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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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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
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 |Â
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
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
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
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edited 1 hour ago
asked 1 hour ago
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asked 1 hour ago
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334
asked 1 hour ago
Java Newbie
334
334
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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.
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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 |Â
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 |Â
2 Answers
2
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.
answered 1 hour ago
Mayuresh L
778216
add a comment |Â
up vote
4
down vote
There are two mistakes:
$0$ 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)- There is no point in using permutations.
answered 1 hour ago
Jaroslaw Matlak
4,077830
add a comment |Â
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.
answered 1 hour ago
Mayuresh L
778216
add a comment |Â
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.
answered 1 hour ago
Mayuresh L
778216
add a comment |Â
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.
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.
answered 1 hour ago
Mayuresh L
778216
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 |Â
add a comment |Â
up vote
4
down vote
There are two mistakes:
$0$ 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)- There is no point in using permutations.
answered 1 hour ago
Jaroslaw Matlak
4,077830
add a comment |Â
up vote
4
down vote
There are two mistakes:
$0$ 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)- There is no point in using permutations.
answered 1 hour ago
Jaroslaw Matlak
4,077830
add a comment |Â
up vote
4
down vote
up vote
4
down vote
There are two mistakes:
$0$ 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)- There is no point in using permutations.
answered 1 hour ago
Jaroslaw Matlak
4,077830
There are two mistakes:
$0$ 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)- There is no point in using permutations.
answered 1 hour ago
Jaroslaw Matlak
4,077830
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 |Â
add a comment |Â
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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