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

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If n is an integer greater than 6, which of the following [#permalink]

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26 Nov 2007, 14:15
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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)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 24 Mar 2012, 01:41, edited 1 time in total.

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26 Nov 2007, 15:08
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A.

All three numbers in product should have different reminders. Only A satisfies this requirement.

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26 Nov 2007, 15:22
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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]

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26 Nov 2007, 15:29
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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]

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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]

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24 Mar 2012, 01:40
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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.

Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html

Hope it helps.
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Re: If n is an integer greater than 6, which of the following [#permalink]

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25 Mar 2012, 10:30
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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.

Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html

Hope 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]

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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.

Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html

Hope 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]

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15 Apr 2012, 21:55
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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]

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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-1

Could 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]

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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]

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04 May 2013, 04:36
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Aristocrat wrote:
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

An integer divided by 3 can have 3 possible remainders: 0, 1, or 2.

Now, consider the product of three numbers a*b*c. If we are told that a, b, and c have different reminders upon division by three, this would mean that one of the numbers yields the remainder of zero, thus it's a multiple of 3. Thus abc is a multiple of 3.

Hope it helps.
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Re: PS - Integer N greater than 6 [#permalink]

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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: PS - Integer N greater than 6 [#permalink]

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08 Jul 2013, 07:28
GMAT TIGER 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)

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.

I think your solution for option A is incorrect. When you write an expression like this n (n+1) ((n-1)-3) , you are multiplying -3 with the whole expression, which eventually will turn out to (-3n+3), that is not equal to what is given. Correct me if I am wrong

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Re: If n is an integer greater than 6, which of the following [#permalink]

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19 Oct 2013, 03:39
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.

Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html

Hope it helps.

Hi Bunuel,

I took the picking numbers method but I want to understand yours as well.
I understand that n (n+1), n (n+2), n (n+3)...etc will all have different remainders, and I understand the concept about the different remainders.
I don't understand how you reached this statement: "As for n-4, it will have the same remainder as (n-4)+3=n-1"
Can you please elaborate on that?

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Re: If n is an integer greater than 6, which of the following [#permalink]

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19 Oct 2013, 04:11
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Skag55 wrote:
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.

Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html

Hope it helps.

Hi Bunuel,

I took the picking numbers method but I want to understand yours as well.
I understand that n (n+1), n (n+2), n (n+3)...etc will all have different remainders, and I understand the concept about the different remainders.
I don't understand how you reached this statement: "As for n-4, it will have the same remainder as (n-4)+3=n-1"
Can you please elaborate on that?

n-1=(n-4)+3. Now, since 3 IS divisible by 3, then the remainder when (n-4)+3 is divided by by 3 is the same as n-4 is divided by 3.

For example, say n=11:
The remainder when n-1=11-1=10 divided by 3 is 1.
The remainder when n-4=11-4=7 divided by 3 is also 1.

Does this make sense?
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Re: If n is an integer greater than 6, which of the following [#permalink]

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19 Oct 2013, 04:18
Bunuel wrote:
n-1=(n-4)+3. Now, since 3 IS divisible by 3, then the remainder when (n-4)+3 is divided by by 3 is the same as n-4 is divided by 3.

For example, say n=11:
The remainder when n-1=11-1=10 divided by 3 is 1.
The remainder when n-4=11-4=7 divided by 3 is also 1.

Does this make sense?

Yes, makes sense now, thanks. So effectively n-4 has the same remainder as n-1, hence we the same result as we get with n, n+1 and n-1 as factors.

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Re: If n is an integer greater than 6, which of the following [#permalink]

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08 Mar 2014, 00:03
I think I solved this by luck, as per below, by choosing the only answer where the sum of the digits was divisible by 3. However, testing my method by using another possible answer where the sum of the two digits is divisible by 3, shows that it is not always correct. Best to use the test method with values of n, that are not divisible by 3, in order to eliminate incorrect answer choices.

A) 1+(-4) = -3 - only answer where the sum of the two is divisible by 3.
B) 2+(-1) = 1
C) 3+(-5) = -2
D) 4+(-2) = 2
E) 5+(-6)= 1

Testing another possible answer where the sum of the two ;

n(n+2)(n-5)
2+(-5)=-3
Testing with values for n that are not divisible by 3;

n=7; 7(7+2)(7-5); 7(9)(2) - has a factor divisible by 3
n=8; 8(8+2)(8-5); 8(10)(3) - has a factor divisible by 3
n=10; 10(12)(5) - has factor divisible by 3
n=11; 11(13)(6) - has factor divisible by 3

n(n-6)(n+3) -6+3=-3
n=7 ; 7(1)(10) not divisible by 3

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Re: If n is an integer greater than 6, which of the following [#permalink]

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25 May 2014, 09:27
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.

Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html

Hope it helps.

Hi Bunuel,

Why is the highlighted part important? Is that just another way of saying that at least one of the numbers in this sequence is divisible by 3. Correct?

Additionally, I see the correlation you made between n-4 and n-1(both leave a remainder of 1) but by that token, shouldn't (n+1) or (n) leave a remainder of 2 and 1 respectively? Meaning, (11+1) = 12/3 = no remainder and n (11) leaves a remainder of 2. So now we have remainders of 1,2,3 and therefore a consecutive set of integers. Is that the reason we want three different remainders?

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Re: If n is an integer greater than 6, which of the following [#permalink]

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19 Jan 2015, 01:24
Simply look for product of 3 consecutive integers....or their equivalents....i.e if a number falls under consecutive category +/- product of 3...then it work too

A )n * (n+1)* (n-4)

equivalent of (n-4)= n-1

and n-1,n,n+1 are 3 consecutive....and this prod is divisible by 3 for any integer value of n

Ans A

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Re: If n is an integer greater than 6, which of the following   [#permalink] 19 Jan 2015, 01:24

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