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Re: Data Sufficiency with remainder [#permalink]
04 Aug 2010, 02:54

3

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Expert's post

2

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kwhitejr wrote:

Can anyone demonstrate the following?

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) 2 is not a factor of n (2) 3 is not a factor of n

Hi, and welcome to Gmat Club! Below is the solution for your problem:

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?

Number plugging 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.

hi all, please help solving the following question...

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 is not divisible by 2 => n is not even.

n could be 1, 3, 5, 7, 9

and (n+1) * (n-1) could be different values.

2) n is not divisible by 3.

n could be 1, 2, 4, 5, 7, 8, 10...

again (n+1) * (n-1) could be different values.

Combine (1) and (2), n could be 1, 5, 7, 11, 13; ((n+1), (n-1)) would contain a multiple of 6 and 4 when n is 5, 7, 11, 13 ..

When n =1, (n-1) is zero and the remainder is zero.

Multiple of 6 and 4 would contain (3*2) and (2^2) = > (3 * 2 * 2 *2) = > 24.

(n-1)*(n+1) could be 4*6 or 6*8 and either way the remainder when (n-1)*(n+1)/24 would be 0.

Hence both statements is sufficient to answer this. _________________

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PS: This way of solving would probably be better in a GMAT hall with a 2 minute time constraint. A rigorous algebraic proof can also be derived otherwise.

Re: Data Sufficiency with remainder [#permalink]
15 Sep 2010, 16:44

I got this problem right (on the GMATPrep #1 CAT) but I'm sure it took me at least 3 minutes, if not closer to 5. At least a minute to set up the problem and figure out how to approach it (the way described above is what I first thought of, but I tried to think of easier ways), another minute to test numbers, and another minute to think of ways on how my test could be wrong.

Any suggestions on how to cut down on timing for unfamiliar problems like these? These divisor/remainder problems always get to me. _________________

Re: Data Sufficiency with remainder [#permalink]
16 Sep 2010, 23:02

Folks,

Consider this approach.

Stmnt 1: N is not divisible by 2 ==> n=2k+1

==> n-1 * n+1 = 2k(2k+2)=4K(k+1) te K(K+1) is always divisible by 2 as they are consecutive integers. hence 4K(k+1) is always divisible by 8, but may not be by 24 (i.e. 8*3)....NOT Suff. Summary from stmnt 1: (n-1)*(n+1) is always divisible by 8

Stmnt 2: N is not divisible by 3 ==> 2 cases -> remainder can be 1 or 2 ==> n=3k+1 or 3K+2

with n=3K+1 ==> ==> n-1 * n+1 = 3k(3k+2)=is divisible by 3 but not by 2 for ODD K: hence not divisible by 24 (i.e. 2^3 * 3) with n=3K+2 ==> ==> n-1 * n+1 = (3k+1)(3k+3)=3(3k+1)(K+1) is divisible by 3 but not by 2 for EVEN K Not suff. Summary from stmnt 2: (n-1)*(n+1) is always divisible by 3

1 & 2 summaries. (n-1)*(n+1) is always divisible by 8*3 = 24..Suff.

Answer "C"

always divisible by 2 as they are consecutive integers. hence 4K(k+1) is always divisible by 8, but may not be by 24 (i.e. 8*3)....NOT Suff.

Divisiblity remainder [#permalink]
10 Feb 2011, 23:37

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) 2 is not a factor of n. (2) 3 is not a factor of

I marked ans e as we cannot find one value while using two other prime 5 & 7. So defainete value of r can be find out. But the ans is c. can anyone confim

Re: Divisiblity remainder [#permalink]
11 Feb 2011, 00:16

1

This post was BOOKMARKED

24 has the factors; 2*2*2*3

If (n-1)(n+1) contains at least the above factors, it will be divisible by 24.

1. Means; n is odd; and n-1 and n+1 are consecutive even; thus one of them will have at least 2 as factor and the other at least 2*2 Thus; we know (n-1)(n+1) contains 2*2*2 as factor. But, it may or may not contain 3 as its factor.

Not sufficient.

2. if n is not divisible by 3; one of (n-1)(n+1) must be divisible by 3. One number of any set of 3 consecutive numbers is always divisible by 3. We know n is not divisible and n-1,n,n+1 are consecutive. if n is not divisible by 3, one of (n-1) and (n+1) must be divisible by 3 or have a factor as 3. However, we don't know whether these numbers have 2*2*2 as factors.

Not sufficient.

Combining both; (n-1)(n+1) has factors 2*2*2*3 and must be divisible by 24. Remainder 0.

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: If n is a positive integer and r is the remainder when (n-1) [#permalink]
02 Oct 2014, 08:06

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Re: If n is a positive integer and r is the remainder when (n-1) [#permalink]
22 Feb 2015, 04:53

Bunuel wrote:

kwhitejr wrote:

Can anyone demonstrate the following?

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) 2 is not a factor of n (2) 3 is not a factor of n

Hi, and welcome to Gmat Club! Below is the solution for your problem:

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?

Number plugging 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 helps.

Dear Bunuel,

Thanks for sharing the alternatives to approaching this problem. I have a question here. What if we choose a number, n, such as 23 (which is neither divisible by 2 nor by 3)? In that case the given expression (n-1)*(n+1) will be 22*24, which when divided by 24 will leave a remainder 22.

Re: If n is a positive integer and r is the remainder when (n-1) [#permalink]
22 Feb 2015, 05:33

1

This post received KUDOS

Expert's post

Smgs wrote:

Bunuel wrote:

kwhitejr wrote:

Can anyone demonstrate the following?

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) 2 is not a factor of n (2) 3 is not a factor of n

Hi, and welcome to Gmat Club! Below is the solution for your problem:

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?

Number plugging 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 helps.

Dear Bunuel,

Thanks for sharing the alternatives to approaching this problem. I have a question here. What if we choose a number, n, such as 23 (which is neither divisible by 2 nor by 3)? In that case the given expression (n-1)*(n+1) will be 22*24, which when divided by 24 will leave a remainder 22.

Kindly advise!

Thanks!

22*24 is a multiple of 24, so when divided by 24 it yields the remainder of 0. _________________

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