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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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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

Originally posted by avdxz on 10 Jul 2006, 07:56.
Last edited by Bunuel on 31 Oct 2013, 00:52, edited 1 time in total.
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 [#permalink]

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New post 10 Jul 2006, 08:24
(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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 [#permalink]

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New post 10 Jul 2006, 10:05
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1
(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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Re: 224. Arithmetic operation [#permalink]

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New post 24 Feb 2011, 04:11
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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.

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

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New post 01 Jul 2011, 01:17
2
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

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Re: n is a positive integer [#permalink]

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New post 01 Jul 2011, 02:43
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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New post 01 Jul 2011, 05:54
2
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: n is a positive integer [#permalink]

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New post 17 Jul 2011, 04:41
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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New post 17 Jul 2011, 05:11
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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New post 17 Jul 2011, 14:04
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.

Answer Choice E.
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.

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

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New post 19 Jul 2011, 10:20
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: consecutive integers product [#permalink]

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New post 30 Aug 2011, 05:18
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.

Hope the explanation was helpful.


Regards,
Raghav.V

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Re: If n is positive integer, then n(n+)(n+2) is [#permalink]

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New post 27 Aug 2013, 02:56
I am assuming the expression is n(n+1)(n+2)

a) This not true. If n is even, (n+1) will be odd and if n is odd (n+1) will be even. It is sufficient for any one of the terms to be even to make the entire expression even.

b) For the same reasons as "a" this is alos not true.

c) This is also not true. It is sufficient for (n+1) alone to be a multiple of 3 for the entire expression to be divisible by 3. eg : n=2

e) Whenever n is even, (n+2) will also be even. Hence the expression will have two even integers and hence is always divisible by 4.
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Re: PS: Odd/Even and Divisibility [#permalink]

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New post 30 Oct 2013, 13:08
avdxz wrote:
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


E..whenever ..the equation is a product of three consecutive numbers and when the first is even the last will also be an even nmber...thus we will have two two's in the final product and thus the number will be divisible by four...
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Re: 224. Arithmetic operation [#permalink]

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New post 22 Apr 2014, 06:07
Bunuel wrote:
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.

Answer: E.


can we consider 0 as positive integer. If yes then whole number will become 0 and we can not divide 0 by 4.

Please clarify
Thanks.
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Re: 224. Arithmetic operation [#permalink]

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New post 22 Apr 2014, 07:16
PathFinder007 wrote:
Bunuel wrote:
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.

Answer: E.


can we consider 0 as positive integer. If yes then whole number will become 0 and we can not divide 0 by 4.

Please clarify
Thanks.


Three remarks here:

1. Zero is neither positive, nor negative integer.

2. Division by zero is not allowed (number/0 is undefined) but we can divide 0 by any non-zero number (0/number=0).

3. Zero is divisible by EVERY integer except zero itself, since 0/integer=0=integer (or, which is the same, zero is a multiple of every integer except zero itself). Thus even if we were told that n is integer (so if n could be a negative integer, zero or a positive integer) n(n + 1)(n + 2) would still be divisible by 4 for any even n.

Hope it's clear.
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Re: If n is a positive integer, then n(n+1)(n+2) is [#permalink]

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New post 15 Jul 2016, 04:53
avdxz wrote:
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


The problem is easiest to solve by substituting numbers for n. We'll try an odd number first and then an even number.

For an odd number, let’s let n = 1: 1(1+1)(1+2) = 1(2)(3) = 6.
We see than choices A and C can’t be the correct choices. Choice A is false because, while the product is even, n is not even. Choice C is false because, while n is odd, the product is even.

For an even number, let’s let n = 2: 2(2+1)(2+2) = 2(3)(4) = 24.

We see than choices B and D can’t be the correct choices. Choice B is false because, while the product is even, n is not odd. Choice D is false because, while the product is divisible by 3, n is even.

Therefore, the only correct answer is choice E. When n is even, the product is indeed divisible by 4.

The answer is E.
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Re: If n is a positive integer, then n(n+1)(n+2) is [#permalink]

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New post 16 Sep 2017, 07:48
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avdxz wrote:
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


There's a nice rule says: The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1
So, for example, the product of any 5 consecutive integers will be divisible by 5, 4, 3, 2 and 1
Likewise, the product of any 11 consecutive integers will be divisible by 11, 10, 9, . . . 3, 2 and 1
NOTE: the product may be divisible by other numbers as well, but these divisors are guaranteed.

Notice that n, n+1, and n+2 are three consecutive integers.
This means the product of n, n+1, and n+2 will be divisible by 3, 2 and 1
Since n(n+1)(n+2) is divisible by 2, then the product is ALWAYS EVEN.
This means we can eliminate answer choices A and B, because they put restrictions on when the product is even.
We can also eliminate C because it suggests that the product can be odd.
Likewise, since n(n+1)(n+2) is ALWAYS divisible by 3, we can eliminate answer choice D, because it puts a restriction on when the product is divisible by 3

Answer:

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

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New post 17 Sep 2017, 00:29
n is a positive integer
n(n+1)(n+2) = ?

(A) even only when n is even
(B) even only when n is odd
(C) odd whenever n is odd

=> Here we have multiple of 3 consecutive integers so we can have => odd*even*odd or even*odd*even. Both cases product will be even.
So Above 3 options does not give any additional information: n even or odd will not matter. Final product is always even

(D) divisible by 3 only when n is odd
As we have product of 3 consecutive integers, it will always be multiple of 3, whether n is even or odd :
n =odd => 1*2*3 <- divisible by 3
n= even => 2*3*4 <-divisible by 3

(E) divisible by 4 whenever n is even
as we have product of 3 numbers, value of n will decide by product is divisible by 4 or not
n =odd => 1*2*3 <- not divisible by 4
n= even => 2*3*4 <-divisible by 4

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

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New post 06 Dec 2017, 14:43
1
Hi All,

This question can be solved by TESTing VALUES (although you'll need to TEST at least 2 options to properly eliminate all of the wrong answers.

IF...
N = 1, then (N)(N+1)(N+2) = (1)(2)(3) = 6
N = 2, then (N)(N+1)(N+2) = (2)(3)(4) = 24

Answer A is eliminated by the 1st option
Answer B is eliminated by the 2nd option
Answer C is eliminated by the 1st option
Answer D is eliminated by the 2nd option
There's only one option remaining...

Final Answer:

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Re: If n is a positive integer, then n(n+1)(n+2) is   [#permalink] 06 Dec 2017, 14:43
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