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# if n is a +ve intiger and r is the remainder when (n-1)(n+1)

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SVP
Joined: 05 Jul 2006
Posts: 1519
Followers: 5

Kudos [?]: 115 [0], given: 39

if n is a +ve intiger and r is the remainder when (n-1)(n+1) [#permalink]  07 Sep 2006, 02:58
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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
Current Student
Joined: 29 Jan 2005
Posts: 5244
Followers: 23

Kudos [?]: 179 [0], given: 0

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
Manager
Joined: 30 Jun 2006
Posts: 87
Followers: 0

Kudos [?]: 4 [0], given: 0

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.
CEO
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 18

Kudos [?]: 121 [0], given: 0

(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

VP
Joined: 02 Jun 2006
Posts: 1267
Followers: 2

Kudos [?]: 42 [0], given: 0

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.

SVP
Joined: 05 Jul 2006
Posts: 1519
Followers: 5

Kudos [?]: 115 [0], given: 39

OA is C

Folks all of you are just great
SVP
Joined: 05 Jul 2006
Posts: 1519
Followers: 5

Kudos [?]: 115 [0], given: 39

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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