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# n is a positive integer. Find the remainder of (n+1)(n-1)/24

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Manager
Joined: 03 Aug 2005
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n is a positive integer. Find the remainder of (n+1)(n-1)/24 [#permalink]

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06 Nov 2005, 13:22
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n is a positive integer. Find the remainder of (n+1)(n-1)/24

(1) 2 is not a factor of n
(2) 3 is not a factor of n

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VP
Joined: 30 Sep 2004
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06 Nov 2005, 13:30
E)...stem asks remainder for (n^2-1)/24...1) and 2) are insuff...1)+2)...n must be prime greater than 3...5,11 yield diff. remainders...so insuff as well...
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Manager
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06 Nov 2005, 19:35
christoph wrote:
E)...stem asks remainder for (n^2-1)/24...1) and 2) are insuff...1)+2)...n must be prime greater than 3...5,11 yield diff. remainders...so insuff as well...

nope.. They yield the same reaminder 0. The answer is C. n=5 then answer is 4*6/24 reaminder 0.Similarly for all the numbers.

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SVP
Joined: 24 Sep 2005
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06 Nov 2005, 20:02
jdtomatito wrote:
I think this question has been posted before, but I couldn't find the thread.

n is a positive integer. Find the remainder of (n+1)(n-1)/24

(1) 2 is not a factor of n
(2) 3 is not a factor of n

As other members chose different choices of C and E. I now consider C
(1)(2)

from (2) we know n divided by 3 has remainder of 1 or 2, in any cases, (n-1)(n+1) is divisible by 3.
In order to be divisible by 24, (n+1)(n-1) must be divisible by 8
+from (1) ,we eliminate those n which divided by 8 have remainders of 0,2,4,6,8 because such n is divisible by 2, which conflicts the statement.
+ if n divided by 8 has r (remainder) of 1, then n-1 must be divisible by 8
+ if r=7 ---> n+1 must be divisible by 8
+ if r=3 --> n= 8x+3 ( x is integer) ---> n+1= 8x+4 is divisible by 4 AND n-1 = 8x+2 is divisible by 2. Thus, (n+1)(n-1) is divisible by 8.
+ if r=5 --->n= 8z+5( z is integer) ----> n+1= 8z+6 is divisible by 2 AND
n-1= 8z+4 is divisible by 4 ------> (n+1)(n-1) is divisible by 8

THUS, with (1)and (2) we always have (n+1)(n-1) is divisible by 8...the product is also divisible by 3, SO (n+1)(n-1) is divisible by 8*3= 24

C it is.

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VP
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07 Nov 2005, 01:08
amy_v wrote:
christoph wrote:
E)...stem asks remainder for (n^2-1)/24...1) and 2) are insuff...1)+2)...n must be prime greater than 3...5,11 yield diff. remainders...so insuff as well...

nope.. They yield the same reaminder 0. The answer is C. n=5 then answer is 4*6/24 reaminder 0.Similarly for all the numbers.

oh mann what a day full of mistakes...

Quote:
E)...stem asks remainder for (n^2-1)/24...1) and 2) are insuff...1)+2)...n must be prime greater than 3...5,11 yield diff. remainders...so insuff as well...

primes 5 as well as 11 give remainder 0...n^2-1/24=24/24 and 120/24...
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Manager
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07 Nov 2005, 08:57
OA is C.

Anyway I still think the right answer is E.

If you pick n=1, 1 is a positive integer and neither 2 nor 3 are fators of 1.

(n-1)(n+1) = 0 and for 0/24 r = 24.

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SVP
Joined: 24 Sep 2005
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07 Nov 2005, 09:02
jdtomatito wrote:
OA is C.

Anyway I still think the right answer is E.

If you pick n=1, 1 is a positive integer and neither 2 nor 3 are fators of 1.

(n-1)(n+1) = 0 and for 0/24 r = 24.

The bold part is incorrect 0/24 has remainder = 0
uhm, if you say the r= 24, as 0/24= 0 , you can write 0= 24*0 +24 ( This is extremely unreasonable)

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Manager
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07 Nov 2005, 10:47
You are absolutely right, I do not know what I was thinking...

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Current Student
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07 Nov 2005, 10:58
late on this...but C it is...

basically we need to look at N+1 and N-1

(1) tells us that N is not even...then either N-1 or N+1 is even...dont know...so lets say N=3, then N-1=2, N+1=4, 8/24 leaves a remainder...

if N is 5 then 6*4=24...leave no remainder... Insuff

(2) tell us that N is not 3...so N-1 or N+1 must be a multiple of 3...

so if N <> 3, then if it is 2 then 1*3/24 leaves are remainder...if N=5 then there is no remainder...

combining them...Sufficient...N is a prime number...but we dont need to look at N; we are asked about N+1 and N-1....Sufficient...remainder is always 0

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Intern
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03 Jan 2006, 15:48
I think that C is correct, but the explanation is not correct.

n does not need to be prime.

(1) If n is not even, then both n-1 and n+1 are even, i.e. n-1 and n+1 can be divided by 2. Even more important either n-1 or n+1 can be divided by 4. In conclusion (n-1)(n+1) can be divided by 2^3.
Combining (1) and (2) means (n-1)(n+2) can be divided by 2^3*3=24. Hence the remainder is always 0 if (1) and (2) are valid.

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03 Jan 2006, 15:48
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