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M04 # 32

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M04 # 32 [#permalink] New post 22 Sep 2008, 18:43
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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

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Answer:
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!
[Reveal] Spoiler: OA
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Re: M04 # 32 [#permalink] New post 23 Sep 2008, 14:40
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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!
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Re: M04 # 32 [#permalink] New post 22 May 2009, 22:22
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,
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Re: M04 # 32 [#permalink] New post 09 Apr 2010, 16:54
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
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Re: M04 # 32 [#permalink] New post 21 May 2010, 21:02
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This question would've taken me way too long to figure out on the real thing; would've started looking at answer choices... :P

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
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Re: M04 # 32 [#permalink] New post 22 May 2010, 01:02
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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?

Answer:
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

Answer: 216.

Hope it helps.
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Re: M04 # 32 [#permalink] New post 08 Oct 2010, 05:49
Answer is 216. Thanks for the explanation
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Re: M04 # 32 [#permalink] New post 08 Oct 2010, 05:59
5! + (5! - 4!) - 216 . its E
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Re: M04 # 32 [#permalink] New post 08 Oct 2010, 12:09
thanks for all the explanations.

started with the sum of digits should be divisible by 3. got stuck in figuring out the factorials :(
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Re: M04 # 32 [#permalink] New post 10 Oct 2010, 07:00
tiruraju wrote:
5! + (5! - 4!) - 216 . its E


How you can come to the this formula? I just calculate with: 5! +4*4!
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Re: M04 # 32 [#permalink] New post 11 Oct 2010, 10:50
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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)

Total i)+ii) = 120+96=216 answer...


I solved using this method and trust me it's really quick .....
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Re: M04 # 32 [#permalink] New post 04 Oct 2011, 17:38
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.
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Re: M04 # 32 [#permalink] New post 12 Oct 2011, 10:21
Nice question
The answer is 5! + 4*4! = 216
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Re: M04 # 32 [#permalink] New post 12 Oct 2011, 20:54
good question.
ans : 5! + 4.4! = 216
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Re: M04 # 32 [#permalink] New post 24 Nov 2011, 14:14
Bunuel

Excellent explanation as always~
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Re: M04 # 32 [#permalink] New post 12 Oct 2012, 04:47
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'.

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Re: M04 # 32   [#permalink] 12 Oct 2012, 04:47
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