prove: if $abmid cd$ and $amid c$ and $abnmid c$ then $bmid d$ [closed]Find $a$, $b$, and $c$ so $a mid bc ~$ but $a nmid b space $ and $a nmid c$Proving by absurd that $d nmid 4,5,10, 20$Divisibility: if $a mid b$ and $b mid c$, then $a mid (b+c)$Prove that $2k+5 mid kin mathbbZ = 2k+3 mid kin mathbbZ$.Show $nexists k:3^7mid k!$ but $3^8nmid k!$Prove if (a,p)=1, then a,2a,3a,…,pa is a complete residue system modulo p.if $5nmid a$ or $5nmid b$, then $5nmid a^2-2b^2$.Prove that if $amid(b-1)$ and $a^2mid(b^2-2b+4)$, then $amid12$.If $12mida$ then $4nmidb$ or $4mid(8+2a+5b)$.

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prove: if $abmid cd$ and $amid c$ and $abnmid c$ then $bmid d$ [closed]


Find $a$, $b$, and $c$ so $a mid bc ~$ but $a nmid b space $ and $a nmid c$Proving by absurd that $d nmid 4,5,10, 20$Divisibility: if $a mid b$ and $b mid c$, then $a mid (b+c)$Prove that $2k+5 mid kin mathbbZ = 2k+3 mid kin mathbbZ$.Show $nexists k:3^7mid k!$ but $3^8nmid k!$Prove if (a,p)=1, then a,2a,3a,…,pa is a complete residue system modulo p.if $5nmid a$ or $5nmid b$, then $5nmid a^2-2b^2$.Prove that if $amid(b-1)$ and $a^2mid(b^2-2b+4)$, then $amid12$.If $12mida$ then $4nmidb$ or $4mid(8+2a+5b)$.













-2












$begingroup$


I'm having a hard time proving the following claim:



if $abmid cd$ and $amid c$ and $abnmid c$ then $bmid d$



Any help would be appreciated










share|cite|improve this question









New contributor




user5842084 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$



closed as off-topic by Eevee Trainer, Clayton, Vinyl_cape_jawa, Shaun, RRL Mar 12 at 22:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Clayton, Vinyl_cape_jawa, Shaun, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 12 at 20:26










  • $begingroup$
    $abnmid c,$ doesn't suffice, we need $(b,c/a)=1$ or equivalently $(ab,c)=a $
    $endgroup$
    – Bill Dubuque
    Mar 12 at 21:07
















-2












$begingroup$


I'm having a hard time proving the following claim:



if $abmid cd$ and $amid c$ and $abnmid c$ then $bmid d$



Any help would be appreciated










share|cite|improve this question









New contributor




user5842084 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$



closed as off-topic by Eevee Trainer, Clayton, Vinyl_cape_jawa, Shaun, RRL Mar 12 at 22:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Clayton, Vinyl_cape_jawa, Shaun, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 12 at 20:26










  • $begingroup$
    $abnmid c,$ doesn't suffice, we need $(b,c/a)=1$ or equivalently $(ab,c)=a $
    $endgroup$
    – Bill Dubuque
    Mar 12 at 21:07














-2












-2








-2





$begingroup$


I'm having a hard time proving the following claim:



if $abmid cd$ and $amid c$ and $abnmid c$ then $bmid d$



Any help would be appreciated










share|cite|improve this question









New contributor




user5842084 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I'm having a hard time proving the following claim:



if $abmid cd$ and $amid c$ and $abnmid c$ then $bmid d$



Any help would be appreciated







elementary-number-theory divisibility






share|cite|improve this question









New contributor




user5842084 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




user5842084 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Mar 12 at 20:19









Jennifer

8,60521837




8,60521837






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user5842084 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Mar 12 at 20:17









user5842084user5842084

1




1




New contributor




user5842084 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





user5842084 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user5842084 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




closed as off-topic by Eevee Trainer, Clayton, Vinyl_cape_jawa, Shaun, RRL Mar 12 at 22:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Clayton, Vinyl_cape_jawa, Shaun, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Eevee Trainer, Clayton, Vinyl_cape_jawa, Shaun, RRL Mar 12 at 22:52


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Clayton, Vinyl_cape_jawa, Shaun, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.











  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 12 at 20:26










  • $begingroup$
    $abnmid c,$ doesn't suffice, we need $(b,c/a)=1$ or equivalently $(ab,c)=a $
    $endgroup$
    – Bill Dubuque
    Mar 12 at 21:07

















  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 12 at 20:26










  • $begingroup$
    $abnmid c,$ doesn't suffice, we need $(b,c/a)=1$ or equivalently $(ab,c)=a $
    $endgroup$
    – Bill Dubuque
    Mar 12 at 21:07
















$begingroup$
What have you tried so far?
$endgroup$
– Vinyl_cape_jawa
Mar 12 at 20:26




$begingroup$
What have you tried so far?
$endgroup$
– Vinyl_cape_jawa
Mar 12 at 20:26












$begingroup$
$abnmid c,$ doesn't suffice, we need $(b,c/a)=1$ or equivalently $(ab,c)=a $
$endgroup$
– Bill Dubuque
Mar 12 at 21:07





$begingroup$
$abnmid c,$ doesn't suffice, we need $(b,c/a)=1$ or equivalently $(ab,c)=a $
$endgroup$
– Bill Dubuque
Mar 12 at 21:07











1 Answer
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$begingroup$

Counterexample: $a=1,,b=4,,c=d=2$.






share|cite|improve this answer









$endgroup$



















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Counterexample: $a=1,,b=4,,c=d=2$.






    share|cite|improve this answer









    $endgroup$

















      3












      $begingroup$

      Counterexample: $a=1,,b=4,,c=d=2$.






      share|cite|improve this answer









      $endgroup$















        3












        3








        3





        $begingroup$

        Counterexample: $a=1,,b=4,,c=d=2$.






        share|cite|improve this answer









        $endgroup$



        Counterexample: $a=1,,b=4,,c=d=2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 12 at 20:30









        J.G.J.G.

        30.5k23149




        30.5k23149













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