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

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If n is an integer greater than 6, which of the following must be divi  [#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. $$n (n+5) (n-6)$$

Originally posted by gregspirited on 26 Nov 2007, 13:15.
Last edited by Bunuel on 30 Jan 2019, 04:12, edited 2 times in total.
Renamed the topic and edited the question.
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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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24 Mar 2012, 00:40
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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 must be divi  [#permalink]

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26 Nov 2007, 14:29
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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 must be divi  [#permalink]

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26 Nov 2007, 14:08
1
2
A.

All three numbers in product should have different reminders. Only A satisfies this requirement.
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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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26 Nov 2007, 14:22
9
9
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: If n is an integer greater than 6, which of the following must be divi  [#permalink]

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13 Sep 2010, 10:57
2
6
seekmba 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)

Is there an easy way to solve this?

Substituting 3 consecutive numbers might be the easiest way to solve this question as shown above.

Algebraic solution:

The product of 3 numbers to be divisible by 3 at least one of them must be divisible by 3. So, to ensure that the product of 3 integers shown is divisible by 3 all 3 numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3. We should have something like $$n(n+1)(n+2)$$ (for example: if n divided by 3 yields remainder of 1, then n+1 yields remainder of 2 and n+2 yields remainder of 0, thus it's divisible by 3 OR if n divided by 3 yields remainder of 2, then n+2 yields remainder of 1 and n+1 yields remainder of 0, thus it's divisible by 3).

Only option A satisfies this, because $$n(n+1)(n-4)=n(n+1)(n-6+2)$$ and $$n-6$$ has the same remainder as $$n$$ upon division by 3 thus we can replace it by $$n$$.

Hope it's clear.
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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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13 Sep 2010, 12:16
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1
Same concept different way of thinking

How do I ensure that a product of three numbers is divisible by 3 ?
Step 1 : I divide the set of integers into three parts such that one of the parts contains exactly and only the multiples of 3
Step 2 : I ensure that I chose my numbers such that each belongs to a different part
Step 3 : If I achieve this, I must have exactly one number which belongs to the set of multiples of three. Hence their product is divisible by three.

How do I achieve this ?
Step 1 : The division I choose is numbers of the form $$3\alpha, 3\alpha+1, 3\alpha+2$$. The union of these sets is all integers and they have no overlap between them, also one of them contains all and only multiples of 3.
Step 2 : Simplest way to choose three numbers such that exactly one belongs to each of the above sets is 3 consecutive numbers. This choice will always satisfy what I am looking for. Hence n*(n-1)*(n+1)

Important Note : If I shift any number by 3, its constituent set amongst $$\alpha, 3\alpha+1, 3\alpha+2$$ does not change

So how do we solve this question ?
Start from the expression n*(n-1)*(n+1) and see if by adding or subtracting 3 or multiples of 3 from one or more of the three terms we can get any target expression.

A) Satisfies
B) n-1 repeats twice
C) n repeats twice
D) n+1 repeats twice
E) n repeats twice

Hence, solution is 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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13 Sep 2010, 16:39
I have seen many questions that derive upon n n-1 n+1 being divisible by 3.. Does this and the logic below.. can both be extended to any sequence? Meaning 4..5..6.. consecutive numbers? It would seem that is the case?

Can you explain what do you mean by n-1 repeats twice in B?
A) Satisfies
B) n-1 repeats twice
C) n repeats twice
D) n+1 repeats twice
E) n repeats twice

shrouded1 wrote:
Same concept different way of thinking

How do I ensure that a product of three numbers is divisible by 3 ?
Step 1 : I divide the set of integers into three parts such that one of the parts contains exactly and only the multiples of 3
Step 2 : I ensure that I chose my numbers such that each belongs to a different part
Step 3 : If I achieve this, I must have exactly one number which belongs to the set of multiples of three. Hence their product is divisible by three.

How do I achieve this ?
Step 1 : The division I choose is numbers of the form $$3\alpha, 3\alpha+1, 3\alpha+2$$. The union of these sets is all integers and they have no overlap between them, also one of them contains all and only multiples of 3.
Step 2 : Simplest way to choose three numbers such that exactly one belongs to each of the above sets is 3 consecutive numbers. This choice will always satisfy what I am looking for. Hence n*(n-1)*(n+1)

Important Note : If I shift any number by 3, its constituent set amongst $$\alpha, 3\alpha+1, 3\alpha+2$$ does not change

So how do we solve this question ?
Start from the expression n*(n-1)*(n+1) and see if by adding or subtracting 3 or multiples of 3 from one or more of the three terms we can get any target expression.

A) Satisfies
B) n-1 repeats twice
C) n repeats twice
D) n+1 repeats twice
E) n repeats twice

Hence, solution is 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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14 Sep 2010, 01:00
mainhoon wrote:
I have seen many questions that derive upon n n-1 n+1 being divisible by 3.. Does this and the logic below.. can both be extended to any sequence? Meaning 4..5..6.. consecutive numbers? It would seem that is the case?

