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:

Re: PS: Number of Integers- Good one [#permalink]
08 Oct 2008, 00:13

Nice approach from leondas.

I used the following approach.

Since, there are 17 integers between 0 and 50 that are divisible by 3 (including 0) and since the first such integer is 0 and the last is 48....hence there should be 17 such integers that have remainder as 1 when divided by 3 as 0+1 = and 48 + 1 = 49 and both these numbers are within the limits of 0 and 50.

Re: PS: Number of Integers- Good one [#permalink]
11 Oct 2008, 07:44

1

This post received KUDOS

rao_1857 wrote:

Tweaking the question little bit, how about this:

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 and have a remainder of 3 when divided by 5 ?

Any idea Gurus?

remainder of 1 when divided by 3: 3k+1 = 1, 4, 7, 10, (13), 16, 19, 22, 25, (28), 31, 34, 37, 40, (43), 46, 49... remainder of 3 when divided by 5: 5k+3= 3, 8, (13), 18, 23, 28, 33, 38, (43), 48...

Here, 13 and 43 are common in the two series. Other numbers will be 58, 73, so on and so forth. The general equation is: (28-13)k+13= 15k+ 13 (where, k=0,1,2,...) Hence, if we want to find the number of integers even upto 1000, we can use the above formula.

Going back to the problem, I see 3 numbers that satisfies the above criteria i.e 13, 28 and 43. (The next one will be 58 which is greater than 50). Any comments? _________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...