Prove ${0 bmod 2, 0 bmod 3, 1 bmod 4, 1 bmod 6, 11 bmod 12}$is a covering system












0














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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    0














    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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      0












      0








      0







      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







      share|cite|improve this question















      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






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      edited Nov 28 '18 at 4:53









      Alex Vong

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      asked Nov 28 '18 at 2:59









      Jingting931015

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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 $1bmod 6$.






          share|cite|improve this answer























          • 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 '18 at 3:20










          • @Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
            – Valborg
            Nov 28 '18 at 3:28










          • Oops, sorry. Thanks for making the correction
            – Jingting931015
            Nov 28 '18 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 '18 at 3:39



















          1














          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.






          share|cite|improve this answer





















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            2 Answers
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            2 Answers
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            active

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            active

            oldest

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            2














            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 $1bmod 6$.






            share|cite|improve this answer























            • 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 '18 at 3:20










            • @Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
              – Valborg
              Nov 28 '18 at 3:28










            • Oops, sorry. Thanks for making the correction
              – Jingting931015
              Nov 28 '18 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 '18 at 3:39
















            2














            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 $1bmod 6$.






            share|cite|improve this answer























            • 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 '18 at 3:20










            • @Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
              – Valborg
              Nov 28 '18 at 3:28










            • Oops, sorry. Thanks for making the correction
              – Jingting931015
              Nov 28 '18 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 '18 at 3:39














            2












            2








            2






            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 $1bmod 6$.






            share|cite|improve this answer














            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 $1bmod 6$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 28 '18 at 3:26

























            answered Nov 28 '18 at 3:07









            Valborg

            542




            542












            • 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 '18 at 3:20










            • @Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
              – Valborg
              Nov 28 '18 at 3:28










            • Oops, sorry. Thanks for making the correction
              – Jingting931015
              Nov 28 '18 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 '18 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 '18 at 3:20










            • @Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
              – Valborg
              Nov 28 '18 at 3:28










            • Oops, sorry. Thanks for making the correction
              – Jingting931015
              Nov 28 '18 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 '18 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 '18 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 '18 at 3:20












            @Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
            – Valborg
            Nov 28 '18 at 3:28




            @Jingting931015 Ah, you have a typo in your post. You lost the $1bmod 6$ class.
            – Valborg
            Nov 28 '18 at 3:28












            Oops, sorry. Thanks for making the correction
            – Jingting931015
            Nov 28 '18 at 3:31




            Oops, sorry. Thanks for making the correction
            – Jingting931015
            Nov 28 '18 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 '18 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 '18 at 3:39











            1














            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.






            share|cite|improve this answer


























              1














              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.






              share|cite|improve this answer
























                1












                1








                1






                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.






                share|cite|improve this answer












                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 28 '18 at 3:33









                fleablood

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