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 k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?

I.3 II.4 III.6

k is product of 3 numbers. there are two possibilities when m is even and when m is odd: 1: if m is even then k will be evenly divided by 4 due to factor m and (m+4) also it will be divided by 3 and 6 because m(m+4) is always divided by 3 [for m>0 and m is integer]and since m is even m(m+4) will always divided by 3 and 6.

2. when m is odd, it is not necessarily divided by 4. but k will be divided by 3 &6. there will always one 2 and one 3. e.g if m=1,3.. k is divided by 4 but if k=5 then it will not.

If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?

I.3 II.4 III.6

k is product of 3 numbers. there are two possibilities when m is even and when m is odd: 1: if m is even then k will be evenly divided by 4 due to factor m and (m+4) also it will be divided by 3 and 6 because m(m+4) is always divided by 3 [for m>0 and m is integer]and since m is even m(m+4) will always divided by 3 and 6.

2. when m is odd, it is not necessarily divided by 4. but k will be divided by 3 &6. there will always one 2 and one 3. e.g if m=1,3.. k is divided by 4 but if k=5 then it will not.

You would be correct if question were: "Which of the following MUST divide k evenly". Then answer is I (3) and III (6).

But question asks: "Which of the following COULD divide k evenly" and in this case answer is I, II and III (m can take all three options).
_________________

Re: If k=m(m+4)(m+5) k and m are positive integers. Which of the [#permalink]

Show Tags

22 Jan 2014, 12:42

utin wrote:

If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?

I.3 II.4 III.6

Let me take a crack at this

m(m+4)(m+5) are really three consecutive integers

(m)(m+1)(m+2)

So I is always true II is NOT always true. Cause if m is odd then m+4 is odd too, and the only even is m+5 which can or cannot be a multiple of 4 III the product of three consecutive integers is always a multiple of 3!

If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?

I.3 II.4 III.6

Let me take a crack at this

m(m+4)(m+5) are really three consecutive integers

(m)(m+1)(m+2)

So I is always true II is NOT always true. Cause if m is odd then m+4 is odd too, and the only even is m+5 which can or cannot be a multiple of 4 III the product of three consecutive integers is always a multiple of 3!

Numbers in each column increase by one. Now you see that each combination is divisible by 3 and 6 but not 4.

A little rough approach, but it works.

Hi, According to your approach, the first combination ain't divisible by 3 and 6. Then how can the combination MUST be evenly divided by 3 and 6 as stated by all here. I think the ans can be all the 3 options only because the question asked is 'WHICH COULD'. But if the question asked 'WHICH MUST', then I think there would have been no solution. Any expert if please could comment on this.

Re: If k=m(m+4)(m+5) k and m are positive integers. Which of the [#permalink]

Show Tags

24 Jan 2014, 05:08

jlgdr wrote:

utin wrote:

If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?

I.3 II.4 III.6

Let me take a crack at this

m(m+4)(m+5) are really three consecutive integers

(m)(m+1)(m+2)

So I is always true II is NOT always true. Cause if m is odd then m+4 is odd too, and the only even is m+5 which can or cannot be a multiple of 4 III the product of three consecutive integers is always a multiple of 3!

Hence the answer here should be I and III

Hope it helps Cheers! J

It is not clear why m, m+4 and m+5 are consecutive integers.

Numbers in each column increase by one. Now you see that each combination is divisible by 3 and 6 but not 4.

A little rough approach, but it works.

Hi, According to your approach, the first combination ain't divisible by 3 and 6. Then how can the combination MUST be evenly divided by 3 and 6 as stated by all here. I think the ans can be all the 3 options only because the question asked is 'WHICH COULD'. But if the question asked 'WHICH MUST', then I think there would have been no solution. Any expert if please could comment on this.

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Term 1 has begun. If you're confused, wondering what my post on the last 2 official weeks was, that was pre-term. What that means is that the school...