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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE

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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE  [#permalink]

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New post 08 Oct 2010, 03:53
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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE DIGITS 0, 1, 2, 3, 4 and 5 WITHOUT REPETITIONS. THAT TOTAL NO OF WAYS IT CAN BE DONE IS?

A. 122
B. 210
C. 216
D. 217
E. 220
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Re: A FIVE DIGIT NUMBER DIVIBLE BY 3  [#permalink]

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New post 08 Oct 2010, 05:20
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anilnandyala wrote:
A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE DIGITS 0,1,2,3,4&5 WITH OUT REPETITIONS .THAT TOTAL NO OF WAYS IT CAN BE DONE IS?

A)122
B)210
C)216
D)217
E)220


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!=96

120+96=216

Answer: C.
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Re: A FIVE DIGIT NUMBER DIVIBLE BY 3  [#permalink]

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New post 08 Oct 2010, 05:43
Quite a good question will multiple application of math's rule.Thanks Bunnel for explanation.
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Re: A FIVE DIGIT NUMBER DIVIBLE BY 3  [#permalink]

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New post 09 Oct 2010, 10:42
Quote:
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!=96

120+96=216

Answer: C.


Hi Bunuel,
I have an approach...but don;t understand where I am going wrong...Here it goes..

Step One: Find out all the five digit numbers without repetition from 0,1,2,3,4,5
Thus there are 5 spots to be filled from a group of 6..... ( something like this _ _ _ _ _)
The 1st spot can be filled in 5 ways (0 not included here so that it reamains a 5 digit number) and the 2nd spot in 5 ways ( 0 included here ) and the rest can be filled in
4 x 3 x 2 ways..therefore the total number amounts to - 5 x 5 x 4 x 3 x 2 = 5 x 5! = 600

Step Two: Divide this 600 by 3 to get the number of 5-digit numbers that are divisible by 3 . Therefore 600/3= 200.
This 2nd step I have done keeping in mind th approach that we use for sums like these -

The number of numbers that are divsible by 7 between 300 and 500 are ?
Ans: Here's what I do...get a multiple of 7 just greater than 300, thts 301 and a mulitple of 7 just less than 500..ths 497..
then 497-301 = 196..then 196/7 =28....the answer is 28 + 1 = 29( +1 for counting one of the left out multiples within the range ).,..

With the above method, what I am basically doing is finding out how many numbers lie in the range and then divide it by 7..and then add 1 to get the answer...
Similarly, when I am using this in the given question on 5 digits, why am I not able to get eh correct answer ?
What I understand is tht in both these questions we are trying to find out all those numbers that are divisible by a particular number...

Please advise..

RJ
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Re: A FIVE DIGIT NUMBER DIVIBLE BY 3  [#permalink]

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New post 27 Oct 2010, 05:31
Bunuel wrote:
anilnandyala wrote:
A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE DIGITS 0,1,2,3,4&5 WITH OUT REPETITIONS .THAT TOTAL NO OF WAYS IT CAN BE DONE IS?

A)122
B)210
C)216
D)217
E)220


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!=96

120+96=216

Answer: C.



Hi Bunuel,
I have an approach...but don;t understand where I am going wrong...Here it goes..

Step One: Find out all the five digit numbers without repetition from 0,1,2,3,4,5
Thus there are 5 spots to be filled from a group of 6..... ( something like this _ _ _ _ _)
The 1st spot can be filled in 5 ways (0 not included here so that it reamains a 5 digit number) and the 2nd spot in 5 ways ( 0 included here ) and the rest can be filled in
4 x 3 x 2 ways..therefore the total number amounts to - 5 x 5 x 4 x 3 x 2 = 5 x 5! = 600

Step Two: Divide this 600 by 3 to get the number of 5-digit numbers that are divisible by 3 . Therefore 600/3= 200.
This 2nd step I have done keeping in mind th approach that we use for sums like these -

The number of numbers that are divsible by 7 between 300 and 500 are ?
Ans: Here's what I do...get a multiple of 7 just greater than 300, thts 301 and a mulitple of 7 just less than 500..ths 497..
then 497-301 = 196..then 196/7 =28....the answer is 28 + 1 = 29( +1 for counting one of the left out multiples within the range ).,..

With the above method, what I am basically doing is finding out how many numbers lie in the range and then divide it by 7..and then add 1 to get the answer...
Similarly, when I am using this in the given question on 5 digits, why am I not able to get eh correct answer ?
What I understand is tht in both these questions we are trying to find out all those numbers that are divisible by a particular number...

Please advise..

RJ
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Joined: 30 Sep 2010
Posts: 55
Re: A FIVE DIGIT NUMBER DIVIBLE BY 3  [#permalink]

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New post 27 Oct 2010, 19:09
RJ,

your approach works when the numbers are in sequence..
like: 1,2,3...100 >> you will have 33 numbers divisible by 3

but in this problem, there is no gurantee that the numbers will be sequence.. so your approach didnt work.
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Re: A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE  [#permalink]

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