Can you explain what do you mean by n-1 repeats twice in B?
A) Satisfies
B) n-1 repeats twice
C) n repeats twice
D) n+1 repeats twice
E) n repeats twice

The important point to note is that adding or subtracting a multiple of 3 does not change the constituent set
And that we need to ensure there is one element from each set, $$3\alpha, 3\alpha+1,3\alpha+2$$

So B for instance is :

$$n*(n+2)*(n-1) \equiv n*(n+2-3)*(n-1) \equiv n*(n-1)*(n-1)$$
(I dont mean to say these expressions are equal, just equal vis-a-vis what their constituent sets are)

So the number of type (n-1) repeats but there is none of type (n+1) hence there is no guarantee this product is divisible by 3. And the counter example would be a case where n+1 is divisible by 3. Eg. n=11, expression = 11*13*10, not divisible by 3.
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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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14 Sep 2010, 01:07
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The best approach is to back-solve, but to learn the concept read the reply of Bunnel. That concept is very important.

If you have a number n , highest number of remainders are 0 to n-1

Since n =3, the remainders are 0,1,2. Since the remainders are consecutive numbers, we should consider 3 consecutive numbers to back solve.
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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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06 Oct 2010, 13:11
Ok, so, MGMAT says that the product of k consec. int's is ALWAYS divisible by k!

Well, this ain't right. Q.82 of the OG 12 -->

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 chose B because B represents 3 consecutive integers, my reasoning was that the product of B MUST be divisble by k!, which is 3! = 6... therefore by the Factor Foundation Rule, it MUST also be divisible by 3 because 3 is a factor of 6, right? WRONG.

Can anyone think why this is NOT the case? The official answer is 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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06 Oct 2010, 13:15
Merging similar topics.

As for your doubt: n(n+2)(n-1) is NOT a product of 3 consecutive integers. If it were: n(n+1)(n-1) or n(n+2)(n+1) then YES, but in its current form it's not.
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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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06 Oct 2010, 13:21
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2
n2739178 wrote:
Ok, so, MGMAT says that the product of k consec. int's is ALWAYS divisible by k!

Well, this ain't right. Q.82 of the OG 12 -->

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 chose B because B represents 3 consecutive integers, my reasoning was that the product of B MUST be divisble by k!, which is 3! = 6... therefore by the Factor Foundation Rule, it MUST also be divisible by 3 because 3 is a factor of 6, right? WRONG.

Can anyone think why this is NOT the case? The official answer is A...

As for the rule you quote it's correct:

• If $$k$$ is odd, the sum of $$k$$ consecutive integers is always divisible by $$k$$. Given $$\{9,10,11\}$$, we have $$k=3$$ consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If $$k$$ is even, the sum of $$k$$ consecutive integers is never divisible by $$k$$. Given $$\{9,10,11,12\}$$, we have $$k=4$$ consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• The product of $$k$$ consecutive integers is always divisible by $$k!$$, so by $$k$$ too. Given $$k=4$$ consecutive integers: $$\{3,4,5,6\}$$. The product of 3*4*5*6 is 360, which is divisible by 4!=24.
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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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06 Oct 2010, 13:22
1
n2739178 wrote:
Ok, so, MGMAT says that the product of k consec. int's is ALWAYS divisible by k!

Well, this ain't right. Q.82 of the OG 12 -->

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 chose B because B represents 3 consecutive integers, my reasoning was that the product of B MUST be divisble by k!, which is 3! = 6... therefore by the Factor Foundation Rule, it MUST also be divisible by 3 because 3 is a factor of 6, right? WRONG.

Can anyone think why this is NOT the case? The official answer is A...

Well, B is not 3 consecutive integers. That's why your answer is wrong.

The trick is that if you divide integers into 3 sets 3K, 3K+1, 3K+2 and figure out a strategy such that 3 numbers you pick are one each from these sets, then the product has to be divisble be three. The easiest way to do this is selecting 3 consecutive numbers

So start with n(n-1)(n+1) which are consecutive. Now adding or subtracting three from any of these numbers doesnt change its source set. so thats how you figure out the answer.

a) n(n+1)(n-4) ---> n(n+1)(n-4+3) --> This is it !
b) n(n+2)(n-1) --> n(n+2-3)(n-1) --> n-1 repeats
c) n(n+3)(n-5) --> n(n+3-3)(n-5+6) --> n repeats
d) n(n+4)(n-2) --> n(n+4-3)(n-2+3) --> n+1 repeats
e) n(n+5)(n-6) --> n(n+5-6)(n-6+6) --> n repeats
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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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15 Apr 2012, 20:55
1
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: If n is an integer greater than 6, which of the following must be divi  [#permalink]

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04 May 2013, 02:20
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 must be divi  [#permalink]

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04 May 2013, 03:36
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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: If n is an integer greater than 6, which of the following must be divi  [#permalink]

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19 Oct 2013, 02:39
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 must be divi  [#permalink]

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19 Oct 2013, 03:11
3
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 must be divi  [#permalink]

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25 May 2014, 08:27
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?
Re: If n is an integer greater than 6, which of the following must be divi   [#permalink] 25 May 2014, 08:27

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