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# ds#positive integer

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30 Sep 2005, 11:14
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OA to be followed....
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hey ya......

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30 Sep 2005, 12:08
C.
r = 0
here is partial explanation:

question stem:
(n-1) * (n+1) = r (mod 24) , 0 <= r < 24, or
n^2 - 1 = r (mod 24) or
24 / n^2 -1 -r where 0 <= r < 24

the only thing i remember is that any number of the form (n^2 - 1) is divisible by 24 if n is prime > 3.

(1) and (2) confirms that n is prime and is > 3

when 24/n^2-1, r will be 0

Last edited by duttsit on 30 Sep 2005, 13:30, edited 1 time in total.
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30 Sep 2005, 12:10
Ok lets see....

if we have (n-1)n(n+1) well one of them is definetly a multiple of 3....

so (1) says 2 is not a factor of n...

well we know that [(n-1)(n)(n+1)]/24= 1/8 [(n-1)(n+1)] well we know then that (n-1)(n+1) both have at least three factors of 2....so the remainder r =0....

sufficient.

(11)
3 is not a factor of n...well then it means either N-1 or N+1 is a factor of 3...so we have n as a factor of 2...but we dont know anything about n+1 or n vice versa....so insufficient....we cant tell how many factors of 2 are there in n+1....

A it is...
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30 Sep 2005, 12:59
fresinha12 wrote:
Ok lets see....

if we have (n-1)n(n+1) well one of them is definetly a multiple of 3....

so (1) says 2 is not a factor of n...

well we know that [(n-1)(n)(n+1)]/24= 1/8 [(n-1)(n+1)] well we know then that (n-1)(n+1) both have at least three factors of 2....so the remainder r =0....

sufficient.

(11)
3 is not a factor of n...well then it means either N-1 or N+1 is a factor of 3...so we have n as a factor of 2...but we dont know anything about n+1 or n vice versa....so insufficient....we cant tell how many factors of 2 are there in n+1....

A it is...

Freshina12, the stem says (n+1) (n-1). thus C it is
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01 Oct 2005, 13:16
Can you guys explain it little more.....
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01 Oct 2005, 13:43
nakib77 wrote:
Can you guys explain it little more.....

I don't know if this is the easiest way to explain .. but from the stem..
For any number n>0 => (n-1)(n+1) has to be a factor of either 2 or 3 or both (plug some random values.. say n =59 => (58)(60) =>2,3,6 factors or say n=26 =>(25)(27) => 3 is a factor)

1) 2 is not a factor of n, out example of 59 shows r will be 0 but [edited: should consider something greater than 24] n=27, (26)(28) r is not zero. Insuff
2) 3 say n=26, (25)(27) divided by 24 will give some remainder r but this value again varies for every even value of n you consider. Insuff.

1&2 ... picking numbers is easy (since it cannot be a multiple of 2,3 or 6) ... say 49.. i.e. (48)(50) => r=0, or n=121 (120)(122) => r=0. So r=0 now for all values on n which is not a multiple of 2,3,6.

May not be the best explanation...But hope that helps.
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01 Oct 2005, 17:06
duttsit wrote:
the only thing i remember is that any number of the form (n^2 - 1) is divisible by 24 if n is prime > 3.

That's a real nice tip. Didn't know that.
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03 Oct 2005, 05:00
Why is n = 1 excluded?

1 is a positive integer and 2 and 3 are not factors of 1.
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04 Oct 2005, 03:15
nakib77 wrote:
OA to be followed....

My answer is A. The equn can be simplified as n^2 -1

From 1, 2 is not a factor of n. which mean n has to be odd.
If n = 3 then n ^2 -1 = 8 when divided by 24 gives the remanider 8.
If n = 5 then n ^ 2 - 1 = 24 gives the remainder 0.
If n = 7 then n ^ 2 - 1 = 48 gives the remainder 0.

So n should be 3.

From 2, 3 is not a factor of n. so either n or n+1 should be a factor of 2. This leaves us with more than one answer to choose. So B is not sufficient.

Re: ds#positive integer   [#permalink] 04 Oct 2005, 03:15
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