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

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Joined: 02 Aug 2009
Posts: 8752
Re: If n is an integer greater than 6, which of the following mu  [#permalink]

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25 Nov 2017, 05:11
1
dave13 wrote:
Bunuel wrote:
carcass 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: http://gmatclub.com/forum/if-x-is-an-in ... 26853.html

Hope it helps.

Bunuel good day! can you please explain this part "it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers" I dont understand the logic behind this (n-4)+3=n-1 have a great day thanks!

Hi dave,

it basically depends on the divisor..

here you are dividing a term by 3 to find the remainder that is why whatever remainder n-4 leaves, n-4+3 or n-4+6 will leave..

say n is 28, so n-4 = 28-4 = 24, remainder will be 0, when div by 3..
now if you add/subtract 3 or a multiple of 3 to 24, remainder will always be 0..
say n was 30, n-4 would be 26 and would leave a remainder of 2 when div by 3..
so 26-3 = 23 or 26-6=20 will also leave remainder of 2 when div by 3
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If n is an integer greater than 6, which of the following mu  [#permalink]

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15 Jan 2018, 16:59
VeritasPrepKarishma Bunuel niks18 chetan2u

Quote:
I know that talking about positive integers, in any set of 3 consecutive positive integers, one integer will be divisible by 3 and the other 2 will not be.

So I am looking for 3 consecutive positive integers e.g. (n-1)n(n+1)

Note that (n-4) is equivalent to (n-1) since if (n-4) is divisible by 3, so is (n-1). If (n-4) is not divisible by 3, neither is (n-1) (because the difference between these two integers is 3)
Hence (A) is equivalent to 3 consecutive integers.
Answer (A)

Excellent approach.

Just to make sure : there is no (n-1) terms in OA and since you knew sum / difference of three consecutive positive integers is
always divisible by 3, you looked for hint to equating characteristics of n-1 and n-4 and arriving that both are divisible by 3
as represented on number line.

Is my interpretation correct?
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Re: If n is an integer greater than 6, which of the following mu  [#permalink]

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15 Jan 2018, 20:44
1
adkikani wrote:
VeritasPrepKarishma Bunuel niks18 chetan2u

Quote:
I know that talking about positive integers, in any set of 3 consecutive positive integers, one integer will be divisible by 3 and the other 2 will not be.

So I am looking for 3 consecutive positive integers e.g. (n-1)n(n+1)

Note that (n-4) is equivalent to (n-1) since if (n-4) is divisible by 3, so is (n-1). If (n-4) is not divisible by 3, neither is (n-1) (because the difference between these two integers is 3)
Hence (A) is equivalent to 3 consecutive integers.
Answer (A)

Excellent approach.

Just to make sure : there is no (n-1) terms in OA and since you knew sum / difference of three consecutive positive integers is
always divisible by 3, you looked for hint to equating characteristics of n-1 and n-4 and arriving that both are divisible by 3
as represented on number line.

Is my interpretation correct?

Yes, you are right.
Take a look here: https://www.veritasprep.com/blog/2011/0 ... h-part-ii/

If (n - 1) is a multiple of 3, so is (n - 4).
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If n is an integer greater than 6, which of the following mu  [#permalink]

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17 Apr 2018, 10:05
Bunuel wrote:
carcass 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: http://gmatclub.com/forum/if-x-is-an-in ... 26853.html

Hope it helps.

Hi Bunuel, pushpitkc

its me again

you know i am trying to undeerstand the concept and have some questions:) could you please explain

so we are given 5 answer choices:

(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)

you say "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" in order for this to be true we need to TO HAVE 3 CONSECUTVE NINTEGERS , am i correct ? please say something

Next you say " 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.

Question 1) why you say " to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3 ?

why you mention remainders, and not that that they must be consecutive numbers ?

Question 2) what is the logic behind this "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"

i mean remainders could be any numbers, why mention 1 , and than 2 ? is it MUST BE TRUE ?

Question 3) you write "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"

How do you figure out that here (A) n(n + 1)(n – 4) n and n+1 have different remainder upon division by 3 , i simply cant understand this, cause i dont see logic

Question 4) why do you think that for n-4, it will have the same remainder as (n-4)+3=n-1 (Also what to you mean by "the same remainder" 0 ? )

Question 5) how about other answer choices ? how you ruled them out ? i am just trying get under the skin of this concept

(B) n(n + 2)(n – 1)
(C) n(n + 3)(n – 5)
(D) n(n + 4)(n – 2)
(E) n(n + 5)(n – 6)

generis generis any comments on my questions above are apprecuated all moderators are attending now indian premier league, but you so i decided to tag youhere because you are not a football lover
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Re: If n is an integer greater than 6, which of the following mu  [#permalink]

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13 Apr 2020, 14:26
carcass wrote:
I attacked the problem in this was, tell me if pron of errors

n is an integers, so we can choose 7, 8, 9 and so on. Now, a number divisible by 3 the sum of number MUST be divisible by 3.

1) n (n+1)(n-4) ---> 8 * 9 * 4 ---> without perform multiplication the SUM of 8 + 9 + 4 = 21 and is divisible by 3 without reminder

The rest of choices do not work if you try.

thanks

carcass
If you put the value of $$n=8$$, the choice $$D$$ is still works!
Thanks__
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Re: If n is an integer greater than 6, which of the following mu   [#permalink] 13 Apr 2020, 14:26

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

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