Three numbers which are co-primes of each other such that the product of the first $2$ is $551$ and that of the last $2$ is 1073. Find the $3$ numbers

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

1

If we take a,b, and c as the three numbers then, I know the answer is got by using the fact that b will be the common factor of $551$ and $1073$. But what I don't understand is why is b taken as the gcd of $551$ and $1073$ as it can easily be just any of the common factors of those two numbers.

greatest-common-divisor

share|cite|improve this question

edited Jul 19 at 15:00

Key Flex

4,346425

asked Jul 19 at 4:52

GRANZER

1869

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @dxiv Yup. But why should that number be taken as the gcd at all?? Where is it told it will be the gcd as it could be just any common factor of 551 and 1073.
  – GRANZER
  Jul 19 at 4:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @how did you come to that conclusion just by looking at the question? that there are just two common factors
  – GRANZER
  Jul 19 at 5:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Denote $A=ab$, $B=bc$, then: if assume that $GCD(A,B) = b g;$ ($g> 1)$, then $A=alphacdot b g, B=beta cdot b g$, where $GCD(alpha,beta)=1$; hence $a=alpha g$, $b=beta g$. So $GCD(a,b)=g>1$ ($a$ and $b$ aren't co-prime).
  – Oleg567
  Jul 19 at 5:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @GRANZER does that make a and b also to be co-primes The title of your question says "co-primes of each other". This would normally read as "mutually co-prime", meaning that each pair of numbers is co-prime i.e. $gcd(a,b)$ $=gcd(b,c)$ $=gcd(c,a)$ $=1$.
  – dxiv
  Jul 19 at 5:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Oleg567 Did you mean $c= beta.g$ and $GCD(a,c)=g>1$?
  – GRANZER
  Jul 19 at 7:18
  
  

  
  
  
  

 | 
show 5 more comments

up vote
1
down vote

favorite

1

If we take a,b, and c as the three numbers then, I know the answer is got by using the fact that b will be the common factor of $551$ and $1073$. But what I don't understand is why is b taken as the gcd of $551$ and $1073$ as it can easily be just any of the common factors of those two numbers.

greatest-common-divisor

share|cite|improve this question

edited Jul 19 at 15:00

Key Flex

4,346425

asked Jul 19 at 4:52

GRANZER

1869

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @dxiv Yup. But why should that number be taken as the gcd at all?? Where is it told it will be the gcd as it could be just any common factor of 551 and 1073.
  – GRANZER
  Jul 19 at 4:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @how did you come to that conclusion just by looking at the question? that there are just two common factors
  – GRANZER
  Jul 19 at 5:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Denote $A=ab$, $B=bc$, then: if assume that $GCD(A,B) = b g;$ ($g> 1)$, then $A=alphacdot b g, B=beta cdot b g$, where $GCD(alpha,beta)=1$; hence $a=alpha g$, $b=beta g$. So $GCD(a,b)=g>1$ ($a$ and $b$ aren't co-prime).
  – Oleg567
  Jul 19 at 5:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @GRANZER does that make a and b also to be co-primes The title of your question says "co-primes of each other". This would normally read as "mutually co-prime", meaning that each pair of numbers is co-prime i.e. $gcd(a,b)$ $=gcd(b,c)$ $=gcd(c,a)$ $=1$.
  – dxiv
  Jul 19 at 5:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Oleg567 Did you mean $c= beta.g$ and $GCD(a,c)=g>1$?
  – GRANZER
  Jul 19 at 7:18
  
  

  
  
  
  

 | 
show 5 more comments

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

If we take a,b, and c as the three numbers then, I know the answer is got by using the fact that b will be the common factor of $551$ and $1073$. But what I don't understand is why is b taken as the gcd of $551$ and $1073$ as it can easily be just any of the common factors of those two numbers.

greatest-common-divisor

share|cite|improve this question

edited Jul 19 at 15:00

Key Flex

4,346425

asked Jul 19 at 4:52

GRANZER

1869

If we take a,b, and c as the three numbers then, I know the answer is got by using the fact that b will be the common factor of $551$ and $1073$. But what I don't understand is why is b taken as the gcd of $551$ and $1073$ as it can easily be just any of the common factors of those two numbers.

