If $12mid{a}$ then $4nmid{b}$ or $4mid{(8+2a+5b)}$.












0












$begingroup$


Prove that if $12mid{a}$ then $4nmid{b}$ or $4mid{(8+2a+5b)}$.



I have gotten thus far:



Since $12mid{a}$, then $a=12k$ for some integer $k$.
Substitute $a=12k$ into $(8+2a+5b)$ to get $8+24k+5b$.



Not sure what to do from here, any help is appreciated, thanks!






elementary-number-theory proof-writing divisibility







share|cite|improve this question











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  • 4




    $begingroup$
    Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
    $endgroup$
    – lulu
    Jan 27 at 22:23












  • $begingroup$
    @lulu Why are you answering in comments? (Hints are answers)
    $endgroup$
    – Arthur
    Jan 27 at 22:30












  • $begingroup$
    $large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
    $endgroup$
    – Bill Dubuque
    Jan 28 at 1:08












  • $begingroup$
    But isn't it that $4nmid{b}$?
    $endgroup$
    – macy
    Jan 28 at 3:07






  • 1




    $begingroup$
    @Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
    $endgroup$
    – Arthur
    Jan 28 at 11:49


















0












$begingroup$


Prove that if $12mid{a}$ then $4nmid{b}$ or $4mid{(8+2a+5b)}$.



I have gotten thus far:



Since $12mid{a}$, then $a=12k$ for some integer $k$.
Substitute $a=12k$ into $(8+2a+5b)$ to get $8+24k+5b$.



Not sure what to do from here, any help is appreciated, thanks!






elementary-number-theory proof-writing divisibility







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
    $endgroup$
    – lulu
    Jan 27 at 22:23












  • $begingroup$
    @lulu Why are you answering in comments? (Hints are answers)
    $endgroup$
    – Arthur
    Jan 27 at 22:30












  • $begingroup$
    $large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
    $endgroup$
    – Bill Dubuque
    Jan 28 at 1:08












  • $begingroup$
    But isn't it that $4nmid{b}$?
    $endgroup$
    – macy
    Jan 28 at 3:07






  • 1




    $begingroup$
    @Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
    $endgroup$
    – Arthur
    Jan 28 at 11:49
















0












0








0


1



$begingroup$


Prove that if $12mid{a}$ then $4nmid{b}$ or $4mid{(8+2a+5b)}$.



I have gotten thus far:



Since $12mid{a}$, then $a=12k$ for some integer $k$.
Substitute $a=12k$ into $(8+2a+5b)$ to get $8+24k+5b$.



Not sure what to do from here, any help is appreciated, thanks!






elementary-number-theory proof-writing divisibility







share|cite|improve this question











$endgroup$




Prove that if $12mid{a}$ then $4nmid{b}$ or $4mid{(8+2a+5b)}$.



I have gotten thus far:



Since $12mid{a}$, then $a=12k$ for some integer $k$.
Substitute $a=12k$ into $(8+2a+5b)$ to get $8+24k+5b$.



Not sure what to do from here, any help is appreciated, thanks!






elementary-number-theory proof-writing divisibility




elementary-number-theory proof-writing divisibility






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 27 at 23:02









Martin Sleziak

44.9k10122276




44.9k10122276










asked Jan 27 at 22:21









macymacy

526




526








  • 4




    $begingroup$
    Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
    $endgroup$
    – lulu
    Jan 27 at 22:23












  • $begingroup$
    @lulu Why are you answering in comments? (Hints are answers)
    $endgroup$
    – Arthur
    Jan 27 at 22:30












  • $begingroup$
    $large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
    $endgroup$
    – Bill Dubuque
    Jan 28 at 1:08












  • $begingroup$
    But isn't it that $4nmid{b}$?
    $endgroup$
    – macy
    Jan 28 at 3:07






  • 1




    $begingroup$
    @Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
    $endgroup$
    – Arthur
    Jan 28 at 11:49
















  • 4




    $begingroup$
    Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
    $endgroup$
    – lulu
    Jan 27 at 22:23












  • $begingroup$
    @lulu Why are you answering in comments? (Hints are answers)
    $endgroup$
    – Arthur
    Jan 27 at 22:30












  • $begingroup$
    $large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
    $endgroup$
    – Bill Dubuque
    Jan 28 at 1:08












  • $begingroup$
    But isn't it that $4nmid{b}$?
    $endgroup$
    – macy
    Jan 28 at 3:07






  • 1




    $begingroup$
    @Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
    $endgroup$
    – Arthur
    Jan 28 at 11:49










4




4




$begingroup$
Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
$endgroup$
– lulu
Jan 27 at 22:23






$begingroup$
Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
$endgroup$
– lulu
Jan 27 at 22:23














$begingroup$
@lulu Why are you answering in comments? (Hints are answers)
$endgroup$
– Arthur
Jan 27 at 22:30






$begingroup$
@lulu Why are you answering in comments? (Hints are answers)
$endgroup$
– Arthur
Jan 27 at 22:30














$begingroup$
$large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
$endgroup$
– Bill Dubuque
Jan 28 at 1:08






$begingroup$
$large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
$endgroup$
– Bill Dubuque
Jan 28 at 1:08














$begingroup$
But isn't it that $4nmid{b}$?
$endgroup$
– macy
Jan 28 at 3:07




$begingroup$
But isn't it that $4nmid{b}$?
$endgroup$
– macy
Jan 28 at 3:07




1




1




$begingroup$
@Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
$endgroup$
– Arthur
Jan 28 at 11:49






$begingroup$
@Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
$endgroup$
– Arthur
Jan 28 at 11:49












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$begingroup$

Assume $12|a$ and $4|b$ , otherwise nothing has to be proven, then , we can write $a=4s$ and $b=4t$ with intgegers $s,t$



Then, we have $$8+2a+5b=8+8s+20t=4(2+2s+5t)$$ hence $4$ divides $8+2a+5b$






share|cite|improve this answer









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    3












    $begingroup$

    Assume $12|a$ and $4|b$ , otherwise nothing has to be proven, then , we can write $a=4s$ and $b=4t$ with intgegers $s,t$



    Then, we have $$8+2a+5b=8+8s+20t=4(2+2s+5t)$$ hence $4$ divides $8+2a+5b$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Assume $12|a$ and $4|b$ , otherwise nothing has to be proven, then , we can write $a=4s$ and $b=4t$ with intgegers $s,t$



      Then, we have $$8+2a+5b=8+8s+20t=4(2+2s+5t)$$ hence $4$ divides $8+2a+5b$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Assume $12|a$ and $4|b$ , otherwise nothing has to be proven, then , we can write $a=4s$ and $b=4t$ with intgegers $s,t$



        Then, we have $$8+2a+5b=8+8s+20t=4(2+2s+5t)$$ hence $4$ divides $8+2a+5b$






        share|cite|improve this answer









        $endgroup$



        Assume $12|a$ and $4|b$ , otherwise nothing has to be proven, then , we can write $a=4s$ and $b=4t$ with intgegers $s,t$



        Then, we have $$8+2a+5b=8+8s+20t=4(2+2s+5t)$$ hence $4$ divides $8+2a+5b$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 28 at 11:54









        PeterPeter

        49.1k1240136




        49.1k1240136






























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