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M04 # 32
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22 Sep 2008, 19:43
How many fivedigit 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 Source: GMAT Club Tests  hardest GMAT questions 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!



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Re: M04 # 32
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23 Sep 2008, 15:40
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
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22 May 2009, 23:22
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, dg
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Re: M04 # 32
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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



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21 May 2010, 22:02
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
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Re: M04 # 32
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22 May 2010, 02:02
dczuchta wrote: How many fivedigit 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 150={1,2,3,4,5} and 153={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!(51)=4!*4=96 120+96=216 Answer: 216. Hope it helps.
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Re: M04 # 32
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08 Oct 2010, 06:49
Answer is 216. Thanks for the explanation



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



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Re: M04 # 32
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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
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Re: M04 # 32
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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!
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Re: M04 # 32
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11 Oct 2010, 11:50
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
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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.



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Re: M04 # 32
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12 Oct 2011, 11:21
Nice question The answer is 5! + 4*4! = 216
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Re: M04 # 32
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12 Oct 2011, 21:54
good question. ans : 5! + 4.4! = 216
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24 Nov 2011, 15:14
Bunuel
Excellent explanation as always~



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Re: M04 # 32
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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'.
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