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 350,000 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:

2 cannot give you an answer, consider 1,2,3,4,5,6 or 3,4,5,6,7,8 both are not divisible by 9 whereas 2,3,4,5,6,7 is divisible and hence we cannot say conclusively.

D is the OA.
But I am unsure how stmt 2 plays out if you have -ve consecutive integers. There isn't really a proper OE but here's my reasoning:

Well, the sum S=6n+15=3(2n+5), where n is the first number

stmt 1) irrespective of the what n is, S will be always be odd. Suff.

stmt 2) says that the 6 digit number formed by each integer is divisible by 9 --> meaning the sum of the numbers is divisible by 9 --> 3(2n+5) = 9x
therefore, (2n+5)=3x . How do we conclude from this whether the sum is even or odd?

D is the OA. But I am unsure how stmt 2 plays out if you have -ve consecutive integers. There isn't really a proper OE but here's my reasoning:

Well, the sum S=6n+15=3(2n+5), where n is the first number

stmt 1) irrespective of the what n is, S will be always be odd. Suff.

stmt 2) says that the 6 digit number formed by each integer is divisible by 9 --> meaning the sum of the numbers is divisible by 9 --> 3(2n+5) = 9x therefore, (2n+5)=3x . How do we conclude from this whether the sum is even or odd?

Tank....

If you let 6n+15 equal to a multiple of 9, you will find out that the first term is an even number in the series. And from (1), we know that the sum is always odd. So (2) is sufficient.

I am not panicking. Nope, Not at all. But I am beginning to wonder what I was thinking when I decided to work full-time and plan my cross-continent relocation...

Over the last week my Facebook wall has been flooded with most positive, almost euphoric emotions: “End of a fantastic school year”, “What a life-changing year it’s been”, “My...