Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

10 Jul 2006, 07:56

3

This post received KUDOS

16

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

80% (00:59) correct
20% (00:58) wrong based on 1007 sessions

HideShow timer Statistics

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.

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.

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.

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}

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

Show Tags

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.
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

(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...

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.

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.

(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.
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

(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

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

Show Tags

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

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...