Let m be the least common multiple of a,b and c be another common multiple. Prove that m divides c....












1












$begingroup$



This question already has an answer here:




  • Is $ d_1,d_2mid niff [d_1,d_2]mid n,$ true? [LCM Universal Property]

    4 answers




I have tried to solve the problem. I have started by assigning m=(ab)/d, where d is the greatest common divisor of a and b. Then I set up c = aq. That is where I am stuck. I am really not sure what to do next.










share|cite|improve this question











$endgroup$



marked as duplicate by Bill Dubuque, Shailesh, José Carlos Santos, ancientmathematician, saz Jan 26 at 19:17


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    $c$ divides $m$? Don't you mean, prove $m$ divides $c$?
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:56






  • 1




    $begingroup$
    I am sorry... you are right it is m divides c.!!
    $endgroup$
    – Kbiir
    Jan 25 at 3:57






  • 1




    $begingroup$
    Let $c=mq+r$ with $0le r<m$ and show that $r$ is also a common multiple.
    $endgroup$
    – Gerry Myerson
    Jan 25 at 4:01
















1












$begingroup$



This question already has an answer here:




  • Is $ d_1,d_2mid niff [d_1,d_2]mid n,$ true? [LCM Universal Property]

    4 answers




I have tried to solve the problem. I have started by assigning m=(ab)/d, where d is the greatest common divisor of a and b. Then I set up c = aq. That is where I am stuck. I am really not sure what to do next.










share|cite|improve this question











$endgroup$



marked as duplicate by Bill Dubuque, Shailesh, José Carlos Santos, ancientmathematician, saz Jan 26 at 19:17


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    $c$ divides $m$? Don't you mean, prove $m$ divides $c$?
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:56






  • 1




    $begingroup$
    I am sorry... you are right it is m divides c.!!
    $endgroup$
    – Kbiir
    Jan 25 at 3:57






  • 1




    $begingroup$
    Let $c=mq+r$ with $0le r<m$ and show that $r$ is also a common multiple.
    $endgroup$
    – Gerry Myerson
    Jan 25 at 4:01














1












1








1





$begingroup$



This question already has an answer here:




  • Is $ d_1,d_2mid niff [d_1,d_2]mid n,$ true? [LCM Universal Property]

    4 answers




I have tried to solve the problem. I have started by assigning m=(ab)/d, where d is the greatest common divisor of a and b. Then I set up c = aq. That is where I am stuck. I am really not sure what to do next.










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Is $ d_1,d_2mid niff [d_1,d_2]mid n,$ true? [LCM Universal Property]

    4 answers




I have tried to solve the problem. I have started by assigning m=(ab)/d, where d is the greatest common divisor of a and b. Then I set up c = aq. That is where I am stuck. I am really not sure what to do next.





This question already has an answer here:




  • Is $ d_1,d_2mid niff [d_1,d_2]mid n,$ true? [LCM Universal Property]

    4 answers








elementary-number-theory least-common-multiple






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 17:36









Bill Dubuque

212k29195650




212k29195650










asked Jan 25 at 3:50









KbiirKbiir

486




486




marked as duplicate by Bill Dubuque, Shailesh, José Carlos Santos, ancientmathematician, saz Jan 26 at 19:17


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Bill Dubuque, Shailesh, José Carlos Santos, ancientmathematician, saz Jan 26 at 19:17


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    $c$ divides $m$? Don't you mean, prove $m$ divides $c$?
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:56






  • 1




    $begingroup$
    I am sorry... you are right it is m divides c.!!
    $endgroup$
    – Kbiir
    Jan 25 at 3:57






  • 1




    $begingroup$
    Let $c=mq+r$ with $0le r<m$ and show that $r$ is also a common multiple.
    $endgroup$
    – Gerry Myerson
    Jan 25 at 4:01














  • 1




    $begingroup$
    $c$ divides $m$? Don't you mean, prove $m$ divides $c$?
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:56






  • 1




    $begingroup$
    I am sorry... you are right it is m divides c.!!
    $endgroup$
    – Kbiir
    Jan 25 at 3:57






  • 1




    $begingroup$
    Let $c=mq+r$ with $0le r<m$ and show that $r$ is also a common multiple.
    $endgroup$
    – Gerry Myerson
    Jan 25 at 4:01








1




1




$begingroup$
$c$ divides $m$? Don't you mean, prove $m$ divides $c$?
$endgroup$
– Gerry Myerson
Jan 25 at 3:56




$begingroup$
$c$ divides $m$? Don't you mean, prove $m$ divides $c$?
$endgroup$
– Gerry Myerson
Jan 25 at 3:56




1




1




$begingroup$
I am sorry... you are right it is m divides c.!!
$endgroup$
– Kbiir
Jan 25 at 3:57




