How many 5 digit numbers are there, whose i digit is divisible by i?

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For example in the number 34567, the second digit (4) can be divided by 2, but the third digit (5) can't be divided by 3.

I thought that the first digit can be divided by any number as 1,2,3,4,5,6,7,8,9 are all divisible by 1.

The second digit can only be divided by 2, which is 2,4,6,8.

The third digit can only be divided by 3, which is 3,6,9.

The fourth digit can only be divided by 4, which is 4 or 8.

The fifth digit can only be divided by 5.

combinatorics

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• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since there are only 90,000 5 digit numbers, your answers will be somewhat too high. For the first digit you have 9 choices, for the second you have 5 (zero is divisible by 2), for the third you have 4, for the fourth you have 3 and for the fifth you have 2 choices. This makes a total of 9*5*4*3*2=1080.
  – Ronald
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does 0 count as being devisible by a digit?
  – Alucard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You don't need the factorial the first digit can have 9 values not 9!... moreover other digits could be 0
  – ALG
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Alucard: $0$ is divisible by everything -- or at least by everything nonzero. Whenever $nne 0$ we have that $0/n=0$ which is an integer. (Number theory and abstract algebra actually favor a slightly different definition according to which $0$ also counts as "divisible by" itself even though $0/0$ is undefined -- but that is a bit subtler).
  – Henning Makholm
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @javaNewbie If you want to ask additional question, do it in another topic.
  – Jaroslaw Matlak
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

For example in the number 34567, the second digit (4) can be divided by 2, but the third digit (5) can't be divided by 3.

I thought that the first digit can be divided by any number as 1,2,3,4,5,6,7,8,9 are all divisible by 1.

The second digit can only be divided by 2, which is 2,4,6,8.

The third digit can only be divided by 3, which is 3,6,9.

The fourth digit can only be divided by 4, which is 4 or 8.

The fifth digit can only be divided by 5.

combinatorics

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edited 1 hour ago

asked 1 hour ago

Java Newbie

334

New contributor

Java Newbie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since there are only 90,000 5 digit numbers, your answers will be somewhat too high. For the first digit you have 9 choices, for the second you have 5 (zero is divisible by 2), for the third you have 4, for the fourth you have 3 and for the fifth you have 2 choices. This makes a total of 9*5*4*3*2=1080.
  – Ronald
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does 0 count as being devisible by a digit?
  – Alucard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You don't need the factorial the first digit can have 9 values not 9!... moreover other digits could be 0
  – ALG
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Alucard: $0$ is divisible by everything -- or at least by everything nonzero. Whenever $nne 0$ we have that $0/n=0$ which is an integer. (Number theory and abstract algebra actually favor a slightly different definition according to which $0$ also counts as "divisible by" itself even though $0/0$ is undefined -- but that is a bit subtler).
  – Henning Makholm
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @javaNewbie If you want to ask additional question, do it in another topic.
  – Jaroslaw Matlak
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

up vote
4
down vote

favorite

For example in the number 34567, the second digit (4) can be divided by 2, but the third digit (5) can't be divided by 3.

I thought that the first digit can be divided by any number as 1,2,3,4,5,6,7,8,9 are all divisible by 1.

The second digit can only be divided by 2, which is 2,4,6,8.

The third digit can only be divided by 3, which is 3,6,9.

The fourth digit can only be divided by 4, which is 4 or 8.

The fifth digit can only be divided by 5.

combinatorics

share|cite|improve this question

edited 1 hour ago

asked 1 hour ago

Java Newbie

334

New contributor

Java Newbie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

For example in the number 34567, the second digit (4) can be divided by 2, but the third digit (5) can't be divided by 3.

I thought that the first digit can be divided by any number as 1,2,3,4,5,6,7,8,9 are all divisible by 1.

The second digit can only be divided by 2, which is 2,4,6,8.

The third digit can only be divided by 3, which is 3,6,9.

The fourth digit can only be divided by 4, which is 4 or 8.

