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:

There are 16 multiples of 3 and hence there are 16 numbers that leave reminder of 1 (Note checked 48+1 = 49 <50)

1 also leaves a remainder 1 when divided by 3!

Hence 16+1 = 17 _________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

sorry, I know this post is dead but I would like to know how 1 divided by 3 has a remainder of 1? I thought the numerator had to be greater than the denom. in order to have an integer plus an integer remainder. What am I missing?

I think I may have answered my own question. (1/3) will have a remainder of 1 when the quotient is 0. (x/y)=q+r ......x=y(q) + r , for integers x and y, with the remainder r. Therefore, 1=3(0) + 1. Is this correctly stated?

I always forget about zero and the tricky roles it plays in questions....

Let me give it a shot (checking my own understanding in the process)...

Formula for Arithmetic Progression:

an = a1 + (n-1)d
where an = nth term; a1 = first term; n = number of terms; d = difference of succesive numbers

using 4 (which leaves a remainder of 1) as the first term and 49 (which leaves a remainder of 1) as the last term, we can solve for n

49 = 4 + (n-1)3 ---> n = 16

HOWEVER, n = 16 when the first term, i.e. the first number leaving a remainder of 1, is 4. But we have to remember that zero is an integer as well. Therefore, the number 1, when divided by 3 leaves a remainder of 1 as well. The expression would look like this:

Find 'x' for which above equation is true or appox true
your will get x = 17 (giving answer 51)

In other words, just remember table of three & check which position gives you number below/equal to 50. 3*16 = 48. So, 16 numbers, but table of 3 doesn't include '1'. So include '1', 16+1 = 17.

Now, why tables of 3? Because each element has remainder of 0 when divided by 3?
This is a trick, each element + 1, will give remainder of 1. This can be used for any such problem