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Re: DS from GMATPrep [#permalink]
27 Jun 2010, 20:26

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This post received KUDOS

Expert's post

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This post was BOOKMARKED

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.

Re: NEED HELP - 2 DATA SUFFICIENCY PROBLEMS [#permalink]
03 Jul 2010, 21:54

Expert's post

Question 2:

To be divisible by 24, a number must be divisible by 8 and by 3.

Statement 1: n is not divisible by 2. This basically says that n is an odd number. So if n is an odd number, both n-1 and n+1 are even numbers, which means that (n-1)(n+1) must be divisible by 4.

n-1 = 2k n+1 = 2p

Between any two consecutive even integers, one has to be divisible by 4, so we know that the given number is divisible by 8.

Statement 2: n is not divisible by 3. However, between any three consecutive numbers one has to be divisible by 3. Considering the consecutive numbers (n-1)(n)(n+1) we know that n isn't divisible by 3, so either (n-1) or (n+1) should be divisible by 3.

Combining both together, we know that (n-1)(n+1) is divisible by 8 and 3 and hence 24. So reminder = 0. Hence answer is C.

Re: NEED HELP - 2 DATA SUFFICIENCY PROBLEMS [#permalink]
04 Jul 2010, 04:21

in question 2, statement 2 alone will not be sufficient unless taken in conjunction with stmt 1 because if we consider ONLY stmt 2 - then 10 is a value which is NOT divisible by 3 and (10-1)*(10+1) = gives 99 which does not leave a remainder 0 when divided by 24. Thus IMO the answer to problem 2 should be A - stmt 1 alone is sufficient... pls correct me if I am wrong.

Re: NEED HELP - 2 DATA SUFFICIENCY PROBLEMS [#permalink]
06 Jul 2010, 22:58

whiplash2411 wrote:

Question 2:

To be divisible by 24, a number must be divisible by 8 and by 3.

Statement 1: n is not divisible by 2. This basically says that n is an odd number. So if n is an odd number, both n-1 and n+1 are even numbers, which means that (n-1)(n+1) must be divisible by 4.

n-1 = 2k n+1 = 2p

Between any two consecutive even integers, one has to be divisible by 4, so we know that the given number is divisible by 8.

Statement 2: n is not divisible by 3. However, between any three consecutive numbers one has to be divisible by 3. Considering the consecutive numbers (n-1)(n)(n+1) we know that n isn't divisible by 3, so either (n-1) or (n+1) should be divisible by 3.

Combining both together, we know that (n-1)(n+1) is divisible by 8 and 3 and hence 24. So reminder = 0. Hence answer is C.

n is not divisble by 2 implies n-1 and n+1 are even , this implies the product of these two numbers i divisible by 4 and not 8. both the statements taken together say that the number is not even and not divisible by 3 , meaning all odd numbers not divisible by 3 . these are 1,5,7,11,13,17--- , if u (square of any of these minus 1 )/24 = 1 and not 0

Re: DS from GMATPrep [#permalink]
07 Jul 2010, 02:46

Expert's post

utin wrote:

@bunuel... 3 divided by 24 yields remainder of 3???

ruchichitral wrote:

bunnel:- i am too confused....how come 3 divided by 24 the remainder is 3?

THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

So when divisor (24 in our case) is more than dividend (3 in our case) then the reminder equals to the dividend:

3 divided by 24 yields a reminder of 3 --> \(3=0*24+3\);

or:

5 divided by 6 yields a reminder of 5 --> \(5=0*6+5\).

Re: If n is a positive integer and r is the remainder when (n-1) [#permalink]
01 Dec 2014, 13:09

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