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If n is a positive Integer and r is the remainder when

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If n is a positive Integer and r is the remainder when [#permalink] New post 10 Dec 2004, 06:30
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B
C
D
E

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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) 2 is not a factor of n

2) 3 is not a factor of n
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 [#permalink] New post 10 Dec 2004, 08:18
I will go with ‘E’ on this one.
From statement 1: n is odd.
From statement 2: The sum of the digits of n is not divisible by 3
Both put together: n can be 5, 7, 11, 13, so depending on n the remainder r varies.
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 [#permalink] New post 10 Dec 2004, 09:04
Dookie style !!

1st statement : n is obviously odd which means (n-1)(n+1) is a multiple of 4 - not enough

2nd statement : 3 is not a factor of n so (n-1)(n+1) is a multiple of 3

2 combined : multiple of 12 so remainder could be either 0 or 12.

I pick E
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 [#permalink] New post 10 Dec 2004, 09:16
:lol:
Gotcha again !

Wanna give it another try ?
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 [#permalink] New post 10 Dec 2004, 09:57
C.

n would be a prime above 3. Remainder is 0.
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 [#permalink] New post 10 Dec 2004, 10:55
N doesn't have to be a prime. It could be.. 85 for example.
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 [#permalink] New post 10 Dec 2004, 12:10
C

[1] n is odd and that both (n+1) and (n-1) are divisible by 2
[2] Either (n-1) or (n+1) is divisible by 3

consequently (n-1)(n+1) is divisible by 12. Since n has to be greater than
24 for (n-1)(n+1) to leave a remainder when divided by 24 it is therfore divisible by 12 and at least its "first" multiple, that is, 24.

so both statements together are sufficient.

The remainder r = 0.
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 [#permalink] New post 10 Dec 2004, 12:11
yes, I tried n^2-1 for the first few (prime) numbers... but it could work for others too.
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 [#permalink] New post 10 Dec 2004, 12:29
oxon wrote:
C

[1] n is odd and that both (n+1) and (n-1) are divisible by 2
[2] Either (n-1) or (n+1) is divisible by 3

consequently (n-1)(n+1) is divisible by 12. Since n has to be greater than
24 for (n-1)(n+1) to leave a remainder when divided by 24 it is therfore divisible by 12 and at least its "first" multiple, that is, 24.

so both statements together are sufficient.

The remainder r = 0.


I think it's C too, but I have a different explanation.

Look at what they're saying. n isn't divisible by 2 or 3. That means n is an odd number that's not divisible by 3.

That means it could be:
5
7
11
13
17
19
23
25

and on and on. In all cases, the numbers above and below each of these will either be divisible by 4 or by 6. It just works out that way. So when we multiply them together, the answer will be divisible by 24.

Even if we start with 5, this will be true.

Great Q dookie!
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 [#permalink] New post 10 Dec 2004, 12:44
Great, C it is.

1st statement : n is odd. (n-1)(n+1) is a multiple of 8

2nd statement : 3 is not a factor of n so (n-1)(n+1) is a multiple of 3

2 combined : multiple of 24. Remainder = 0.
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 [#permalink] New post 10 Dec 2004, 12:46
ian777 - great stuff!
what I meant in my post is that if you pick n<24 (n<23 in fact) (n-1)(n+1)
will be divisible by both 4 and 3. If you pick n>23 then (n-1)(n+1) will be divisible by at least one multiple of 12 hence 24.

Dookie - You rock!
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 [#permalink] New post 10 Dec 2004, 13:46
Dookie wrote:
Great, C it is.

1st statement : n is odd. (n-1)(n+1) is a multiple of 8

2nd statement : 3 is not a factor of n so (n-1)(n+1) is a multiple of 3

2 combined : multiple of 24. Remainder = 0.


Dookie, Can you explain how from statement 1 you got that (n-1)(n+1) is a multiple of 8
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 [#permalink] New post 10 Dec 2004, 14:00
rthothad wrote:
Dookie wrote:
Great, C it is.

1st statement : n is odd. (n-1)(n+1) is a multiple of 8

2nd statement : 3 is not a factor of n so (n-1)(n+1) is a multiple of 3

2 combined : multiple of 24. Remainder = 0.


Dookie, Can you explain how from statement 1 you got that (n-1)(n+1) is a multiple of 8


Since n is odd, n-1 and n+1 are both even, and, in fact, they are consecutive even number. Consecutive even numbers alternate between being multiples of 4 and not being. So in any two, one will be divisible by 2 only, and the other will be divisible by 4. So multiplying them together gives a number with 3 two's in it, or a multiple of 8.
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 [#permalink] New post 10 Dec 2004, 14:50
Great question Dookie, as usual !

In fact simply picking numbers provided the answer... my mistake...

Than you Ian for your insight.
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 [#permalink] New post 10 Dec 2004, 19:23
Thanks ian7777 for your explanation
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 [#permalink] New post 12 Dec 2004, 13:17
This is a very good one.

1) If 2 is not a factor of n then n could be 1,3,5,7... (set of odd numbers)
(n-1)(n+1) ranges from 0, 8, 24 ...
We can't get the same remainder if (n-1)(n+1) is divided by 24

2) If 3 is not a factor of n then n could be 1,2,4, 5...
(n-1)(n+1) ranges from 0, 3, 15 ...
We can't get the same remainder if (n-1)(n+1) is divided by 24

When both statements are combined
n could be 1 5 7 11 ..

All these numbers or odd. Any odd number is preceeded and followed by an even number. With n starting from 5, it turns out that (n-1)(n-2) will have the factor 3*2*2*2 = 24 (proof for this might be complicated; by trial). So remainder when divided by 24 will be 0.
For n =1, (n-1)(n+1) = 0, so remainder is still 0.

C works Ok.

This question seems a bit hard to appear on a GMAT and to be solved in 2 minutes, because we have to extrapolate the result by substitution w/o a clear proof.
You never know!
  [#permalink] 12 Dec 2004, 13:17
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