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:

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi,

Can u help me solve the below mentioned prob:

If the sum of four consecutive positive integers a three digit multiple of 50, the mean of the these integers must be one of k possible values, where k=

I tried to come up w/ a test case to solve this.. I knew that the ones digits of the 4 consecutive digits had to add up to either 5 or 0 for it to be a multiple of 50..

I went w/ 61, 62, 63, and 64.. all consectuive..and add up to 250 ..

and thus the mean is greater than 10... so anything other than E has to be wrong...

Since a is an integer, you have to eliminate 150, 250,....etc. You have 100, 200,...,900 left. Count them. It is 9. So only 9 possibilities are there. Ans C. _________________

4 consecutive integers cannot add up to 100, try it.

therefore we have 150,250,350 ... 950, so 9

some notes about consecutives:
the average of an even number of consecutives is never a whole number

the average of an odd number of consecutives is a whole number

the sum of an odd number of consecutives is divisible by the number of terms (1,2,3 = 6/3 = 2 or 10,11,12,13,14,15 = 65/5 = 13). this DOES not apply to an even number of terms.

the product of any n consecutives has n as a factor.