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How many 5 digit numbers can be formed which are divisible

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How many 5 digit numbers can be formed which are divisible [#permalink] New post 15 Dec 2011, 17:59
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How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0, 1, 2, 3, 4, 5 (WITHOUT REPETITION)

A. 216
B. 3152
C. 240
D. 600
E. 305

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-five-digit-number-divisible-by-3-is-to-be-formed-using-136900.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Sep 2013, 06:28, edited 2 times in total.
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Re: digit [#permalink] New post 15 Dec 2011, 18:07
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Divisibilty rule for 3: Sum of all digits is a multiple of 3

Selections for the above to be valid
1,2,3,4,5
No of possible ways = 5! = 120

0,1,2,4,5
No of possible ways = 4*4! = 96


Total = 120+96 = 216
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Re: digit [#permalink] New post 06 Sep 2013, 04:06
No of possible ways = 4*4! = 96 - shouldn't this be subtracted by the number of combination that has 0 at the start because the the number is technically 4 digits long!

Please, can you explain rijul007
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Re: digit [#permalink] New post 06 Sep 2013, 05:56
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arakban99 wrote:
No of possible ways = 4*4! = 96 - shouldn't this be subtracted by the number of combination that has 0 at the start because the the number is technically 4 digits long!

Please, can you explain rijul007


The first spot has 4 possibilities (1,2,4,5). Then, after that spot has been chosen, 4 possibilities remain for the next spot, and so forth.

Hence, 4*4!.
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Re: How many 5 digit numbers can be formed which are divisible [#permalink] New post 06 Sep 2013, 05:58
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To form a 5 digit number from available 6 digits 0,1,2,3,4,5
Case 1: Exclude 0 and form a 5 digit number from 1,2,3,4,5..since the sum of these digits is multiple of 3
as per divisibility rule for 3..A number is divisible by 3 if sum of the digits is divisible by 3
hence all 5 digit numbers formed from 1,2,3,4,5 are divisible by 3
5 digit number formed using 1,2,3,4,5 and divisible by 3 is 5! ways=120
Case 2:Include 0 and form a 5 digit number from 0,1,2,3,4,5
possible ways (0,1,2,3,4) (0,1,2,3,5)(0,1,2,4,5)(0,1,3,4,5)(0,2,3,4,5)...of which only a 5 digit number formed from (0,1,2,4,5) is divisible by 3
Since we need a 5 digit number...first digit can be selected from (1,2,4,5) but not 0..which is 4 ways
rest of the digits can be selected from the remaining 4 digits..in 4! ways
So no of possible ways will be 4*4! ways

Total number of ways to form a 5 digit no divisible by 3 is 5!+4*4!=120+96=216 ways
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Re: How many 5 digit numbers can be formed which are divisible [#permalink] New post 06 Sep 2013, 06:27
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ashiima wrote:
How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0, 1, 2, 3, 4, 5 (WITHOUT REPETITION)

A. 216
B. 3152
C. 240
D. 600
E. 305


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 these 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, desired # would be 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: A.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-five-digit-number-divisible-by-3-is-to-be-formed-using-136900.html
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Re: How many 5 digit numbers can be formed which are divisible   [#permalink] 06 Sep 2013, 06:27
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