It is currently 23 Oct 2017, 19:24

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M04 # 32

Author Message
Intern
Joined: 10 Sep 2008
Posts: 36

Kudos [?]: 44 [3], given: 0

### Show Tags

22 Sep 2008, 19:43
3
KUDOS
9
This post was
BOOKMARKED
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?

(A) 15
(B) 96
(C) 120
(D) 181
(E) 216

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

For an integer to be divisible by 3 sum of its digits must be divisible by 3. Then you can have only 15= 1+2+3+4+5 or
12= 0+1+2+4+5 as the sum. In the first case there are 5! permutations, and in the second there are 4*4! because 0 can not be the first digit. Therefore, the answer is 5! + (4*4!) - 216.

My only question here is: How do we come up with 4*4! for the second sum permutations? The 0 that can't be in first place has only four options so I think the logic is..how many times can we arrange one number into four spots? 4 differnt ways. Then 4! comes from how do we arrange 4 numbers into five spots? Help with the 4*4! explanation. I really wish these answers were more explanatory.

Thank you!

Kudos [?]: 44 [3], given: 0

Intern
Joined: 17 Sep 2008
Posts: 6

Kudos [?]: 9 [3], given: 0

Schools: Chicago Booth, Wharton, MIT, Haas

### Show Tags

23 Sep 2008, 15:40
3
KUDOS
1
This post was
BOOKMARKED
Start from the leftmost digit :

1st digit - can be filled in 4 ways (1,2,4,5).
2nd digit- can then be filled in 4 ways again - 0 , and 3 of the remaining digits from(1,2,4,5).
3rd digit - 3 ways
4th digit - 2 ways
5th digit - 1 way

So the total number of possibilities is 4*4*3*2*1, which is 4*4!

Kudos [?]: 9 [3], given: 0

Manager
Joined: 12 Apr 2009
Posts: 210

Kudos [?]: 245 [1], given: 4

### Show Tags

22 May 2009, 23:22
1
KUDOS
mmond4 wrote:
Start from the leftmost digit :

1st digit - can be filled in 4 ways (1,2,4,5).
2nd digit- can then be filled in 4 ways again - 0 , and 3 of the remaining digits from(1,2,4,5).
3rd digit - 3 ways
4th digit - 2 ways
5th digit - 1 way

So the total number of possibilities is 4*4*3*2*1, which is 4*4!

I believe the explanation above is wrong.

For an integer to be divisible by 3 the sum of its digits must be divisible by 3. Then you can have only $$15 = 1+2+3+4+5$$ or $$12 = 0+1+2+4+5$$ as the sum. In the first case there are $$5!$$ permutations, and in the second - $$4*4!$$ since 0 can not be the first digit. Therefore, the answer is $$5!+4*4!=216$$ .

che,
dg
_________________

-talent is the desire to practice-

Kudos [?]: 245 [1], given: 4

Intern
Joined: 15 Feb 2010
Posts: 11

Kudos [?]: 13 [0], given: 0

### Show Tags

09 Apr 2010, 17:54
the questionn "...divisible by 3, without repeating the digits." <- does this mean that 3 shouldn't be repeated more than once for each digit?
and why do multiply by 4

Kudos [?]: 13 [0], given: 0

Intern
Joined: 23 Dec 2009
Posts: 46

Kudos [?]: 47 [1], given: 7

Schools: HBS 2+2
WE 1: Consulting
WE 2: Investment Management

### Show Tags

21 May 2010, 22:02
1
KUDOS
This question would've taken me way too long to figure out on the real thing; would've started looking at answer choices...

1) Need the digits to sum to a multiple of 3 (to satisfy "divisible by 3" rule):

12345 does the job
01425 does the job

2) Need to make a 5 digit number:

5! counts the number of ways of arranging 1, 2, 3, 4, 5
5!-4! counts the number of ways of arranging 0, 1, 2, 4, 5 taking out numbers that start with 0 (otherwise we would be counting 4 digit numbers as well!)

3) 5! + (5!-4!) = 120 + 96 = 216
_________________

My GMAT quest...

...over!

Kudos [?]: 47 [1], given: 7

Math Expert
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129535 [10], given: 12201

### Show Tags

22 May 2010, 02:02
10
KUDOS
Expert's post
9
This post was
BOOKMARKED
dczuchta wrote:
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?

For an integer to be divisible by 3 sum of its digits must be divisible by 3. Then you can have only 15= 1+2+3+4+5 or
12= 0+1+2+4+5 as the sum. In the first case there are 5! permutations, and in the second there are 4*4! because 0 can not be the first digit. Therefore, the answer is 5! + (4*4!) - 216.

My only question here is: How do we come up with 4*4! for the second sum permutations? The 0 that can't be in first place has only four options so I think the logic is..how many times can we arrange one number into four spots? 4 differnt ways. Then 4! comes from how do we arrange 4 numbers into five spots? Help with the 4*4! explanation. I really wish these answers were more explanatory.

Thank you!