The fifth digit can only be divided by 5.

combinatorics

combinatorics

share|cite|improve this question

edited 1 hour ago

asked 1 hour ago

Java Newbie

334

New contributor

Java Newbie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question

edited 1 hour ago

asked 1 hour ago

Java Newbie

334

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share|cite|improve this question

share|cite|improve this question

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

asked 1 hour ago

Java Newbie

334

New contributor

Java Newbie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Java Newbie

334

asked 1 hour ago

Java Newbie

334

334

New contributor

Java Newbie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Java Newbie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Java Newbie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since there are only 90,000 5 digit numbers, your answers will be somewhat too high. For the first digit you have 9 choices, for the second you have 5 (zero is divisible by 2), for the third you have 4, for the fourth you have 3 and for the fifth you have 2 choices. This makes a total of 9*5*4*3*2=1080.
  – Ronald
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does 0 count as being devisible by a digit?
  – Alucard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You don't need the factorial the first digit can have 9 values not 9!... moreover other digits could be 0
  – ALG
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Alucard: $0$ is divisible by everything -- or at least by everything nonzero. Whenever $nne 0$ we have that $0/n=0$ which is an integer. (Number theory and abstract algebra actually favor a slightly different definition according to which $0$ also counts as "divisible by" itself even though $0/0$ is undefined -- but that is a bit subtler).
  – Henning Makholm
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @javaNewbie If you want to ask additional question, do it in another topic.
  – Jaroslaw Matlak
  1 hour ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since there are only 90,000 5 digit numbers, your answers will be somewhat too high. For the first digit you have 9 choices, for the second you have 5 (zero is divisible by 2), for the third you have 4, for the fourth you have 3 and for the fifth you have 2 choices. This makes a total of 9*5*4*3*2=1080.
  – Ronald
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Does 0 count as being devisible by a digit?
  – Alucard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You don't need the factorial the first digit can have 9 values not 9!... moreover other digits could be 0
  – ALG
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Alucard: $0$ is divisible by everything -- or at least by everything nonzero. Whenever $nne 0$ we have that $0/n=0$ which is an integer. (Number theory and abstract algebra actually favor a slightly different definition according to which $0$ also counts as "divisible by" itself even though $0/0$ is undefined -- but that is a bit subtler).
  – Henning Makholm
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @javaNewbie If you want to ask additional question, do it in another topic.
  – Jaroslaw Matlak
  1 hour ago
  

  
  
  
  

1

1

Since there are only 90,000 5 digit numbers, your answers will be somewhat too high. For the first digit you have 9 choices, for the second you have 5 (zero is divisible by 2), for the third you have 4, for the fourth you have 3 and for the fifth you have 2 choices. This makes a total of 9*5*4*3*2=1080.
– Ronald
1 hour ago

Since there are only 90,000 5 digit numbers, your answers will be somewhat too high. For the first digit you have 9 choices, for the second you have 5 (zero is divisible by 2), for the third you have 4, for the fourth you have 3 and for the fifth you have 2 choices. This makes a total of 9*5*4*3*2=1080.
– Ronald
1 hour ago

Does 0 count as being devisible by a digit?
– Alucard
1 hour ago

Does 0 count as being devisible by a digit?
– Alucard
1 hour ago

You don't need the factorial the first digit can have 9 values not 9!... moreover other digits could be 0
– ALG
1 hour ago

You don't need the factorial the first digit can have 9 values not 9!... moreover other digits could be 0
– ALG
1 hour ago

1

1

@Alucard: $0$ is divisible by everything -- or at least by everything nonzero. Whenever $nne 0$ we have that $0/n=0$ which is an integer. (Number theory and abstract algebra actually favor a slightly different definition according to which $0$ also counts as "divisible by" itself even though $0/0$ is undefined -- but that is a bit subtler).
– Henning Makholm
1 hour ago

@Alucard: $0$ is divisible by everything -- or at least by everything nonzero. Whenever $nne 0$ we have that $0/n=0$ which is an integer. (Number theory and abstract algebra actually favor a slightly different definition according to which $0$ also counts as "divisible by" itself even though $0/0$ is undefined -- but that is a bit subtler).
– Henning Makholm
1 hour ago

@javaNewbie If you want to ask additional question, do it in another topic.
– Jaroslaw Matlak
1 hour ago

@javaNewbie If you want to ask additional question, do it in another topic.
– Jaroslaw Matlak
1 hour ago

add a comment | 

2 Answers
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up vote
1
down vote



accepted

For the first position we have 9 choices 1,2,3,.....,9 (we cannot include 0 as we want 5 digit number).

For the second postion we have 5 choices 0,2,4,6,8 (here we don’t have problem in including 0).

Similarly for third, forth and fifth position we have 4, 3 and 2 choices respectively.

