Let m be the least common multiple of a,b and c be another common multiple. Prove that m divides c....
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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.
elementary-number-theory least-common-multiple
edited Jan 25 at 17:36
Bill Dubuque
212k29195650
asked Jan 25 at 3:50
KbiirKbiir
486
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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.
add a comment |
$begingroup$
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.
elementary-number-theory least-common-multiple
edited Jan 25 at 17:36
Bill Dubuque
212k29195650
asked Jan 25 at 3:50
KbiirKbiir
486
$endgroup$
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
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$c$ divides $m$? Don't you mean, prove $m$ divides $c$?
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– Gerry Myerson
Jan 25 at 3:56
1
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I am sorry... you are right it is m divides c.!!
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– 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.
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– Gerry Myerson
Jan 25 at 4:01
add a comment |
$begingroup$
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.
elementary-number-theory least-common-multiple
edited Jan 25 at 17:36
Bill Dubuque
212k29195650
asked Jan 25 at 3:50
KbiirKbiir
486
$endgroup$
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
elementary-number-theory least-common-multiple
edited Jan 25 at 17:36
Bill Dubuque
212k29195650
asked Jan 25 at 3:50
KbiirKbiir
486
edited Jan 25 at 17:36
Bill Dubuque
212k29195650
asked Jan 25 at 3:50
KbiirKbiir
486
edited Jan 25 at 17:36
Bill Dubuque
212k29195650
edited Jan 25 at 17:36
Bill Dubuque
212k29195650
edited Jan 25 at 17:36
Bill Dubuque
212k29195650
212k29195650
asked Jan 25 at 3:50
KbiirKbiir
486
asked Jan 25 at 3:50
KbiirKbiir
486
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
add a comment |
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
add a comment |
1 Answer
1
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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 $d$?
answered Jan 25 at 4:01
MikeMike
4,301412
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1
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then d must be equal to 0!!!
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– Kbiir
Jan 25 at 4:03
1
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...Exactly! You got it
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– Mike
Jan 25 at 4:04
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $d$?
answered Jan 25 at 4:01
MikeMike
4,301412
$endgroup$
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
add a comment |
$begingroup$
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 $d$?
answered Jan 25 at 4:01
MikeMike
4,301412
$endgroup$
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
add a comment |
$begingroup$
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 $d$?
answered Jan 25 at 4:01
MikeMike
4,301412
$endgroup$
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 $d$?
answered Jan 25 at 4:01
MikeMike
4,301412
edited Jan 25 at 4:03
answered Jan 25 at 4:01
MikeMike
4,301412
answered Jan 25 at 4:01
MikeMike
4,301412
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
add a comment |
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
add a comment |
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