greatest-common-divisor

share|cite|improve this question

edited Jul 19 at 15:00

Key Flex

4,346425

asked Jul 19 at 4:52

GRANZER

1869

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 15:00

Key Flex

4,346425

edited Jul 19 at 15:00

Key Flex

4,346425

edited Jul 19 at 15:00

Key Flex

4,346425

4,346425

asked Jul 19 at 4:52

GRANZER

1869

asked Jul 19 at 4:52

GRANZER

1869

asked Jul 19 at 4:52

GRANZER

1869

1869

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @dxiv Yup. But why should that number be taken as the gcd at all?? Where is it told it will be the gcd as it could be just any common factor of 551 and 1073.
  – GRANZER
  Jul 19 at 4:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @how did you come to that conclusion just by looking at the question? that there are just two common factors
  – GRANZER
  Jul 19 at 5:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Denote $A=ab$, $B=bc$, then: if assume that $GCD(A,B) = b g;$ ($g> 1)$, then $A=alphacdot b g, B=beta cdot b g$, where $GCD(alpha,beta)=1$; hence $a=alpha g$, $b=beta g$. So $GCD(a,b)=g>1$ ($a$ and $b$ aren't co-prime).
  – Oleg567
  Jul 19 at 5:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @GRANZER does that make a and b also to be co-primes The title of your question says "co-primes of each other". This would normally read as "mutually co-prime", meaning that each pair of numbers is co-prime i.e. $gcd(a,b)$ $=gcd(b,c)$ $=gcd(c,a)$ $=1$.
  – dxiv
  Jul 19 at 5:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Oleg567 Did you mean $c= beta.g$ and $GCD(a,c)=g>1$?
  – GRANZER
  Jul 19 at 7:18
  
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @dxiv Yup. But why should that number be taken as the gcd at all?? Where is it told it will be the gcd as it could be just any common factor of 551 and 1073.
  – GRANZER
  Jul 19 at 4:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @how did you come to that conclusion just by looking at the question? that there are just two common factors
  – GRANZER
  Jul 19 at 5:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Denote $A=ab$, $B=bc$, then: if assume that $GCD(A,B) = b g;$ ($g> 1)$, then $A=alphacdot b g, B=beta cdot b g$, where $GCD(alpha,beta)=1$; hence $a=alpha g$, $b=beta g$. So $GCD(a,b)=g>1$ ($a$ and $b$ aren't co-prime).
  – Oleg567
  Jul 19 at 5:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @GRANZER does that make a and b also to be co-primes The title of your question says "co-primes of each other". This would normally read as "mutually co-prime", meaning that each pair of numbers is co-prime i.e. $gcd(a,b)$ $=gcd(b,c)$ $=gcd(c,a)$ $=1$.
  – dxiv
  Jul 19 at 5:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Oleg567 Did you mean $c= beta.g$ and $GCD(a,c)=g>1$?
  – GRANZER
  Jul 19 at 7:18
  
  

  
  
  
  

@dxiv Yup. But why should that number be taken as the gcd at all?? Where is it told it will be the gcd as it could be just any common factor of 551 and 1073.
– GRANZER
Jul 19 at 4:57

@dxiv Yup. But why should that number be taken as the gcd at all?? Where is it told it will be the gcd as it could be just any common factor of 551 and 1073.
– GRANZER
Jul 19 at 4:57

@how did you come to that conclusion just by looking at the question? that there are just two common factors
– GRANZER
Jul 19 at 5:00

@how did you come to that conclusion just by looking at the question? that there are just two common factors
– GRANZER
Jul 19 at 5:00

1

1

Denote $A=ab$, $B=bc$, then: if assume that $GCD(A,B) = b g;$ ($g> 1)$, then $A=alphacdot b g, B=beta cdot b g$, where $GCD(alpha,beta)=1$; hence $a=alpha g$, $b=beta g$. So $GCD(a,b)=g>1$ ($a$ and $b$ aren't co-prime).
– Oleg567
Jul 19 at 5:04

