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:

60% (02:18) correct
40% (01:38) wrong based on 26 sessions

Hello does anyone has a clue on this one I thought 40 was the answer but i was wrong plz Explains your works and reasoning thanks

How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits? (1) 15 2(40) (3) 96 (4) 216 (5) 120

Topic locked. In case of any questions please continue the discussion here: ps-91597.html

Last edited by Bunuel on 13 Jan 2012, 06:19, edited 2 times in total.

Topic locked. In case of any questions please continue the discussion here: http://gmatclub.com/forum/ps-91597.html

I followed qpoo's logic
Only the following sets of numbers add up to a number that is divisible by 3
{54321} and (54210}

The answer is the factorial of the first set plus a little less than the factorial of the second set. If 0 is in the first digit from the left then you don't really have a 5 digit number do you.

This first set is 5! = 120. At this point you can solve the problem because there is only one answer larger than 120.

If you want to take the problem farther than when 0 is all the way to the left you have 4! so my answer is 2*5! - 4! = 216

0 can't be the leftmost digit, otherwise it will be a four digit number
So total numbers = 5! = 120
Numbers with 0 as leftmost digit = 4! =24
So total 5 digit numbers = 120 - 24 =96
Total 5 digit numbers = Total numbers (non 0) + Total numbers (including 0 but not at first place) = 120 + 96 =216

we can have two scenarios with a sum that is divisible by 13.

sum of 15
12345

or

sum of 12
01245

We can permute the first scenario in 5! ways.
5! = 120

We can permute the second scenario in 4*4! ways.
The first slot cannot be zero. Remove zero from the options. 4 ways to fill first slot. Replace zero in the second slot and account for the number we placed in the first slot. We have 5-1+1= 4 or 4! to permute the rest of the slots.

Hello does anyone has a clue on this one I thought 40 was the answer but i was wrong plz Explains your works and reasoning thanks

How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits? (1) 15 2(40) (3) 96 (4) 216 (5) 120

= non zero -5 digit + include zerio in five digit ={12345} +{01245}

= 5! + 4*4*3*2 = 120+96 =216 _________________

Your attitude determines your altitude Smiling wins more friends than frowning

I'd think its E - 120 Explanation: # of 5 digit numbers using (0-5) = 6P5 # of the above, that has 0 in the beginning = 6P4

Total legit 5-digit numbers = 6P5- 6P4 = 360

I kind of at this point, reckoned a third of these must be divisible by 3 = 120 (Can someone give a better way of adding in divisibility by 3 as a criteria?)

RULE: When sum of digits of a number is divisible by 3, that number is divisible by 3. In 0,1,2,3,4,5, we can form 5 digits out of 1,2,3,4,5 and 0,1,2,4,5 that are divisible by 3.

1. 1,2,3,4,5 -> 5! = 120 2. 0,1,2,4,5 -> 5! = 120. However, can't have leftmost digit = 0. So, numbers to be removed = 4! = 24 -> 120 - 24 = 96

Total = 120 + 96 = 216 -> D. _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

I´ve done an interview at Accepted.com quite a while ago and if any of you are interested, here is the link . I´m through my preparation of my second...

It’s here. Internship season. The key is on searching and applying for the jobs that you feel confident working on, not doing something out of pressure. Rotman has...