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# How many five-digit numbers can be formed using digits

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Re: digit counting [#permalink]

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08 Jun 2014, 08:12
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Bunuel wrote:
TheRob wrote:
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?

A. 15
B. 96
C. 120
D. 181
E. 216

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

I tried to do as follows:
take all 5 digit numbers possible : 5 *5*4*3*2
divide by 3 to get all numbers divisible by 3.

What is wrong with this logic?
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Re: digit counting [#permalink]

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08 Jun 2014, 10:46
ronr34 wrote:
Bunuel wrote:
TheRob wrote:
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?

A. 15
B. 96
C. 120
D. 181
E. 216

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

I tried to do as follows:
take all 5 digit numbers possible : 5 *5*4*3*2
divide by 3 to get all numbers divisible by 3.

What is wrong with this logic?

Because the numbers divisible by 3 are not 1/3rd of all possible numbers.

{0, 1, 2, 3, 4} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 3, 5} --> 96 5-digit numbers possible with this set.

{0, 1, 2, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 1, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 2, 3, 4, 5} --> 96 5-digit numbers possible with this set.

{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set.

Total = 5*5*4*3*2 = 600 but the numbers which are divisible by 3 come from third and sixth sets: 96 + 120 = 216.
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Re: digit counting [#permalink]

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16 Jun 2014, 13:52
Bunuel wrote:
ronr34 wrote:
I tried to do as follows:
take all 5 digit numbers possible : 5 *5*4*3*2
divide by 3 to get all numbers divisible by 3.

What is wrong with this logic?

Because the numbers divisible by 3 are not 1/3rd of all possible numbers.

{0, 1, 2, 3, 4} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 3, 5} --> 96 5-digit numbers possible with this set.

{0, 1, 2, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 1, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 2, 3, 4, 5} --> 96 5-digit numbers possible with this set.

{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set.

Total = 5*5*4*3*2 = 600 but the numbers which are divisible by 3 come from third and sixth sets: 96 + 120 = 216.

How did you see this?
If 600 is all options possible, than why isn't it logical to think that 1/3 of the numbers will be multiples of 3?
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Re: digit counting [#permalink]

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16 Jun 2014, 22:11
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Expert's post
ronr34 wrote:
I tried to do as follows:
take all 5 digit numbers possible : 5 *5*4*3*2
divide by 3 to get all numbers divisible by 3.

What is wrong with this logic?

We cannot do this because we have the asymmetric 0 as one of the digits. The number of 5 digit numbers that can be formed with 0, 1, 2, 3 and 4 is different from the number of 5 digit numbers that can be formed with 1, 2, 3, 4 and 5 (because 0 cannot be the first digit).

Had the digits been 1, 2, 3, 4, 5 and 6, then your method would have been correct.

If 0 is included:
{0, 1, 2, 3, 4} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 3, 5} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 4, 5} --> 96 5-digit numbers possible with this set. - All these numbers are divisible by 3
{0, 1, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 2, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
The number of 5 digit numbers in these sets is not the same - Sets with 0 have fewer numbers

If 0 is not included:
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set.
{1, 2, 3, 4, 6} --> 120 5-digit numbers possible with this set.
{1, 2, 3, 5, 6} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
{1, 2, 4, 5, 6} --> 120 5-digit numbers possible with this set.
{1, 3, 4, 5, 6} --> 120 5-digit numbers possible with this set.
{2, 3, 4, 5, 6} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
Here exactly 1/3rd of the numbers will be divisible by 3.
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Re: How many five-digit numbers can be formed using digits [#permalink]

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25 Jun 2015, 11:37
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Re: How many five-digit numbers can be formed using digits [#permalink]

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17 Sep 2015, 08:42
TheRob wrote:
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?

A. 15
B. 96
C. 120
D. 181
E. 216

Key Knowledge:
(Divisibility Rules, Number Theory) math-number-theory-88376.html
For a number to be divisible by 3, the sum of it's digits must be a multiple of 3

Case Method:

Case#1:
$$\sum 0,1,2,4,5 = 12$$

$$5!=120$$ = Number of arrangements of 5 digits
Less:
$$4!=24$$ = Number of arrangements of 4 digits
(Removes arrangements where 0 is the leading digit (i.e. fix 0 as leading, arrange the remaining 4 digits 4! ways))

$$5!-4!=96$$

Case #2:
$$\sum 1,2,3,4,5 = 15$$

$$5!=120$$ = Number of arrangements of 5 digits

$$120+96=216$$
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Re: How many five-digit numbers can be formed using digits [#permalink]

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22 Dec 2015, 00:44
Awesome question to practise.
Re: How many five-digit numbers can be formed using digits   [#permalink] 22 Dec 2015, 00:44

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# How many five-digit numbers can be formed using digits

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