This question was posted in PS subforum. Below is my solution from there:

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 this 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, total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=4!(5-1)=4!*4=96

120+96=216

Hope it helps.
_________________

Kudos [?]: 129535 [10], given: 12201

Manager
Joined: 24 Aug 2010
Posts: 187

Kudos [?]: 98 [0], given: 18

Location: Finland
Schools: Admitted: IESE(),HEC, RSM,Esade
WE 1: 3.5 years international

### Show Tags

08 Oct 2010, 06:49
Answer is 216. Thanks for the explanation

Kudos [?]: 98 [0], given: 18

Manager
Joined: 17 Apr 2010
Posts: 97

Kudos [?]: 58 [0], given: 12

### Show Tags

08 Oct 2010, 06:59
5! + (5! - 4!) - 216 . its E

Kudos [?]: 58 [0], given: 12

Manager
Joined: 07 Aug 2010
Posts: 75

Kudos [?]: 19 [0], given: 9

### Show Tags

08 Oct 2010, 13:09
thanks for all the explanations.

started with the sum of digits should be divisible by 3. got stuck in figuring out the factorials
_________________

Click that thing - Give kudos if u like this

Kudos [?]: 19 [0], given: 9

Manager
Joined: 27 Jul 2010
Posts: 190

Kudos [?]: 46 [0], given: 15

Location: Prague
Schools: University of Economics Prague

### Show Tags

10 Oct 2010, 08:00
tiruraju wrote:
5! + (5! - 4!) - 216 . its E

How you can come to the this formula? I just calculate with: 5! +4*4!
_________________

You want somethin', go get it. Period!

Kudos [?]: 46 [0], given: 15

Intern
Joined: 01 Oct 2010
Posts: 4

Kudos [?]: 10 [1], given: 1

### Show Tags

11 Oct 2010, 11:50
1
KUDOS
Here is the way I solved it hope it helps.....

for numbers 0,1,2,3,4,5 we have to find how many five digit numbers can be formed that's divisible by 3

For a number to be divisible by 3 sum of the digits have to be divisible by 3

let take 1,2,3,4,5 we leave the zero for now.

we see the sum of digits are 1+2+3+4+5=15 all numbers that have these no will be divisible by 3

So no of ways these numbers can be arranged is 5!=120 ...i)

Now again we take zero we have to see the sum of digits must be divisible by 3

for zero only be possible for 0,1,2,4,5 => 0+1+2+4+5=12 rest does not add up to mutiple of 3

No of ways =5!=120 however we have to see the 0 is not the first digit.(then its a 4 digit no.)

[so when 0 is first digit no of permutations are 4!=24
numbers formed without 0 being first digit = 120 -24= 96 ...ii)

I solved using this method and trust me it's really quick .....

Kudos [?]: 10 [1], given: 1

Manager
Joined: 21 Nov 2010
Posts: 127

Kudos [?]: 5 [0], given: 12

### Show Tags

04 Oct 2011, 18:38
mmond4 wrote:
Start from the leftmost digit :

1st digit - can be filled in 4 ways (1,2,4,5).
2nd digit- can then be filled in 4 ways again - 0 , and 3 of the remaining digits from(1,2,4,5).
3rd digit - 3 ways
4th digit - 2 ways
5th digit - 1 way

So the total number of possibilities is 4*4*3*2*1, which is 4*4!

Kudos! I had the same question and that was a great explanation.

Kudos [?]: 5 [0], given: 12

Manager
Joined: 20 Nov 2010
Posts: 214

Kudos [?]: 41 [0], given: 38

### Show Tags

12 Oct 2011, 11:21
Nice question
The answer is 5! + 4*4! = 216
_________________

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
MGMAT 6 650 (51,31) on 31/8/11
MGMAT 1 670 (48,33) on 04/9/11
MGMAT 2 670 (47,34) on 07/9/11
MGMAT 3 680 (47,35) on 18/9/11
GMAT Prep1 680 ( 50, 31) on 10/11/11

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
CR notes
http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142
http://gmatclub.com/forum/1001-ds-questions-file-106193.html#p832133
http://gmatclub.com/forum/gmat-prep-critical-reasoning-collection-106783.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html?hilit=chineseburned

Kudos [?]: 41 [0], given: 38

Senior Manager
Joined: 01 Nov 2010
Posts: 284

Kudos [?]: 85 [0], given: 44

Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)

### Show Tags

12 Oct 2011, 21:54
good question.
ans : 5! + 4.4! = 216
_________________

kudos me if you like my post.

Attitude determine everything.
all the best and God bless you.

Kudos [?]: 85 [0], given: 44

Director
Joined: 28 Jun 2011
Posts: 879

Kudos [?]: 244 [0], given: 57

### Show Tags

24 Nov 2011, 15:14

Kudos [?]: 244 [0], given: 57

Intern
Status: Preparing for GMAT
Joined: 19 Sep 2012
Posts: 19

Kudos [?]: 10 [0], given: 8

Location: India
GMAT Date: 01-31-2013
WE: Information Technology (Computer Software)

### Show Tags

12 Oct 2012, 05:47
when 1 2 3 4 5
5! = 120

when 0 1 2 3 5 excluding zero starting = 120 - 4! =96.

So total will be

120+96=216 . 'E'.

_________________

Rajeev Nambyar
Chennai, India.

Kudos [?]: 10 [0], given: 8

Re: M04 # 32   [#permalink] 12 Oct 2012, 05:47
Display posts from previous: Sort by

# M04 # 32

Moderator: Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.