Prove ${0 bmod 2, 0 bmod 3, 1 bmod 4, 1 bmod 6, 11 bmod 12}$is a covering system
I'm trying to figure out why ${0 bmod 2, 0 bmod 3, 1 bmod 4, 1 bmod 6, 11 bmod 12}$ is a covering system. Is there a neat way to prove it?
modular-arithmetic
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I'm trying to figure out why ${0 bmod 2, 0 bmod 3, 1 bmod 4, 1 bmod 6, 11 bmod 12}$ is a covering system. Is there a neat way to prove it?
modular-arithmetic
add a comment |
I'm trying to figure out why ${0 bmod 2, 0 bmod 3, 1 bmod 4, 1 bmod 6, 11 bmod 12}$ is a covering system. Is there a neat way to prove it?
modular-arithmetic
I'm trying to figure out why ${0 bmod 2, 0 bmod 3, 1 bmod 4, 1 bmod 6, 11 bmod 12}$ is a covering system. Is there a neat way to prove it?
modular-arithmetic
modular-arithmetic
edited Nov 28 at 4:53
Alex Vong
1,229819
1,229819
asked Nov 28 at 2:59
Jingting931015
828
828
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Write out what all of the equivalence classes are modulo the LCM of the moduli. In this case, $0bmod 2$ is the same thing as $0,2,4,6,8$, and $10bmod 12$, $0bmod3$ is the same thing as $0,3,6$, and $9bmod 12$, $1bmod4$ is the same thing as $1,5$, and $9bmod12$. Add to that $11bmod12$ and...
Unfortunately, you seem to be missing $7bmod12$, so I don't think this is actually a covering system?
This appears to have been found in a lovely little paper, but was reproduced here with a typo! You want to include $1bmod 6$.
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 at 3:39
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If $n equiv 0,2,4,6,8,10 pmod {12}$ then $n equiv 0 pmod 2$ and is covered.
If $nequiv 1,5,9 pmod {12}$ then $nequiv 1pmod {4}$ and $n$ is covered.
If $n equiv 0, 3, 6, 9 pmod{12}$ then $nequiv 0 pmod {3}$ and $n$ is covered.
If $n equiv 1,7 pmod {12}$ then $n equiv 1 pmod {6}$ and $n$ is covered.
If $n equiv 11 pmod{12}$ then it is covered.
So $n$ is covered if $n equiv 0,1,2,3,4,5,6,7,8,9,10,11 pmod {12}$. And all natural numbers fall into one of those categories. So all natural numbers are covered.
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2 Answers
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2 Answers
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Write out what all of the equivalence classes are modulo the LCM of the moduli. In this case, $0bmod 2$ is the same thing as $0,2,4,6,8$, and $10bmod 12$, $0bmod3$ is the same thing as $0,3,6$, and $9bmod 12$, $1bmod4$ is the same thing as $1,5$, and $9bmod12$. Add to that $11bmod12$ and...
Unfortunately, you seem to be missing $7bmod12$, so I don't think this is actually a covering system?
This appears to have been found in a lovely little paper, but was reproduced here with a typo! You want to include $1bmod 6$.
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 at 3:39
add a comment |
Write out what all of the equivalence classes are modulo the LCM of the moduli. In this case, $0bmod 2$ is the same thing as $0,2,4,6,8$, and $10bmod 12$, $0bmod3$ is the same thing as $0,3,6$, and $9bmod 12$, $1bmod4$ is the same thing as $1,5$, and $9bmod12$. Add to that $11bmod12$ and...
Unfortunately, you seem to be missing $7bmod12$, so I don't think this is actually a covering system?
This appears to have been found in a lovely little paper, but was reproduced here with a typo! You want to include $1bmod 6$.
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 at 3:39
add a comment |
Write out what all of the equivalence classes are modulo the LCM of the moduli. In this case, $0bmod 2$ is the same thing as $0,2,4,6,8$, and $10bmod 12$, $0bmod3$ is the same thing as $0,3,6$, and $9bmod 12$, $1bmod4$ is the same thing as $1,5$, and $9bmod12$. Add to that $11bmod12$ and...
Unfortunately, you seem to be missing $7bmod12$, so I don't think this is actually a covering system?
This appears to have been found in a lovely little paper, but was reproduced here with a typo! You want to include $1bmod 6$.