Denote $A=ab$, $B=bc$, then: if assume that $GCD(A,B) = b g;$ ($g> 1)$, then $A=alphacdot b g, B=beta cdot b g$, where $GCD(alpha,beta)=1$; hence $a=alpha g$, $b=beta g$. So $GCD(a,b)=g>1$ ($a$ and $b$ aren't co-prime).
– Oleg567
Jul 19 at 5:04

1

1

@GRANZER does that make a and b also to be co-primes The title of your question says "co-primes of each other". This would normally read as "mutually co-prime", meaning that each pair of numbers is co-prime i.e. $gcd(a,b)$ $=gcd(b,c)$ $=gcd(c,a)$ $=1$.
– dxiv
Jul 19 at 5:42

@GRANZER does that make a and b also to be co-primes The title of your question says "co-primes of each other". This would normally read as "mutually co-prime", meaning that each pair of numbers is co-prime i.e. $gcd(a,b)$ $=gcd(b,c)$ $=gcd(c,a)$ $=1$.
– dxiv
Jul 19 at 5:42

1

1

@Oleg567 Did you mean $c= beta.g$ and $GCD(a,c)=g>1$?
– GRANZER
Jul 19 at 7:18

@Oleg567 Did you mean $c= beta.g$ and $GCD(a,c)=g>1$?
– GRANZER
Jul 19 at 7:18

 | 
show 5 more comments

2 Answers
2

active

oldest

votes

up vote
4
down vote



accepted

Assuming $a,b,c$ are all positive.

$$ab=551$$
$$bc=1073$$

$b$ clearly is a common divisor.

Suppose it is not the greatest common divisor, then $a$ and $c$ would share some common factors that are bigger than $1$ which contradicts to the fact that they are coprime.

share|cite|improve this answer

answered Jul 19 at 5:00

Siong Thye Goh

77.6k134795

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry but I think I am missing some crucial point of knowledge that is making me not grasp the answer. If b is not the gcd, how would that make a and c have factors bigger than 1? Also, a,b, and c are together suppose to be coprime so that doesn't have to mean a and c should also be co-primes
  – GRANZER
  Jul 19 at 5:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I was just failing with this one fact that Every common divisor of two integers divides their greatest common divisor.
  – GRANZER
  Jul 19 at 5:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let $s$ be a common divisor of $a$ and $b$ and $d$ be the largest common divisor of $a$ and $b$. By Bezout's identity there exists $x,y in mathbbZ$ such that $d=xa+yb$, since $s$ divides $a$ and $b$, it divides $xa+yb$ and it divides $d$.
  – Siong Thye Goh
  Jul 19 at 5:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $b$ is not the gcd then the gcd must have some factors that don't come from $b$. Where do they come from if they don't come from the $b $? They can't come from $a$ or $c $. What's left? They can't come from anywhere.
  – fleablood
  Jul 19 at 6:06
  

  
  
  
  

add a comment | 

up vote
2
down vote

If $a,b,c $ are the numbers then $ab $ and $bc $ are the products. Those have $b $ as a common factor. But $a$ and $c $ are relatively prime and have no factors in common. So $ab$ and $bc$ can't have any factors in common that aren't a factor of $b $.

So $b$ is the greatest common factor of $ab $ and $bc $. So we can find $b $. Just divide $ab$ and $bc $ by $b$ to get $a $ and $c $.

share|cite|improve this answer

answered Jul 19 at 6:03

fleablood

60.5k22575

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2856277%2fthree-numbers-which-are-co-primes-of-each-other-such-that-the-product-of-the-fir%23new-answer', 'question_page');

);

Post as a guest

Name Email

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Assuming $a,b,c$ are all positive.

$$ab=551$$
$$bc=1073$$

$b$ clearly is a common divisor.

