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

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

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New post 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.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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New post 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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New post 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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New post 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

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

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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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New post 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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New post 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.

Answer: A.

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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New post 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.

Answer: A.

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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New post 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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New post 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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New post 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.

Answer: A.

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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New post 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 must be divi [#permalink]

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New post 07 Mar 2015, 19:33
Hi,

Another approach is plug n=7 and n=8 in answer choices. Only A is divisible by 3 for both values of n.

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If n is an integer greater than 6... [#permalink]

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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)

I came across this question while studying the OG13, and answered it correctly, but I'm a little unsure if my process is sound. I initially scanned all the answer choices. I noticed A) n(n+1)(n-4) jumped out to me because 1+(-4)=-3. My thought was this might force the whole expression to always have a hidden value of 3 in it. I created a table and let n=7, 8 ,9, 10, and sure enough, it did:

n (n+1) (n-4)
7 8 3
8 9 4
9 10 5
10 11 6

Notice there is a repeating pattern of a multiple of 3 on the diagonal. Can anyone weigh in if this is a good method to apply to this sort of problem?

Thanks! :)
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Re: If n is an integer greater than 6... [#permalink]

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New post 21 Mar 2015, 17:36
Hi dcbark01,

Yes, your approach works quite nicely on this question. It's essentially a Number Property rule - about how numbers relate to other numbers. Notice though that you were able to PROVE that you were correct by TESTing your idea. You'll come to find that THAT part of the process is quite helpful in getting many Quant questions correct.

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Re: If n is an integer greater than 6... [#permalink]

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New post 21 Mar 2015, 19:21
dcbark01 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 came across this question while studying the OG13, and answered it correctly, but I'm a little unsure if my process is sound. I initially scanned all the answer choices. I noticed A) n(n+1)(n-4) jumped out to me because 1+(-4)=-3. My thought was this might force the whole expression to always have a hidden value of 3 in it. I created a table and let n=7, 8 ,9, 10, and sure enough, it did:

n (n+1) (n-4)
7 8 3
8 9 4
9 10 5
10 11 6

Notice there is a repeating pattern of a multiple of 3 on the diagonal. Can anyone weigh in if this is a good method to apply to this sort of problem?

Thanks! :)


hi dcbark01,
you are absolutely correct in your approach and the point mentioned about hidden value is correct...
however it may not have a hidden value in one scenario n(n-1)(n+1)..
the logic behind the answer is that all three numbers have to be in a pattern as consecutive numbers are to be certain that their product is div by 3..
lets take the answer choice which is correct... A) n(n+1)(n-4)
here n(n+1) are consecutive .. the consecutive number to them is either (n-1) or (n+2)... but we have (n-4), which is same as (n-1-3)..
(n-1-3) will have the same property of div as (n-1)... so in a way we have three numbers having same properties as three consecutive number when divided by three..
you will not find the same in all other choices...
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Re: If n is an integer greater than 6, which of the following [#permalink]

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New post 22 Mar 2015, 09:23
very good discussion and as usual Math God Bunuels' method stands tallest among all .

Bunuel, am i right to conclude that for any integer N if there exists N integers with different remainders when divided by N then their product will always be divisible by N .

eg: 5
1,2,3,4,5 product will be divisible by 5 .
-2,-3,1,4,5 product will be divisible by 5
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Re: If n is an integer greater than 6, which of the following [#permalink]

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New post 23 Mar 2015, 00:51
A.

Just plug in some test numbers and check the divisibility, that might be a faster way.

Cheers !
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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)

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

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New post 27 Jun 2015, 12:58
russ9 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.

Answer: A.

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?


In case you are still active here, here's a take on your questions.

The unknown is what \(n\) is, so without knowing \(n\), how could what's divisible by 3 be concluded? Find an answer choice that expresses the three different remainders. If each integer has a different remainder of three and since there are three possible remainders for 3, \(R1, R2, R0\), then the answer must be divisible by three.

The second question assumes \(n\) is already multiple of three, and the answer choices took advantage of that assumption, because \(n\) was not a multiple of 3. Had the question stated, "if \(n\) is not a multiple of three, which of the below is divisible by three", I'm sure you would have approached the problem differently. Without it, another layer of complexity was added.


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

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

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