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If n is an integer greater than 6, which of the following [#permalink]
 26 Nov 2007, 14:15
Question Stats:
69% (01:50) correct
30% (01:01) wrong based on 47 sessions
If n is an integer greater than 6, which of the following must be divisible by 3? A. n (n+1) (n-4) B. n (n+2) (n-1) C. n (n+3) (n-5) D. n (n+4) (n-2) E. n (n+5) (n-6)
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Last edited by Bunuel on 24 Mar 2012, 01:41, edited 1 time in total.
Added the OA
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A.
All three numbers in product should have different reminders. Only A satisfies this requirement.
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question testing multiples of 3.
plug in 7 and 8 as test values.
A. n (n+1) (n-4) = 7*8*3 and if n = 8 --> 8*9*4
B. n (n+2) (n-1) = 7*9*6 and if n = 8 --> 8*10*7
C. n (n+3) (n-5) = 7*10*5; eliminate as there are no multiples of 3
D. n (n+4) (n-2) = 7*11*5; eliminate as there are no multiples of 3
E. n (n+5) (n-6) = 7*12*1 and and if n = 8 --> 8*13*2
only A works.
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Re: PS - Integer N greater than 6 [#permalink]
26 Nov 2007, 15:29
gregspirited wrote: If n is an integer greater than 6, which of the following must be divisible by 3?
A. n (n+1) (n-4)
B. n (n+2) (n-1)
C. n (n+3) (n-5)
D. n (n+4) (n-2)
E. n (n+5) (n-6)
anything in the form of (n-1) (n) (n+1) is divvisible by 3. in other word, a product of any 3 consecutie intevers is divisible by 3.
A. n (n+1) (n-4) = n (n+1) ((n-1)-3) is equivalant to (n-1) (n) (n+1)
B. n (n+2) (n-1) is equivalant to (n+1) missing.
C. n (n+3) (n-5) is equivalant to (n-1) missing and n repeating.
D. n (n+4) (n-2) is equivalant to odd/even consqcutive integers
E. n (n+5) (n-6) is equivalant to (n+1) missing and n repeating.
So A is good.
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Re: If n is an integer greater than 6, which of the following [#permalink]
23 Mar 2012, 21:04
IMO A. I used an arbitrary number greater than 6 and then filled each equation out. if you happen to have chosen a number that makes more than 1 answer correct, choose a different number and check the ones that were previously correct.
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Re: If n is an integer greater than 6, which of the following [#permalink]
24 Mar 2012, 01:40
gregspirited wrote: If n is an integer greater than 6, which of the following must be divisible by 3?
A. n (n+1) (n-4)
B. n (n+2) (n-1)
C. n (n+3) (n-5)
D. n (n+4) (n-2)
E. n (n+5) (n-6) Since 3 is a prime number then in order the product to be divisible by 3 either of the multiples must be divisible by 3. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3. For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers. Answer: A. Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.htmlHope it helps.
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Re: If n is an integer greater than 6, which of the following [#permalink]
25 Mar 2012, 10:30
Bunuel wrote: gregspirited wrote: If n is an integer greater than 6, which of the following must be divisible by 3?
A. n (n+1) (n-4)
B. n (n+2) (n-1)
C. n (n+3) (n-5)
D. n (n+4) (n-2)
E. n (n+5) (n-6) Since 3 is a prime number then in order the product to be divisible by 3 either of the multiples must be divisible by 3. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3. For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers. Answer: A. Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.htmlHope it helps. I liked the approach - Bookmaking it. Thank you Bunuel...
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Re: If n is an integer greater than 6, which of the following [#permalink]
15 Apr 2012, 17:12
Bunuel wrote: gregspirited wrote: If n is an integer greater than 6, which of the following must be divisible by 3?
A. n (n+1) (n-4)
B. n (n+2) (n-1)
C. n (n+3) (n-5)
D. n (n+4) (n-2)
E. n (n+5) (n-6) Since 3 is a prime number then in order the product to be divisible by 3 either of the multiples must be divisible by 3. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3. For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers. Answer: A. Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.htmlHope it helps. Bunuel, could you please explain how you arrive at the conclusion that "As for n-4, it will have the same remainder as (n-4)+3=n-1"? Also, is it implied that n will have a remainder of either 0 or 1, n+1 will have either 1 or 2? Thanks!
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Re: If n is an integer greater than 6, which of the following [#permalink]
15 Apr 2012, 21:55
gregspirited wrote: If n is an integer greater than 6, which of the following must be divisible by 3?
A. n (n+1) (n-4)
B. n (n+2) (n-1)
C. n (n+3) (n-5)
D. n (n+4) (n-2)
E. n (n+5) (n-6) I went by the substitution method since n > 6 i took 7 and A,B and E were satisfied i took 8 and only A was satsified, hence the answer is A
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Re: PS - Integer N greater than 6 [#permalink]
24 Nov 2012, 21:56
GMAT TIGER wrote:
A. n (n+1) (n-4) = n (n+1) ((n-1)-3) is equivalant to (n-1) (n) (n+1)
So A is good.
Hi Bunuel, Why is n (n+1) (n-4) = n (n+1) ((n-1)-3) is equivalant to (n-1) (n) (n+1)? Also, you said.. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3.
For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1Could you please elaborate the 2 statements above?
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Re: If n is an integer greater than 6, which of the following [#permalink]
04 May 2013, 03:20
Hi Bunuel, Can you explain why the remainder should be different upon division by 3? I atill dint understood uproach Posted from my mobile device 
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Re: If n is an integer greater than 6, which of the following [#permalink]
04 May 2013, 04:36
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Re: PS - Integer N greater than 6 [#permalink]
13 May 2013, 17:24
Sachin9 wrote: GMAT TIGER wrote:
A. n (n+1) (n-4) = n (n+1) ((n-1)-3) is equivalant to (n-1) (n) (n+1)
So A is good.
Hi Bunuel, Why is n (n+1) (n-4) = n (n+1) ((n-1)-3) is equivalant to (n-1) (n) (n+1)? For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1[/i] i also dont understand this part, could you elaborate why ((n-1)-3) is equivalant to (n-1) (n) (n+1)? is it because the extra 3 remaining means it has zero remainder?
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Re: If n is an integer greater than 6, which of the following [#permalink]
14 May 2013, 13:27
I actually hit the answer by mistake before i got it. but i found this one easy
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Re: If n is an integer greater than 6, which of the following
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14 May 2013, 13:27
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