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Ps arrangement : 5 LETTERS

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Ps arrangement : 5 LETTERS [#permalink] New post 06 Sep 2005, 16:02
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

62% (02:10) correct 38% (01:36) wrong based on 20 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
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 [#permalink] New post 06 Sep 2005, 18:23
I got (4) 216. If this is right I will explain.
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 [#permalink] New post 06 Sep 2005, 20:54
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)
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 [#permalink] New post 06 Sep 2005, 20:56
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.
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 [#permalink] New post 07 Sep 2005, 02:22
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
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 [#permalink] New post 14 Nov 2007, 08:23
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
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 [#permalink] New post 25 Dec 2007, 13:40
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
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Re: Ps arrangement : 5 LETTERS [#permalink] New post 26 Aug 2008, 08:16
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
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Re: Ps arrangement : 5 LETTERS [#permalink] New post 05 Jan 2012, 08:58
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?)
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Re: Ps arrangement : 5 LETTERS [#permalink] New post 05 Jan 2012, 14:01
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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.
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Re: Ps arrangement : 5 LETTERS [#permalink] New post 05 Jan 2012, 20:16
I got the answer as D

5! + 4*4! = 216
Re: Ps arrangement : 5 LETTERS   [#permalink] 05 Jan 2012, 20:16
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