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# If n is a positive integer, then n(n+1)(n+2) is

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If n is a positive integer, then n(n+1)(n+2) is [#permalink]

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10 Jul 2006, 06:56
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If n is a positive integer, then n(n+1)(n+2) is

(A) even only when n is even
(B) even only when n is odd
(C) odd whenever n is odd
(D) divisible by 3 only when n is odd
(E) divisible by 4 whenever n is even
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Oct 2013, 23:52, edited 1 time in total.
RENAMED THE TOPIC.
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10 Jul 2006, 07:24
(E)

Assume: n is even, then either n or n+2 is a multiple of 4. Hence, n(n+1)(n+2) is divisible by 4.
Therefore: whenever n is even, the term above is divisble by 4.
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10 Jul 2006, 09:05
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(E) divisible by 4 whenever n is even

If n is even => even x odd x even (Prod of two even numbers always divisible by 2x2)
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24 Feb 2011, 03:11
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Baten80 wrote:
224. If n is a positive integer, then n(n + 1)(n + 2) is
(A) even only when n is even
(B) even only when n is odd
(C) odd whenever n is odd
(D) divisible by 3 only when n is odd
(E) divisible by 4 whenever n is even

n(n + 1)(n + 2) is the product of 3 consecutive integers. The product of 3 consecutive integers is ALWAYS divisible by 2 and 3 (generally the product of k consecutive integers is always divisible by k!, check this: defined-functions-108309.html), so n(n + 1)(n + 2) is always even and always divisible by 3: A, B, C and D are out.

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Re: n is a positive integer [#permalink]

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01 Jul 2011, 00:07
You can use number plugging to do this. Eliminate answers by choosing 1 and 2 as test numbers. I'm not solving it here, perhaps you can give this a try now.
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Re: n is a positive integer [#permalink]

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01 Jul 2011, 00:17
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Manhattan NP covers these well.

Will be of the form

Odd, even, odd = even
Even, odd, even = even

Can quickly rule out all but E

Posted from my mobile device
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Re: n is a positive integer [#permalink]

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01 Jul 2011, 00:36
pike wrote:
Manhattan NP covers these well.

Will be of the form

Odd, even, odd = even
Even, odd, even = even

Can quickly rule out all but E

Posted from my mobile device

Hi Pike,

can you please explain in detail how can you rule out other options without calculating? It will be good to know and might save some time.
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Re: n is a positive integer [#permalink]

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01 Jul 2011, 01:43
siddhans wrote:
How to solve this?

If n is a positive integer, then n(n+1)(n+2) is

A)even only when n is even
B)even only when n is odd
C)odd whenever n is odd
D)divisible by 3 only when n is odd
E)divisible by 4 whenever n is even

n(n+1)(n+2) will always be even as n is a +ve integer so that rules out A, B & C. Atleast one of n, n+1 & n+2 will be even as they are consecutive integers.

even * even is always even e.g 2*4 = 8 or 6*10 = 60 always even
even * odd is always even e.g 2*3 = 6 or 5 * 8 = 40 always even

Either of n, n+1 & n+2 will always be divisible by 3 till the time n is a +ve integer and they are consecutive integers.

Hence that leaves us with E as answer.

Also we can prove it like this way also,
First +ve even integer is 2 and not 0 (0 is neither +ve nor -ve).

so n*n+1*n+2 = 2*3*4 divisible by 4.
or if n=6 then 6*7*8 again divisible by 4.

so is E.
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Re: n is a positive integer [#permalink]

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01 Jul 2011, 04:54
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ankushjain wrote:
siddhans wrote:
How to solve this?

If n is a positive integer, then n(n+1)(n+2) is

A)even only when n is even
B)even only when n is odd
C)odd whenever n is odd
D)divisible by 3 only when n is odd
E)divisible by 4 whenever n is even

Also we can prove it like this way also,
First +ve even integer is 2 and not 0 (0 is neither +ve nor -ve).

so n*n+1*n+2 = 2*3*4 divisible by 4.
or if n=6 then 6*7*8 again divisible by 4.

so is E.

We can generalize:

If n is even, then n + 2 is also even and n and n + 2 are consecutive even numbers.

