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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

elementary-number-theory divisibility

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edited Mar 12 at 20:19

Jennifer

8,60521837

asked Mar 12 at 20:17

user5842084user5842084

1

New contributor

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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
  
  

  
  
  
  

add a comment | 

-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

elementary-number-theory divisibility

share|cite|improve this question

edited Mar 12 at 20:19

Jennifer

8,60521837

asked Mar 12 at 20:17

user5842084user5842084

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.

$endgroup$

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
  
  

  
  
  
  

add a comment | 

$begingroup$

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

elementary-number-theory divisibility

share|cite|improve this question

edited Mar 12 at 20:19

Jennifer

8,60521837

asked Mar 12 at 20:17

user5842084user5842084

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.

$endgroup$

share|cite|improve this question

edited Mar 12 at 20:19

Jennifer

8,60521837

asked Mar 12 at 20:17

user5842084user5842084

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.

share|cite|improve this question

edited Mar 12 at 20:19

Jennifer

8,60521837

asked Mar 12 at 20:17

user5842084user5842084

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.

edited Mar 12 at 20:19

Jennifer

8,60521837

edited Mar 12 at 20:19

Jennifer

8,60521837

asked Mar 12 at 20:17

user5842084user5842084

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.

asked Mar 12 at 20:17

user5842084user5842084

1

1 Answer
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3

$begingroup$

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

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answered Mar 12 at 20:30

J.G.J.G.

30.5k23149

$endgroup$

add a comment | 

3

$begingroup$

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

share|cite|improve this answer

answered Mar 12 at 20:30

J.G.J.G.

30.5k23149

$endgroup$

add a comment | 

3

$begingroup$

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

share|cite|improve this answer

answered Mar 12 at 20:30

J.G.J.G.

30.5k23149

$endgroup$

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered Mar 12 at 20:30

J.G.J.G.

30.5k23149

$endgroup$

share|cite|improve this answer

answered Mar 12 at 20:30

J.G.J.G.

30.5k23149

answered Mar 12 at 20:30

J.G.J.G.

30.5k23149

answered Mar 12 at 20:30

J.G.J.G.

30.5k23149