If n is a positive integer and r is the remainder when (n - : GMAT Data Sufficiency (DS)
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If n is a positive integer and r is the remainder when (n -

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If n is a positive integer and r is the remainder when (n - [#permalink]

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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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Sep 2012, 10:53, edited 1 time in total.
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Re: Data sufficiency remainder problem [#permalink]

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New post 26 Apr 2010, 13:56
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ksharma12 wrote:
9.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.


I know this is not new to the forum, but I couldnt find an explanation that was clear to me. Can someone provide detailed reasoning?

Thank you


For statement 1. take n = 5 the number becomes 4*6 when divided by 24 gives remainder 0
take n = 7 the number becomes 6*8 when divided by 24 gives remainder 0
take n = 9 the number becomes 8*10 when divided by 24 gives remainder different than 0. Thus not sufficient.

For statement 2. take n= 5 and n=7 it will give remainder as 0
if you take n = 12 14 number becomes 13*15 when divided by 24 gives remainder other than 0. Thus not sufficient.

So only answers left are either C or E

Now you must have observed that when n is not multiple of 2 or not multiple of 3 independently we are unable to answer the question. But when n is neither multiple of 2 nor 3 then we are able to answer the question as r=0 in all those cases.

Hence C

Another way of looking at it is.... take n-1 , n , n+1 these are 3 consecutive integers, we know we will have both the multiple of 2 and 3 in three consecutive integer. If that integer is same i.e. n is multiple of both 2 and 3 , then n-1 and n+1 will never have any factor for 24.

But if n is not divisible by both 2 and 3 that means n-1 and n+1 will be both be even with one of them having 3 as factor. One another point to be noted is , since n-1 and n+1 are consecutive even numbers that implies one of them must be divisible by 4.

so (n-1)(n+1) will be always divisible by 2*4*3 = 24 hence r will always remain as 0
(When n is neither divisible by 2 nor by 3)
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Re: If n is a positive integer and r is the remainder when (n - [#permalink]

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New post 26 Sep 2012, 10:38
Gurpreet for statement 2 you chose n = 12 which is divisible by 3.

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Re: If n is a positive integer and r is the remainder when (n - [#permalink]

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New post 26 Sep 2012, 10:52
manwiththeharmonica wrote:
Gurpreet for statement 2 you chose n = 12 which is divisible by 3.

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:lol: u r right.... but 2.5 years back.. :)

just take n = 14.. so (n-1 )*(n+1) = 13*15 which does not give r =0
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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.

Answer: C.

Hope it's clear.
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Re: If n is a positive integer and r is the remainder when (n - [#permalink]

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New post 26 Sep 2012, 12:44
ksharma12 wrote:
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.



(1) n must be odd. Try n = 1 and n = 3.
Not sufficient.

(2) Try n = 1 and n = 2.
Not sufficient.

(1) and (2) together: When divided by 6, the remainders can be 0, 1, 2, 3, 4, 5. Since n is not divisible neither by 2, nor by 3, when divide by 6, n can give remainder 1 or 5, which means n is of the form 6k + 1 or 6k - 1, for some positive integer k.
If n = 6k + 1, then (n - 1)(n + 1) = 6k(6k + 2) = 12k(3k + 1) = 12k(k + 2k + 1). Since 2k + 1 is odd, k and 3k + 1 are of different parities, so their product is for sure divisible by 2. Altogether, (n - 1)(n + 1) is divisible by 24.
If n = 6k - 1, then (n - 1)(n + 1) = (6k - 2)6k)= 12k(3k - 1) = 12k(k + 2k - 1). Since 2k - 1 is odd, k and 3k - 1 are of different parities, so their product is for sure divisible by 2. Altogether, (n - 1)(n + 1) is divisible by 24.
The remainder must be 0.
Sufficient.

Answer C.
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Re: If n is a positive integer and r is the remainder when (n - [#permalink]

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Re: If n is a positive integer and r is the remainder when (n -   [#permalink] 03 Feb 2016, 10:47
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