Now by multiplication rule in combinatorics (no need to take factorial)

We have total 9*5*4*3*2=1080 such numbers.

Answer to the second question is correct.

share|cite|improve this answer

answered 1 hour ago

Mayuresh L

778216

add a comment | 

up vote
4
down vote

There are two mistakes:

1. 
   $0$ is divisible by each number, therefore it can be selected on every position (except the first one, because then we wouldn't get 5-digit number)


2. There is no point in using permutations.

share|cite|improve this answer

answered 1 hour ago

Jaroslaw Matlak

4,077830

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

For the first position we have 9 choices 1,2,3,.....,9 (we cannot include 0 as we want 5 digit number).

For the second postion we have 5 choices 0,2,4,6,8 (here we don’t have problem in including 0).

Similarly for third, forth and fifth position we have 4, 3 and 2 choices respectively.

Now by multiplication rule in combinatorics (no need to take factorial)

We have total 9*5*4*3*2=1080 such numbers.

Answer to the second question is correct.

share|cite|improve this answer

answered 1 hour ago

Mayuresh L

778216

add a comment | 

up vote
1
down vote



accepted

For the first position we have 9 choices 1,2,3,.....,9 (we cannot include 0 as we want 5 digit number).

For the second postion we have 5 choices 0,2,4,6,8 (here we don’t have problem in including 0).

Similarly for third, forth and fifth position we have 4, 3 and 2 choices respectively.

Now by multiplication rule in combinatorics (no need to take factorial)

We have total 9*5*4*3*2=1080 such numbers.

Answer to the second question is correct.

share|cite|improve this answer

answered 1 hour ago

Mayuresh L

778216

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

For the first position we have 9 choices 1,2,3,.....,9 (we cannot include 0 as we want 5 digit number).

For the second postion we have 5 choices 0,2,4,6,8 (here we don’t have problem in including 0).

Similarly for third, forth and fifth position we have 4, 3 and 2 choices respectively.

Now by multiplication rule in combinatorics (no need to take factorial)

We have total 9*5*4*3*2=1080 such numbers.

Answer to the second question is correct.

share|cite|improve this answer

answered 1 hour ago

Mayuresh L

778216

For the first position we have 9 choices 1,2,3,.....,9 (we cannot include 0 as we want 5 digit number).

For the second postion we have 5 choices 0,2,4,6,8 (here we don’t have problem in including 0).

Similarly for third, forth and fifth position we have 4, 3 and 2 choices respectively.

Now by multiplication rule in combinatorics (no need to take factorial)

We have total 9*5*4*3*2=1080 such numbers.

Answer to the second question is correct.

share|cite|improve this answer

answered 1 hour ago

Mayuresh L

778216

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

Mayuresh L

778216

answered 1 hour ago

Mayuresh L

778216

answered 1 hour ago

Mayuresh L

778216

778216

add a comment | 

add a comment | 

up vote
4
down vote

There are two mistakes:

1. 
   $0$ is divisible by each number, therefore it can be selected on every position (except the first one, because then we wouldn't get 5-digit number)


2. There is no point in using permutations.

share|cite|improve this answer

answered 1 hour ago

Jaroslaw Matlak

4,077830

add a comment | 

up vote
4
down vote

There are two mistakes:

1. 
   $0$ is divisible by each number, therefore it can be selected on every position (except the first one, because then we wouldn't get 5-digit number)


2. There is no point in using permutations.

share|cite|improve this answer

answered 1 hour ago

Jaroslaw Matlak

4,077830

add a comment | 

up vote
4
down vote

up vote
4
down vote

There are two mistakes:

1. 
   $0$ is divisible by each number, therefore it can be selected on every position (except the first one, because then we wouldn't get 5-digit number)


2. There is no point in using permutations.

share|cite|improve this answer

answered 1 hour ago

Jaroslaw Matlak

4,077830

There are two mistakes:

1. 
   $0$ is divisible by each number, therefore it can be selected on every position (except the first one, because then we wouldn't get 5-digit number)


2. There is no point in using permutations.

share|cite|improve this answer

answered 1 hour ago

Jaroslaw Matlak

4,077830

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

Jaroslaw Matlak

4,077830

answered 1 hour ago

Jaroslaw Matlak

4,077830

answered 1 hour ago

Jaroslaw Matlak

4,077830

4,077830

add a comment | 

add a comment | 

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