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If n is an integer greater than 6, which of the following

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SVP
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If n is an integer greater than 6, which of the following [#permalink] New post 25 Jul 2008, 08:48
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A
B
C
D
E

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If n is an integer greater than 6, which of the following must be divisible by 3?

a) n(n+2)(n+3)

b) n(n+5)(n-3)

c) n(n+2)(n+5)

d) n(n+1)(n-4)

e) n(n+1)(n-2)




I chose A as my answer, but it's wrong. How come??? whenever we have 3 consecutive numbers, obviously one of them will always be a multiple of 3. I really don't understand why it is wrong this time! :?
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Re: PS: In [#permalink] New post 25 Jul 2008, 08:56
tarek99 wrote:
If n is an integer greater than 6, which of the following must be divisible by 3?

a) n(n+2)(n+3)

b) n(n+5)(n-3)

c) n(n+2)(n+5)

d) n(n+1)(n-4)

e) n(n+1)(n-2)


you want to ensure that the ans choice contains at least a 3 and an even number

if n>6, then look at 7-4=3, 8-4=4 but 8+1=9.. 11-4=7 but 11+1=12 which is a multiple of 3..

so based on this D is the only choice that will always ensure that there is a 3 and an even number..
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Re: PS: In [#permalink] New post 25 Jul 2008, 08:57
choice A is NOT 3 consec integers

tarek99 wrote:
If n is an integer greater than 6, which of the following must be divisible by 3?

a) n(n+2)(n+3)

b) n(n+5)(n-3)

c) n(n+2)(n+5)

d) n(n+1)(n-4)

e) n(n+1)(n-2)




I chose A as my answer, but it's wrong. How come??? whenever we have 3 consecutive numbers, obviously one of them will always be a multiple of 3. I really don't understand why it is wrong this time! :?
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Re: PS: In [#permalink] New post 25 Jul 2008, 09:06
D: n(n+1)(n-4) Image(n-1)(n)(n+1) (mod 3)
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Re: PS: In [#permalink] New post 25 Jul 2008, 09:20
I say D...I don't even see why we would need the info that n>6 to answer this question....
do we?
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Re: PS: In [#permalink] New post 25 Jul 2008, 09:23
tarek99 wrote:
If n is an integer greater than 6, which of the following must be divisible by 3?

a) n(n+2)(n+3)

b) n(n+5)(n-3)

c) n(n+2)(n+5)

d) n(n+1)(n-4)

e) n(n+1)(n-2)




I chose A as my answer, but it's wrong. How come??? whenever we have 3 consecutive numbers, obviously one of them will always be a multiple of 3. I really don't understand why it is wrong this time! :?


I chose D too..
here is the strategy


...(n-3)(n-2) (n-1) ..... n (n+1) (n+2) .... (n+3)(n+4)(n+5)..
--> multiple of any three consecutive positive numbers defnitiely divsiable by 3.

n (n+1) (n+2) --- > must divisible by 3
(n-6)(n-5)(n-4)...(n-3)(n-2) (n-1) ..... n (n+1) (n+2) .... (n+3)(n+4)(n+5)..

for any one number if substract or add 3(or multiples of 3) ..then it must divisible by 3.
n (n+1) (n+2 -3)= n (n+1) (n-1)
n (n+1) (n+2 -3*2)= n (n+1) (n-4)
n (n+1) (n+2 +3)= n (n+1) (n+5)
n (n+1) (n+2 +3*2)= n (n+1) (n+8)

---------
By above strategy (substraction or addition .. to any numbers) I try to see if any answer choice will lead to three consecutive numbers..

a) n(n+2)(n+3) -- out not consecutive numbers...
b) n(n+5)(n-3) --> n (n+2) n --out not consecutive numbers...
subtract 3 from (n+5) add 3 to (n-3)
c) n(n+2)(n+5) -->n(n+2)(n+2) --out not consecutive numbers
subtract 3 from (n+5)
d) n(n+1)(n-4) -->n(n+1)(n-1) -- consecutive numbers
add 3 to (n-4)
e) n(n+1)(n-2) --> n(n+1)(n+1) out not consecutive numbers
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Re: PS: In   [#permalink] 25 Jul 2008, 09:23
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