2 * 2 = 4, so any two even numbers multiplied together will yield a multiple of 4.

Therefore, any two consecutive even numbers multiplied together will yield a multiple of 4.

THEREFORE, if n is even, n(n + 2) is always a multiple of 4.

But actually, we can go a step further (this may be useful for some future problems):

Every second even number is a multiple of 4. Therefore, given any two consecutive even numbers, one of them will always be a multiple of 4.

4 * 2 is 8, so any multiple of 4 multiplied by another even number will yield a multiple of 8.

Therefore, any two consecutive even numbers multiplied together will yield a multiple of 8.

THEREFORE, if n is even, n(n + 2) is always a multiple of 8.
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Re: A simple math problem, pls help me explain,thx [#permalink]

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16 Jul 2011, 01:57
tracyyahoo wrote:
(1) If n is a positive integer, then n(n+1)(n+2) is

a) Even only when n is even
b) Even only when n is odd
c) Odd whenever n is odd
d) Divisible by 3 only when n is odd
e) Divisible by 4 only whenever n is even

Why A isn't correct since I used the plug in to calculate and I know e is correct. why a) isn't correct?

Whats the source?
E is right .
A is wrong becasue even when N is odd, it can be even ,ex: 3*4*5 ( n=3)
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Re: A simple math problem, pls help me explain,thx [#permalink]

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17 Jul 2011, 02:57
n(n+1)(n+2) will be either

ODD * EVEN * ODD = EVEN
EVEN * ODD * EVEN = EVEN

A says that the result will be even, only when n is even, which we have shown is not the case.
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Re: n is a positive integer [#permalink]

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17 Jul 2011, 03:41
siddhans wrote:
How to solve this?

If n is a positive integer, then n(n+1)(n+2) is

A)even only when n is even
B)even only when n is odd
C)odd whenever n is odd
D)divisible by 3 only when n is odd
E)divisible by 4 whenever n is even

n(n+1)(n+2) is the product of three consecutive integers because n is an integer.

0,1,2
-200,-199,-198
100,101,102
-1,0,1

In any set of three consecutive numbers, there must be at least one odd and one even.

odd,even,odd
OR
even,odd,even

A)even only when n is even
The product of three or more consecutive integers will always be EVEN. To make the product even, we just need one even. It really doesn't matter whether n is even or n+1.

If n is even, say 0
0,1,2. product=0=even

If n is odd, say -1
-1,0,1. product=0=even

Saying that n(n+1)(n+2) will be even ONLY if n=even is NOT correct.

B)even only when n is odd

We just saw that the product will always be even irrespective of whether n is even or odd.

C)odd whenever n is odd

Product will never be odd.

D)divisible by 3 only when n is odd
Rule: Product of n consecutive number will always be divisible be n!

{1,2}: Two numbers. n=2
1*2 will be divisible by 2!=2

{45,46,47,48,49,50}: Six numbers. n=6
45*46*47*48*49*50 will be divisible by 6!=720

Similarly,
3 consecutive numbers: {1,2,3}
1*2*3 will be divisible by 3!=6
If the product is divisible by 6, it must be divisible by its factor, which is 3.

Thus, "n" can be even/odd.
FALSE.

E)divisible by 4 whenever n is even
n=2
2,3,4. Product=24

TRUE.

Ans: "E"
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Re: n is a positive integer [#permalink]

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17 Jul 2011, 04:11
This can be solved easily by process of elimination, it's important to see this as the multiplication of consecutive numbers. Please note the following properties of three Consecutive numbers
They will always be divisible by 3
Irrespective of n, the answer will always be even, because any n multiples by an even number yields an even number.
Hence out of all the options only E makes sense.
And now the icing on the cake, any three consecutive numbers have atleast 2 2's in their prime factors.

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Re: n is a positive integer [#permalink]

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17 Jul 2011, 13:04
n * n+1 * n+2 is always even irrespective of whether n is odd or even.

Answer choice D would have been good if there is no "Only" in it. product of 3 consecutive integers is always divisible by 3 irrespective of whether n is odd or even.

i.e when n is even =>n+1 is odd => n+2 is even . As we have two even numbers in the product this will always be
divisible by 4.

