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# If n is a positive integer and r is the remainder when

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Intern
Joined: 23 Jun 2009
Posts: 47
If n is a positive integer and r is the remainder when [#permalink]

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26 Jul 2009, 22:03
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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.

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Senior Manager
Joined: 25 Jun 2009
Posts: 287

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27 Jul 2009, 01:04
uzonwagba 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.

IMO it should be C,

1. Is not suff because it only tells us that the N is odd and we can diff possibilities for it for e.g n= 9 which means (n-1)(n+1)=8x10=80 which is gives a remainder of 8 and we can also have N= 5 where (n-1)(n+1)= 4x6=24 remainder =0

2. Similarly this is not suff; and we can have diff values like n=8 , (n-1)(n+1)=7x9 = 63, hence r= 15 and we have n=5 as well where r =0.
But we combine both these 2 we would get values with the pattern as N=5, 7, 11, 13, 17, 19, 23, .... and in all these cases (n-1)(n+1) will give us the remainder as 0 only ((n-1)(n+1) will have multiples of 2, and 3).

Cheers
Intern
Joined: 23 Jun 2009
Posts: 47

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27 Jul 2009, 01:22
nitishmahajan wrote:
uzonwagba 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.

IMO it should be C,

1. Is not suff because it only tells us that the N is odd and we can diff possibilities for it for e.g n= 9 which means (n-1)(n+1)=8x10=80 which is gives a remainder of 8 and we can also have N= 5 where (n-1)(n+1)= 4x6=24 remainder =0

2. Similarly this is not suff; and we can have diff values like n=8 , (n-1)(n+1)=7x9 = 63, hence r= 15 and we have n=5 as well where r =0.
But we combine both these 2 we would get values with the pattern as N=5, 7, 11, 13, 17, 19, 23, .... and in all these cases (n-1)(n+1) will give us the remainder as 0 only ((n-1)(n+1) will have multiples of 2, and 3).

Cheers

The answer is C. You are correct! Could you please explain the logic/method behind finding out that all cases in the pattern 5, 7, 11, 13, 17, 19, 23 will give us remainder 0? Did you just use trial and error (i.e. trying each value), or was there any logic you figured out with which one could make a general conclusion like that? Also, would it be right to say that any number that has multiples of 2 and 3 would be completely divisible by 24? How? Thx.
Senior Manager
Joined: 25 Jun 2009
Posts: 287

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27 Jul 2009, 03:05
uzonwagba wrote:
nitishmahajan wrote:
uzonwagba 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.

IMO it should be C,

1. Is not suff because it only tells us that the N is odd and we can diff possibilities for it for e.g n= 9 which means (n-1)(n+1)=8x10=80 which is gives a remainder of 8 and we can also have N= 5 where (n-1)(n+1)= 4x6=24 remainder =0

2. Similarly this is not suff; and we can have diff values like n=8 , (n-1)(n+1)=7x9 = 63, hence r= 15 and we have n=5 as well where r =0.
But we combine both these 2 we would get values with the pattern as N=5, 7, 11, 13, 17, 19, 23, .... and in all these cases (n-1)(n+1) will give us the remainder as 0 only ((n-1)(n+1) will have multiples of 2, and 3).

Cheers

The answer is C. You are correct! Could you please explain the logic/method behind finding out that all cases in the pattern 5, 7, 11, 13, 17, 19, 23 will give us remainder 0? Did you just use trial and error (i.e. trying each value), or was there any logic you figured out with which one could make a general conclusion like that? Also, would it be right to say that any number that has multiples of 2 and 3 would be completely divisible by 24? How? Thx.

From both the statements its evident now that (n-1) or (n+1) will be the multiple of either or 3 ( I mean one of them will be a multiple of 2 and second will be a multiple of 3 ) And now we factorize 24 we get , 2,2,2,3. Now we look at the equation the minimum value which N can have is 5, So, (n-1) will be 4 (2, 2) and (N+1) will be 6 (2,3). Hence the minimum factors which we need is there in the minimum value of N ( i.e 5) and now any no greater than 5 will be having the factors atleast these factors i.e 2,2,2 and 3.

Also, would it be right to say that any number that has multiples of 2 and 3 would be completely divisible by 24? How?

No it wont be right to say that any number having multiples of 2 and 3 would be completely divisible by 24 take for eg 18. it as multiples as 2,3 and but not 2,2,2, and 3.

Let me know it that helps else I will try to elaborate more
Intern
Joined: 23 Jun 2009
Posts: 47

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27 Jul 2009, 11:19
whatthehell wrote:
IMO C

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Manager
Joined: 21 Jun 2009
Posts: 137

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27 Jul 2009, 11:47
n n-1 n+1 P R (when /24)
5 4 6 24 0
6 5 7 35 11
7 6 8 48 0
8 7 9 63 15
9 8 10 80 8
10 9 11 99 3
11 10 12 120 0
13 12 14 168 0

1) when n is not divisible by 2 -> n=5,7,9,11(note that 9 will give a remainder of 8 while others will give a 0 reminder). Therefore it is not sufficient.
2) when n is not divisible by 3 ->n =5,7,8,10,11 (remainders are all over the place therefore not sufficient)
1+2) will eliminate->(5,7,8,9,10), look a the others- >5,7-> both has 0 as the reminder. Hence appears sufficient.
I tried n =13 for further confirmation. since it satisfies both 1 and 2 and it gives remainder = 0. Hence suffienciency verified.

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Re: DS Problem #8: Please take a stab, anyone! Thx   [#permalink] 27 Jul 2009, 11:47
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