Write out what all of the equivalence classes are modulo the LCM of the moduli. In this case, $0bmod 2$ is the same thing as $0,2,4,6,8$, and $10bmod 12$, $0bmod3$ is the same thing as $0,3,6$, and $9bmod 12$, $1bmod4$ is the same thing as $1,5$, and $9bmod12$. Add to that $11bmod12$ and...
Unfortunately, you seem to be missing $7bmod12$, so I don't think this is actually a covering system?
This appears to have been found in a lovely little paper, but was reproduced here with a typo! You want to include $1bmod 6$.
edited Nov 28 at 3:26
answered Nov 28 at 3:07
Valborg
792
792
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 at 3:39
add a comment |
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 at 3:39
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 at 3:20
I came across this system of congruence in an article and it does not include 7 mod 12. The article is here: arxiv.org/pdf/1705.04372.pdf
– Jingting931015
Nov 28 at 3:20
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 at 3:28
@Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
– Valborg
Nov 28 at 3:28
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 at 3:31
Oops, sorry. Thanks for making the correction
– Jingting931015
Nov 28 at 3:31
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 at 3:39
In your article: "Each of the moduli in C divides 12, so we can check that C covers the integers by verifying that each residue class modulo 12 is a subset of one of the residue c lasses in C" That's how you verify it. In my answer I did that explicitely.
– fleablood
Nov 28 at 3:39
add a comment |
If $n equiv 0,2,4,6,8,10 pmod {12}$ then $n equiv 0 pmod 2$ and is covered.
If $nequiv 1,5,9 pmod {12}$ then $nequiv 1pmod {4}$ and $n$ is covered.
If $n equiv 0, 3, 6, 9 pmod{12}$ then $nequiv 0 pmod {3}$ and $n$ is covered.
If $n equiv 1,7 pmod {12}$ then $n equiv 1 pmod {6}$ and $n$ is covered.
If $n equiv 11 pmod{12}$ then it is covered.
So $n$ is covered if $n equiv 0,1,2,3,4,5,6,7,8,9,10,11 pmod {12}$. And all natural numbers fall into one of those categories. So all natural numbers are covered.
add a comment |
If $n equiv 0,2,4,6,8,10 pmod {12}$ then $n equiv 0 pmod 2$ and is covered.
If $nequiv 1,5,9 pmod {12}$ then $nequiv 1pmod {4}$ and $n$ is covered.
If $n equiv 0, 3, 6, 9 pmod{12}$ then $nequiv 0 pmod {3}$ and $n$ is covered.
If $n equiv 1,7 pmod {12}$ then $n equiv 1 pmod {6}$ and $n$ is covered.
If $n equiv 11 pmod{12}$ then it is covered.
So $n$ is covered if $n equiv 0,1,2,3,4,5,6,7,8,9,10,11 pmod {12}$. And all natural numbers fall into one of those categories. So all natural numbers are covered.
add a comment |
If $n equiv 0,2,4,6,8,10 pmod {12}$ then $n equiv 0 pmod 2$ and is covered.
If $nequiv 1,5,9 pmod {12}$ then $nequiv 1pmod {4}$ and $n$ is covered.
If $n equiv 0, 3, 6, 9 pmod{12}$ then $nequiv 0 pmod {3}$ and $n$ is covered.
If $n equiv 1,7 pmod {12}$ then $n equiv 1 pmod {6}$ and $n$ is covered.
If $n equiv 11 pmod{12}$ then it is covered.
So $n$ is covered if $n equiv 0,1,2,3,4,5,6,7,8,9,10,11 pmod {12}$. And all natural numbers fall into one of those categories. So all natural numbers are covered.
If $n equiv 0,2,4,6,8,10 pmod {12}$ then $n equiv 0 pmod 2$ and is covered.
If $nequiv 1,5,9 pmod {12}$ then $nequiv 1pmod {4}$ and $n$ is covered.
If $n equiv 0, 3, 6, 9 pmod{12}$ then $nequiv 0 pmod {3}$ and $n$ is covered.
If $n equiv 1,7 pmod {12}$ then $n equiv 1 pmod {6}$ and $n$ is covered.
If $n equiv 11 pmod{12}$ then it is covered.
So $n$ is covered if $n equiv 0,1,2,3,4,5,6,7,8,9,10,11 pmod {12}$. And all natural numbers fall into one of those categories. So all natural numbers are covered.
answered Nov 28 at 3:33
fleablood
68.1k22684
68.1k22684
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