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Re: GMATPrep - Good one. [#permalink]
21 Apr 2008, 21:11

3

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kyatin wrote:

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

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

C

(1) n is odd if n is odd, then pick some numbers... (n, (n-1)*(n+1), r) (1, 0, 0) (3, 8, 8) INSUFFICIENT

(2) (1, 0, 0) (2, 3, 3) INSUFFICIENT

Together, find some more examples... (1, 0, 0) (3, 8, 8) (5, 24, 0) (7, 48, 0) (9, 56, 8) So the pattern is if n is divisible by 3, then remainder is 8, if not remainder is 0. Together is SUFFICIENT

Re: GMATPrep - Good one. [#permalink]
01 May 2008, 00:47

1

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Quote:

(3, 8, 8) So the pattern is if n is divisible by 3, then remainder is 8, if not remainder is 0. Together is SUFFICIENT

Do you misconceive the circumstance? Cos' (n-1)(n+1) is divided by 24, the remainder is r, thus (n-1)(n+1)>24. Further, should we try many numbers to answer whether or not the given information is sufficient? I think we'll conclude only that one case is not sufficient if we find a number counts against the case, but we can not conclude that one case is sufficient if we find a number does not count against the case. That why I cast doubt on C.

Re: GMATPrep - Good one. [#permalink]
01 May 2008, 05:55

lexis wrote:

Quote:

(3, 8, 8) So the pattern is if n is divisible by 3, then remainder is 8, if not remainder is 0. Together is SUFFICIENT

Do you misconceive the circumstance? Cos' (n-1)(n+1) is divided by 24, the remainder is r, thus (n-1)(n+1)>24. Further, should we try many numbers to answer whether or not the given information is sufficient? I think we'll conclude only that one case is not sufficient if we find a number counts against the case, but we can not conclude that one case is sufficient if we find a number does not count against the case. That why I cast doubt on C.

Need a general solution!

(n-1)*(n+1) can be less than 24. For example, fraction 1/4 has a remainder of 1. Regarding how many numbers we should try, if you are able to figure out the pattern, that should be sufficient.

Re: GMATPrep - Good one. [#permalink]
02 May 2008, 18:10

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my thoughts -

N is not divisible by 2: Means n-1 and n+1 are even. And - As they are consecutive numbers one of them is divisible by 2 and other by 4. So: (n-1)* (N+1) is divisible by 8 - not sufficient to conclude that the product is divisible by 24

N is not divisible by 3: means as n-1,n,n+1 are consecutive , one of them: (n-1) or (n+1) is divisible by 3. Again, not sufficient for 24.

Together - (n-1)(n+1) is divisible by 2,4,3 i.e: 2*4*3 = 24 . Hence, r=0

Re: GMATPrep - Good one. [#permalink]
05 May 2008, 17:47

HefAR wrote:

my thoughts -

N is not divisible by 2: Means n-1 and n+1 are even. And - As they are consecutive numbers one of them is divisible by 2 and other by 4. So: (n-1)* (N+1) is divisible by 8 - not sufficient to conclude that the product is divisible by 24

N is not divisible by 3: means as n-1,n,n+1 are consecutive , one of them: (n-1) or (n+1) is divisible by 3. Again, not sufficient for 24.

Together - (n-1)(n+1) is divisible by 2,4,3 i.e: 2*4*3 = 24 . Hence, r=0

does it make sense?

I think this very much makes sense, thanks.

I got this question on GMATPrep (it's roughly a Q50 question, give or take a few), and was able to solve it by just plugging in numbers, but it's very helpful to know where the logic comes from - much easier and quicker to solve!

Re: GMATPrep - Good one. [#permalink]
24 Nov 2010, 07:06

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1

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This is how I solved it From statement 1) 2 is not a factor of n ( we can conclude that n is odd ) so by plugging in numbers like n=15 remainder=1 and for n=17 remainder =0 i.e not sufficient From statement 2 it says that 3 is not a factor of n so we can have n as an even number or it can be prime number( not #0 again by plugging in values we can find that the remainders are different.i.e insufficient.

So by combining both we can conclude that the value for n will be a prime number which is not 2 and 3 for n=19 remainder =0 ; for n=13 remainder =0 ( just testing cases )

Re: GMATPrep - Good one. [#permalink]
19 Dec 2010, 15:21

This was my approach. Did you actually calculate out the remainders when you tried n=15 and 17? I just used n=2 or n=5 to arrive at my examples. C gives you that n has to be prime other than 2 and 3 and you test a few cases to find that all non 2 and 3 primes will provide a 0 remainder for N.

Re: GMATPrep - Good one. [#permalink]
19 Dec 2010, 15:37

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Expert's post

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kyatin wrote:

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

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

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: If n is a positive integer and r is remainder, when [#permalink]
14 May 2015, 00:57

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Re: If n is a positive integer and r is remainder, when [#permalink]
17 May 2015, 20:51

rite2deepti wrote:

This is how I solved it From statement 1) 2 is not a factor of n ( we can conclude that n is odd ) so by plugging in numbers like n=15 remainder=1 and for n=17 remainder =0 i.e not sufficient From statement 2 it says that 3 is not a factor of n so we can have n as an even number or it can be prime number( not #0 again by plugging in values we can find that the remainders are different.i.e insufficient.

So by combining both we can conclude that the value for n will be a prime number which is not 2 and 3 for n=19 remainder =0 ; for n=13 remainder =0 ( just testing cases )

SO answer is C

Guys let me know if my solution is good ?

Great....That,s how I solved this question.... _________________

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