oct 2013, 07:38, edited 1 time in total.
Renamed the topic and edited the question.
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Re: GMATPrep: division by 24 again [#permalink]  31 Aug 2009, 04:48
Yes its C
if they are not divisible by 2 and 3 ...then no can be 5 , 7 , 11, ....so on......

caculate n-1 and n+1 , it will give ( product will ) mulitiple of 24 , thus R will be 0....
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Re: GMATPrep: division by 24 again [#permalink]  31 Aug 2009, 16:36
Again,plugging numbers is the best approach..

1.) n is not divisible by 2 => n is odd

Case1: n = 1
(n-1)(n+1) = 0/24 ; r = 0

Case2: n = 3
(n-1)(n+1) /24 = 8/24; r = 8

Case3: n = 5
(n-1)(n+1) /24 = 24/24; r = 0
Hence, insufficient.

2.) n is not divisible by 3

case1: n = 1
(n-1)(n+1) = 0/24 ; r = 0

Case2: n = 2
(n-1)(n+1) /24 = 3/24; r = 3

Case3: n = 5
(n-1)(n+1) /24 = 24/24; r = 0
Hence, insufficient.

Combining both, n is neither divisible by 2 nor 3 => n =1, 5, 7,11

for all these numbers , r = 0 .. hence, sufficient..

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Re: GMATPrep: division by 24 again [#permalink]  03 Sep 2009, 22:50
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Expert's post
Here is another way:
Given (n-1)(n+1)=24x+r; r<24 (n-1)(n+1)=n^2-1 (n^2-1 is divisible by 8 if n is odd, worse to remember )

(1) n is not divisible by 2 => n odd => (n-1)(n+1) divisible by 8. But 24x is already multiple of 8, so r also should be multiple of 8 or 0. (Well it can be only 0 and 8 r<24 it can't be 16 but it doesn't matter here). Not suff

(2) n is not divisible by 3 => ether n-1 or n+1 is divisible by 3, so multiplication of these numbers (24x+r) should also be divisible by 3, but if r=8 24x+8 is not divisible by 3 => r=0.
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Re: GMATPrep: division by 24 again [#permalink]  27 Mar 2011, 03:42
Bunuel wrote:
Here is another way:
Given (n-1)(n+1)=24x+r; r<24 (n-1)(n+1)=n^2-1 (n^2-1 is divisible by 8 if n is odd, worse to remember )

(1) n is not divisible by 2 => n odd => (n-1)(n+1) divisible by 8. But 24x is already multiple of 8, so r also should be multiple of 8 or 0. (Well it can be only 0 and 8 r<24 it can't be 16 but it doesn't matter here). Not suff

(2) n is not divisible by 3 => ether n-1 or n+1 is divisible by 3, so multiplication of these numbers (24x+r) should also be divisible by 3, but if r=8 24x+8 is not divisible by 3 => r=0.

nice explanation. I had to pluging number which took me 5 minutes to solve out. This is a great reasoning
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Re: GMATPrep: division by 24 again [#permalink]  04 Oct 2013, 07:07
Hi Bunuel,

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Re: GMATPrep: division by 24 again [#permalink]  04 Oct 2013, 07:36
Expert's post
irda wrote:
Hi Bunuel,

$$n-1$$ and $$n+1$$ are consecutive even integers --> $$(n-1)(n+1)$$ must be divisible by 8 (as both multiples are even and one of them will be divisible by 4. From consecutive even integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8, 10); (10, 12), ...).

If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?

Plug-in method:

$$(n-1)(n+1)=n^2-1$$

(1) n is not divisible by 2 --> pick two odd numbers: let's say 1 and 3 --> if $$n=1$$, then $$n^2-1=0$$ and as zero is divisible by 24 (zero is divisible by any integer except zero itself) so remainder is 0 but if $$n=3$$, then $$n^2-1=8$$ and 8 divided by 24 yields remainder of 8. Two different answers, hence not sufficient.

(2) n is not divisible by 3 --> pick two numbers which are not divisible by 3: let's say 1 and 2 --> if $$n=1$$, then $$n^2-1=0$$, so remainder is 0 but if $$n=2$$, then $$n^2-1=3$$ and 3 divided by 24 yields remainder of 3. Two different answers, hence not sufficient.

(1)+(2) Let's check for several numbers which are not divisible by 2 or 3:
$$n=1$$ --> $$n^2-1=0$$ --> remainder 0;
$$n=5$$ --> $$n^2-1=24$$ --> remainder 0;
$$n=7$$ --> $$n^2-1=48$$ --> remainder 0;
$$n=11$$ --> $$n^2-1=120$$ --> remainder 0.
Well it seems that all appropriate numbers will give remainder of 0. Sufficient.

Algebraic approach:

(1) n is not divisible by 2. Insufficient on its own, but this statement says that $$n=odd$$ --> $$n-1$$ and $$n+1$$ are consecutive even integers --> $$(n-1)(n+1)$$ must be divisible by 8 (as both multiples are even and one of them will be divisible by 4. From consecutive even integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8, 10); (10, 12), ...).

(2) n is not divisible by 3. Insufficient on its own, but form this statement either $$n-1$$ or $$n+1$$ must be divisible by 3 (as $$n-1$$, $$n$$, and $$n+1$$ are consecutive integers, so one of them must be divisible by 3, we are told that it's not $$n$$, hence either $$n-1$$ or $$n+1$$).

(1)+(2) From (1) $$(n-1)(n+1)$$ is divisible by 8, from (2) it's also divisible by 3, therefore it must be divisible by $$8*3=24$$, which means that remainder upon division $$(n-1)(n+1)$$ by 24 will be 0. Sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-n-is-a-positive-integer-and-r-is-the-remainder-when-n-126393.html
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Re: GMATPrep: division by 24 again   [#permalink] 04 Oct 2013, 07:36
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