$begingroup$
I am sorry... you are right it is m divides c.!!
$endgroup$
– Kbiir
Jan 25 at 3:57




1




1




$begingroup$
Let $c=mq+r$ with $0le r<m$ and show that $r$ is also a common multiple.
$endgroup$
– Gerry Myerson
Jan 25 at 4:01




$begingroup$
Let $c=mq+r$ with $0le r<m$ and show that $r$ is also a common multiple.
$endgroup$
– Gerry Myerson
Jan 25 at 4:01










1 Answer
1






active

oldest

votes


















2












$begingroup$

Hint: $c = rm+d$ where $r$ is some integer and $d < m$. So $d = c-rm$. As $a$ divides $c$ and $m$, and $-r$ is an integer, $a$ divides $d$. Likewise as $b$ divides both $c$ and $m$, it follows that $b$ divides $d$. So both $a$ and $b$ divide $d$. But $m$ is LCM of $a$ and $b$ and $d< m$. What can you say about $d$?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    then d must be equal to 0!!!
    $endgroup$
    – Kbiir
    Jan 25 at 4:03






  • 1




    $begingroup$
    ...Exactly! You got it
    $endgroup$
    – Mike
    Jan 25 at 4:04




















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Hint: $c = rm+d$ where $r$ is some integer and $d < m$. So $d = c-rm$. As $a$ divides $c$ and $m$, and $-r$ is an integer, $a$ divides $d$. Likewise as $b$ divides both $c$ and $m$, it follows that $b$ divides $d$. So both $a$ and $b$ divide $d$. But $m$ is LCM of $a$ and $b$ and $d< m$. What can you say about $d$?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    then d must be equal to 0!!!
    $endgroup$
    – Kbiir
    Jan 25 at 4:03






  • 1




    $begingroup$
    ...Exactly! You got it
    $endgroup$
    – Mike
    Jan 25 at 4:04


















2












$begingroup$

Hint: $c = rm+d$ where $r$ is some integer and $d < m$. So $d = c-rm$. As $a$ divides $c$ and $m$, and $-r$ is an integer, $a$ divides $d$. Likewise as $b$ divides both $c$ and $m$, it follows that $b$ divides $d$. So both $a$ and $b$ divide $d$. But $m$ is LCM of $a$ and $b$ and $d< m$. What can you say about $d$?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    then d must be equal to 0!!!
    $endgroup$
    – Kbiir
    Jan 25 at 4:03






  • 1




    $begingroup$
    ...Exactly! You got it
    $endgroup$
    – Mike
    Jan 25 at 4:04
















2












2








2





$begingroup$

Hint: $c = rm+d$ where $r$ is some integer and $d < m$. So $d = c-rm$. As $a$ divides $c$ and $m$, and $-r$ is an integer, $a$ divides $d$. Likewise as $b$ divides both $c$ and $m$, it follows that $b$ divides $d$. So both $a$ and $b$ divide $d$. But $m$ is LCM of $a$ and $b$ and $d< m$. What can you say about $d$?






share|cite|improve this answer











$endgroup$



Hint: $c = rm+d$ where $r$ is some integer and $d < m$. So $d = c-rm$. As $a$ divides $c$ and $m$, and $-r$ is an integer, $a$ divides $d$. Likewise as $b$ divides both $c$ and $m$, it follows that $b$ divides $d$. So both $a$ and $b$ divide $d$. But $m$ is LCM of $a$ and $b$ and $d< m$. What can you say about $d$?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 25 at 4:03

























answered Jan 25 at 4:01









MikeMike

4,301412




4,301412








  • 1




    $begingroup$
    then d must be equal to 0!!!
    $endgroup$
    – Kbiir
    Jan 25 at 4:03






  • 1




    $begingroup$
    ...Exactly! You got it
    $endgroup$
    – Mike
    Jan 25 at 4:04
















  • 1




    $begingroup$
    then d must be equal to 0!!!
    $endgroup$
    – Kbiir
    Jan 25 at 4:03






  • 1




    $begingroup$
    ...Exactly! You got it
    $endgroup$
    – Mike
    Jan 25 at 4:04










1




1




$begingroup$
then d must be equal to 0!!!
$endgroup$
– Kbiir
Jan 25 at 4:03




$begingroup$
then d must be equal to 0!!!
$endgroup$
– Kbiir
Jan 25 at 4:03




1




1




$begingroup$
...Exactly! You got it
$endgroup$
– Mike
Jan 25 at 4:04






$begingroup$
...Exactly! You got it
$endgroup$
– Mike
Jan 25 at 4:04





Popular posts from this blog

Mario Kart Wii

The Binding of Isaac: Rebirth/Afterbirth

What does “Dominus providebit” mean?