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:

No of possible ways = 4*4! = 96 - shouldn't this be subtracted by the number of combination that has 0 at the start because the the number is technically 4 digits long!

No of possible ways = 4*4! = 96 - shouldn't this be subtracted by the number of combination that has 0 at the start because the the number is technically 4 digits long!

Please, can you explain rijul007

The first spot has 4 possibilities (1,2,4,5). Then, after that spot has been chosen, 4 possibilities remain for the next spot, and so forth.

Re: How many 5 digit numbers can be formed which are divisible [#permalink]

Show Tags

06 Sep 2013, 06:58

3

This post received KUDOS

To form a 5 digit number from available 6 digits 0,1,2,3,4,5 Case 1: Exclude 0 and form a 5 digit number from 1,2,3,4,5..since the sum of these digits is multiple of 3 as per divisibility rule for 3..A number is divisible by 3 if sum of the digits is divisible by 3 hence all 5 digit numbers formed from 1,2,3,4,5 are divisible by 3 5 digit number formed using 1,2,3,4,5 and divisible by 3 is 5! ways=120 Case 2:Include 0 and form a 5 digit number from 0,1,2,3,4,5 possible ways (0,1,2,3,4) (0,1,2,3,5)(0,1,2,4,5)(0,1,3,4,5)(0,2,3,4,5)...of which only a 5 digit number formed from (0,1,2,4,5) is divisible by 3 Since we need a 5 digit number...first digit can be selected from (1,2,4,5) but not 0..which is 4 ways rest of the digits can be selected from the remaining 4 digits..in 4! ways So no of possible ways will be 4*4! ways

Total number of ways to form a 5 digit no divisible by 3 is 5!+4*4!=120+96=216 ways

Re: How many 5 digit numbers can be formed which are divisible [#permalink]

Show Tags

06 Sep 2013, 07:27

4

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

ashiima wrote:

How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0, 1, 2, 3, 4, 5 (WITHOUT REPETITION)

A. 216 B. 3152 C. 240 D. 600 E. 305

First step:

We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.

We have six digits: 0, 1, 2, 3, 4, 5. Their sum=15.

For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.

Second step:

We have two set of numbers: 1, 2, 3, 4, 5 and 0, 1, 2, 4, 5. How many 5 digit numbers can be formed using these two sets:

1, 2, 3, 4, 5 --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.

0, 1, 2, 4, 5 --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, desired # would be total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=4!(5-1)=4!*4=96

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

With the limited financing options available to NZ citizens (especially those comme moi who aren't planning to return to work in NZ), I've really had to...

Strategy, innovation, marketing, finance... The second module has been pretty engaging. Though, no lack of memorable times. There is no lack of high profile guest speakers. One of the...