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SVP
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if n is a +ve intiger and r is the remainder when (n-1)(n+1) [#permalink]
 07 Sep 2006, 03:58
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if n is a +ve intiger and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?
1) n is not devisible by 2
2) n is not devisible by 3
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Current Student
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yezz wrote: jainvik7 can u plz explain
(n+1)(n-1)/24 = remainder r ( to get a remainder n>/= to 24)
solve for r
1. pick 5, 7, 9 , 11.... all different remainders INSUFF
2. pick 5, 7, 8, 11 ... all differeent remainders INSUFF
Together, 7 and 11 ...and all subsequent primes result in a remainder of 0
(C) sweet formula 
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Manager
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Lets take the smallest possible number that satisfies the condition mentioned in the above statements. The number would be 5.
Now, (n + 1) (n - 1) = 4 * 6 = 24
Lets take a step further and choose another number that satisfies the mentioned conditions - The number would be 7
Now, (n + 1) (n - 1) = 6 * 8 = 48 = 24 * 2
Another number --> 11
Now, (n + 1) (n - 1) = 10 * 12 = 120 = 24 * 5
All we are doing here is picking up 2 consecutive even numbers which is greater than or equal to 4 and one of them is also divisible by 3.
( because n-1, n , n + 1 are three consecutive numbers and one of them should be divisible by 3.)
I am not sure if this is the best explanation. Some one else would definitely have a better explanation.
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CEO
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(n-1) (n+1) = n^2 -1
n^2 -1 is always divisible by 3 where n is a integer not equal to 0 and
not a multiple of 3.
n^2 -1 could be 3, 8 ,15, 24 ,35 , 48, 63, 80, 99, 120, 143, 168.......
St1: n is odd.: n^2 -1 could be 8, 24 , 48, 80, 120, 168....... remainder could be 8,0: INSUFF
St2: n is not divisible by 3. n^2 -1 could be 3, 15 , 24, 48, 63, 99, 120, 168.......: Remainder could be 3, 15, 1, 0 :INSUFF
Together:
n^2 -1 could be 24, 48, 120, 168. Remainder is 0.: SUFF
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
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VP
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Answer: C
S1: n not divisible by 2
n is odd, n^2 -1 is even
But divide by 24 gives different remainders...
Not sufficient.
S2: n not divisible by 3
n = 11, n^-1 = 120, Remainder = 0
n= 10, n^-1 = 99, Remainder )non zero..
Not Sufficient.
S1 & S2:
n = {5, 7, 11, 13, ..}
Remainder is zero in all cases.
sufficient.
Answer: C
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SVP
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OA is C
Folks all of you are just great
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SVP
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Folks this is How the legendary brilliant Kevincan analysed this problem
I think it is a good lesson to note
Enjoy 
Both together
Either n=1, in which case n^2-1=0
or n is greater than from 1, in which case n-1 and n+1 are even and nonzero, because n is odd from (1). One of {n-1,n+1} will be a multiple of 2 and the other a multiple of 4, so (n-1)*(n+1) is divisible by 8.
Also, in three consecutive nonzero numbers, one will always be a multiple of 3. Since n is not a multiple of 3 (from (2)), either n-1 or n+1 is a multiple of 3 and so (n-1)*(n+1) is a multiple of 3 and thus of 24.
In either case, when n is divided by 24, the remainder is 0 SUFF
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