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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE 08 Oct 2010, 04:53
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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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C
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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE 08 Oct 2010, 06:20
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anilnandyala
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
Show more

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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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE 08 Oct 2010, 06:43
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Quite a good question will multiple application of math's rule.Thanks Bunnel for explanation.
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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE 09 Oct 2010, 11:42
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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.
Show more

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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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE 27 Oct 2010, 06:31
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anilnandyala
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.
Show more


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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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE 27 Oct 2010, 20:09
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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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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE 23 Jul 2019, 07:40
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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
Show more

At first I thought on the same lines..without properly understanding the question..
bt first we had go through the question and analyse.
the question is abt the numbers divisible by 3.

So quickly reverted and gone to basics..
Divisibility rule by 3. Sum has to be divisible by 3.
Quckly checking over the numbers given..
1 2 3 4 5.. sum is 15... divisible..
then again checking for any other combinations..
remocing three abnd adding 0... divisible by 3..

So calculating permutations for the first group- 5!=120
for second group, with a bit caution as zero shd not be there in the first place(As it would become four digit number)
4*4*3*2=96
total 120+96=216
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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE 23 Jul 2019, 08:08
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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
Show more

Digits to be used = { 0, 1, 2, 3, 4, 5}

Divisibility test of 3: The sum of the digits should be divisible by 3

Sum of the available digits = 0+1+2+3+4+5 = 15

But only 5 of these 6 digits have to be used at any point of time for forming a five digit number

So the digit that can be left unused are 0 as one case and 5 as other case

Case 1: Digits used are 1, 2, 3, 4, 5 : Total 5 digit numbers using these digits = 5*4*3*2*1 = 5! = 120

Case 2: Digits used are 0, 1, 2, 3, 4 : Total 5 digit numbers using these digits = 4*4*3*2*1 = 96

Total 5 digit numbers = 120+96 = 216

Answer: Option C

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A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE 27 Aug 2025, 14:55
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