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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, 07:56

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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

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.

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.

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. _________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

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.

Re: A simple math problem, pls help me explain,thx [#permalink]

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16 Jul 2011, 02:57

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)

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

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.

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.

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.

Re: A simple math problem, pls help me explain,thx [#permalink]

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21 Jul 2011, 06:29

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!!

Re: A simple math problem, pls help me explain,thx [#permalink]

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21 Jul 2011, 14:48

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.

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

Consider kudos if my post was helpful.

gmatclubot

Re: consecutive integers product
[#permalink]
30 Aug 2011, 05:18

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