It is currently 18 Oct 2017, 12:04

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

Ps arrangement : 5 LETTERS

Author Message
TAGS:

Hide Tags

Senior Manager
Joined: 30 May 2005
Posts: 275

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

Ps arrangement : 5 LETTERS [#permalink]

Show Tags

06 Sep 2005, 17:02
3
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

66% (01:45) correct 34% (01:33) wrong based on 91 sessions

HideShow timer Statistics

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, 07:19, edited 2 times in total.
Topic locked. In case of any questions please continue the discussion here: http://gmatclub.com/forum/ps-91597.html

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

Senior Manager
Joined: 27 Aug 2005
Posts: 331

Kudos [?]: 194 [0], given: 0

Show Tags

06 Sep 2005, 19:23
I got (4) 216. If this is right I will explain.

Kudos [?]: 194 [0], given: 0

Senior Manager
Joined: 04 May 2005
Posts: 278

Kudos [?]: 79 [0], given: 0

Location: CA, USA

Show Tags

06 Sep 2005, 21:54
choice 4

there are two possibilities:
1) five digits all none 0
there are 5! = 120

2) five digits with one 0
there are > 0 one possibility

from above analysis, we can already choose E

But for 2), we can refine it as:
when choosing 0, we have to give up 3, otherwise, the number can't be
divided by 3. That leaves us with:

5P5 - 4P4 = 96 ( 4P4 is for the case where 0 is at the left most digit)

Kudos [?]: 79 [0], given: 0

Senior Manager
Joined: 27 Aug 2005
Posts: 331

Kudos [?]: 194 [0], given: 0

Show Tags

06 Sep 2005, 21:56
I did P(6,5) to get 720 permutations of 6 digits in five locations.

Divided by 3 to get multiples of 3, so got 240.

Then subtracted those that had 0 as the first digit to get to 216.

Kudos [?]: 194 [0], given: 0

Intern
Joined: 04 Sep 2005
Posts: 13

Kudos [?]: 1 [1], given: 0

Show Tags

07 Sep 2005, 03:22
1
KUDOS
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

Kudos [?]: 1 [1], given: 0

Manager
Joined: 26 Sep 2007
Posts: 65

Kudos [?]: 26 [0], given: 5

Show Tags

14 Nov 2007, 09:23
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

Kudos [?]: 26 [0], given: 5

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1046 [0], given: 4

Location: New York City

Show Tags

25 Dec 2007, 14:40
alright. finally understood this question.

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.

5! + 4*4! = 216

Kudos [?]: 1046 [0], given: 4

SVP
Joined: 07 Nov 2007
Posts: 1792

Kudos [?]: 1059 [1], given: 5

Location: New York
Re: Ps arrangement : 5 LETTERS [#permalink]

Show Tags

26 Aug 2008, 09:16
1
KUDOS
mandy wrote:
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
_________________

Smiling wins more friends than frowning

Kudos [?]: 1059 [1], given: 5

Intern
Joined: 31 Dec 2011
Posts: 17

Kudos [?]: 8 [0], given: 7

Location: United States
Concentration: Technology, Marketing
Schools: IIM Calcutta - Class of 2001
GMAT 1: 750 Q50 V40
GPA: 3.4
WE: Information Technology (Consulting)
Re: Ps arrangement : 5 LETTERS [#permalink]

Show Tags

05 Jan 2012, 09:58
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?)

Kudos [?]: 8 [0], given: 7

Manager
Joined: 29 Jul 2011
Posts: 104

Kudos [?]: 69 [6], given: 6

Location: United States
Re: Ps arrangement : 5 LETTERS [#permalink]

Show Tags

05 Jan 2012, 15:01
6
KUDOS
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

Kudos [?]: 69 [6], given: 6

Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 172

Kudos [?]: 98 [0], given: 1

Location: India
WE: Information Technology (Investment Banking)
Re: Ps arrangement : 5 LETTERS [#permalink]

Show Tags

05 Jan 2012, 21:16
I got the answer as D

5! + 4*4! = 216

Kudos [?]: 98 [0], given: 1

Re: Ps arrangement : 5 LETTERS   [#permalink] 05 Jan 2012, 21:16
Display posts from previous: Sort by