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

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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 21 Mar 2015, 14:34
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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, which of the following must be divi  [#permalink]

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New post 21 Mar 2015, 16:36
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, which of the following must be divi  [#permalink]

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New post 21 Mar 2015, 18:21
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 must be divi  [#permalink]

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New post 22 Mar 2015, 08:23
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 must be divi  [#permalink]

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New post 27 Jun 2015, 11:58
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 must be divi  [#permalink]

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New post 13 Oct 2016, 03:48
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)


for any to be divisible by 3 it must be the product of 3 consecutive numbers or their equivalent.

E re written in terms of equivalent is n*n*(n-1) no consecutive
d re written in terms of equivalent is n ( n+1) (n+1) non
c ...................................................... is n*n*(n+1)
b ..........................................................n * (n-1) (n-1)
a........................................................... n (n+1) (n-1) ..... answer
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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 25 Jan 2017, 07:46
1) Every third integer is divisible by 3, consequently the option that contains three consecutive integers or integers that are +/-3n from the three consecutive integers, then their product will be divisible by 3.
2) The desireable combination is n(n+1)(n+2)
3) A) n(n+1)(n-4) - if we add 2*3 to the last term, we get the desired outcome - n(n+1)(n+2)

A must be divisible by 3, and it must be the correct answer.

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

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New post 27 Feb 2017, 03:06
Solution:
The best way to do this problem is plugging the values and checking which option will satisfy the condition given:

As given ‘n’ is an integer greater than 6;
Case1: Lets plug in n = 7.
We get:
Option A: n (n+1) (n-4) = 7×8 ×3 ;Satisfies
Option B: n (n+2) (n-1) = 7×9 ×6;Satisfies
Option C: n (n+3) (n-5) = 7×10 ×2;Does not satisfies
Option D: n (n+4) (n-2) = 7×11 ×5;Does not satisfies
Option E: n (n+5) (n-6) = 7×12 ×1; Satisfies
From this we can eliminate the options C and D.
Case 2: Let's plug in n = 8.
We get:
Option A: n (n+1) (n-4) = 8×9 ×4 ;Satisfies
Option B: n (n+2) (n-1) = 8×10 ×7;Does not Satisfies
Option E: n (n+5) (n-6) = 8×13 ×2; Does not Satisfies
From this we can eliminate the options B and E.

So, the correct answer option is “A”.
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