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Re: n is a positive integer [#permalink]

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19 Jul 2011, 09:20
Firstly, we can see that n,(n+1) and (n+2) are consecutive integers.
Consecutive integers alternate in an Even-odd fashion. i.e., if n is even, (n+1) is odd, and (n+2) is even. Similarly, when n is odd,(n+1) is even and (n+2) is odd.
In any case, we notice that the product MUST be even. (even*any number = even)
Also,
There is a rule that 'n' consecutive integers are divisible by 'n!'
Here, n=3 => n(n+1)(n+2) div. by 3! = 3.2.1
Let us check the options:
A)even only when n is even --- wrong. Since, it is even when n is both even AND odd.
B)even only when n is odd ----wrong. Same reason as above.
C)odd whenever n is odd ----wrong. Even when n is odd.
D)divisible by 3 only when n is odd ----wrong. div. by 3 when n is even or odd
E)divisible by 4 whenever n is even---Correct. when n is even, (n+1) is odd and (n+2) is even. PRODUCT of two even no.s(here, n & n+2) is ALWAYS div.by 4.
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Re: A simple math problem, pls help me explain,thx [#permalink]

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20 Jul 2011, 21:36
I think even E is wrong. Even if n=3, odd, then the product is 3*4*5 and is divisible by 4.
I question the validity of this question!!!
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Re: A simple math problem, pls help me explain,thx [#permalink]

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20 Jul 2011, 22:26
Claudia777 wrote:
I think even E is wrong. Even if n=3, odd, then the product is 3*4*5 and is divisible by 4.
I question the validity of this question!!!

Divisible by 4 only whenever n is even

Hence n can't equal 3, as n must be even.
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Re: A simple math problem, pls help me explain,thx [#permalink]

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21 Jul 2011, 05:29
Many thanks. I thougth the problem said " only when n is even" , but it actually said "whenever n is even"...i got it now!
Thanks

On the same note, I found the problem elsewhere and the E. answer was : e) Divisible by 4 whenever n is even ( while here E is e) Divisible by 4 only whenever n is even) Was a bit confusing!!

In any case,I got it now!!:)
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Re: A simple math problem, pls help me explain,thx [#permalink]

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21 Jul 2011, 13:48
Claudia777 wrote:
Many thanks. I thougth the problem said " only when n is even" , but it actually said "whenever n is even"...i got it now!
Thanks

On the same note, I found the problem elsewhere and the E. answer was : e) Divisible by 4 whenever n is even ( while here E is e) Divisible by 4 only whenever n is even) Was a bit confusing!!

In any case,I got it now!!:)

Right, but to divide by four, n must be even. So it is E in both questions.
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30 Aug 2011, 04:18
This revolves around two principles.

1. if one number in a product of two or more is even then the number is always EVEN

2. A product of three consicutive POSITIVE integers is always divisible by 3.

IF N is even then the least possible product is 2*3*4 which is divisible by 4 . Holds true for any higher even value for N.

a) even only when n is even

even when N is odd the product is even because N+1 is even .

b) even only when n is odd

even when N is even the product is even because (if one number in a product of two or more is even then the number is always EVEN).

c) odd whenever n is odd
THe product of two or more consecutive positive integers is never ODD

d) divisible by 3 only when n is odd

Does not matter if N is even or ODD

Every third poitive integer is divisible by three. Does not matter if N is ODD or EVEN

Example:
1. N= 2 set S= {2,3,4} product is divisible by 3
2. N = 4 set S = {4,5, 6} product is divisible by 3.

Note 3 has a cyclicity of {0,1,2} as reminder for all Positive integers.

e) divisible by 4 whenever n is even

True: if N is even then N and N+2 are necessarily even hence divisible by 4 :

Consider least even positive integer 2

2*3*4 is divisible by 4 {true for all values of N as even because divisibility by 4 means the number must be divisible by 2 twice. In this scenario we would have N and N+2 as even}

Hence option E.

Regards,
Raghav.V

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Re: consecutive integers product   [#permalink] 30 Aug 2011, 04:18

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