Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 30 May 2005
Posts: 276

Ps arrangement : 5 LETTERS [#permalink]
Show Tags
06 Sep 2005, 17:02
3
This post was BOOKMARKED
Question Stats:
63% (02:37) correct
37% (01:33) wrong based on 85 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: ps91597.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/ps91597.html



Senior Manager
Joined: 27 Aug 2005
Posts: 331
Location: Montreal, Canada

I got (4) 216. If this is right I will explain.



Senior Manager
Joined: 04 May 2005
Posts: 279
Location: CA, USA

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)



Senior Manager
Joined: 27 Aug 2005
Posts: 331
Location: Montreal, Canada

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.



Intern
Joined: 04 Sep 2005
Posts: 13

1
This post received 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



Manager
Joined: 26 Sep 2007
Posts: 65

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



CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City

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 51+1= 4 or 4! to permute the rest of the slots.
5! + 4*4! = 216



SVP
Joined: 07 Nov 2007
Posts: 1799
Location: New York

Re: Ps arrangement : 5 LETTERS [#permalink]
Show Tags
26 Aug 2008, 09:16
1
This post received 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
_________________
Your attitude determines your altitude Smiling wins more friends than frowning



Intern
Joined: 31 Dec 2011
Posts: 17
Location: United States
Concentration: Technology, Marketing
Schools: IIM Calcutta  Class of 2001
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 (05) = 6P5 # of the above, that has 0 in the beginning = 6P4
Total legit 5digit 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?)



Manager
Joined: 29 Jul 2011
Posts: 107
Location: United States

Re: Ps arrangement : 5 LETTERS [#permalink]
Show Tags
05 Jan 2012, 15:01
6
This post received 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 = 34 min



Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 176
Location: India
Concentration: Finance, International Business
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




Re: Ps arrangement : 5 LETTERS
[#permalink]
05 Jan 2012, 21:16