Suppose it is not the greatest common divisor, then $a$ and $c$ would share some common factors that are bigger than $1$ which contradicts to the fact that they are coprime.

share|cite|improve this answer

answered Jul 19 at 5:00

Siong Thye Goh

77.6k134795

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry but I think I am missing some crucial point of knowledge that is making me not grasp the answer. If b is not the gcd, how would that make a and c have factors bigger than 1? Also, a,b, and c are together suppose to be coprime so that doesn't have to mean a and c should also be co-primes
  – GRANZER
  Jul 19 at 5:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I was just failing with this one fact that Every common divisor of two integers divides their greatest common divisor.
  – GRANZER
  Jul 19 at 5:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let $s$ be a common divisor of $a$ and $b$ and $d$ be the largest common divisor of $a$ and $b$. By Bezout's identity there exists $x,y in mathbbZ$ such that $d=xa+yb$, since $s$ divides $a$ and $b$, it divides $xa+yb$ and it divides $d$.
  – Siong Thye Goh
  Jul 19 at 5:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $b$ is not the gcd then the gcd must have some factors that don't come from $b$. Where do they come from if they don't come from the $b $? They can't come from $a$ or $c $. What's left? They can't come from anywhere.
  – fleablood
  Jul 19 at 6:06
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

Assuming $a,b,c$ are all positive.

$$ab=551$$
$$bc=1073$$

$b$ clearly is a common divisor.

Suppose it is not the greatest common divisor, then $a$ and $c$ would share some common factors that are bigger than $1$ which contradicts to the fact that they are coprime.

share|cite|improve this answer

answered Jul 19 at 5:00

Siong Thye Goh

77.6k134795

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry but I think I am missing some crucial point of knowledge that is making me not grasp the answer. If b is not the gcd, how would that make a and c have factors bigger than 1? Also, a,b, and c are together suppose to be coprime so that doesn't have to mean a and c should also be co-primes
  – GRANZER
  Jul 19 at 5:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I was just failing with this one fact that Every common divisor of two integers divides their greatest common divisor.
  – GRANZER
  Jul 19 at 5:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let $s$ be a common divisor of $a$ and $b$ and $d$ be the largest common divisor of $a$ and $b$. By Bezout's identity there exists $x,y in mathbbZ$ such that $d=xa+yb$, since $s$ divides $a$ and $b$, it divides $xa+yb$ and it divides $d$.
  – Siong Thye Goh
  Jul 19 at 5:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $b$ is not the gcd then the gcd must have some factors that don't come from $b$. Where do they come from if they don't come from the $b $? They can't come from $a$ or $c $. What's left? They can't come from anywhere.
  – fleablood
  Jul 19 at 6:06
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Assuming $a,b,c$ are all positive.

$$ab=551$$
$$bc=1073$$

$b$ clearly is a common divisor.

Suppose it is not the greatest common divisor, then $a$ and $c$ would share some common factors that are bigger than $1$ which contradicts to the fact that they are coprime.

share|cite|improve this answer

answered Jul 19 at 5:00

Siong Thye Goh

77.6k134795

Assuming $a,b,c$ are all positive.

$$ab=551$$
$$bc=1073$$

$b$ clearly is a common divisor.

Suppose it is not the greatest common divisor, then $a$ and $c$ would share some common factors that are bigger than $1$ which contradicts to the fact that they are coprime.

share|cite|improve this answer

answered Jul 19 at 5:00

Siong Thye Goh

77.6k134795

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 5:00

Siong Thye Goh

77.6k134795

answered Jul 19 at 5:00

Siong Thye Goh

77.6k134795

answered Jul 19 at 5:00

Siong Thye Goh

77.6k134795

77.6k134795

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry but I think I am missing some crucial point of knowledge that is making me not grasp the answer. If b is not the gcd, how would that make a and c have factors bigger than 1? Also, a,b, and c are together suppose to be coprime so that doesn't have to mean a and c should also be co-primes
  – GRANZER
  Jul 19 at 5:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I was just failing with this one fact that Every common divisor of two integers divides their greatest common divisor.
  – GRANZER
  Jul 19 at 5:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let $s$ be a common divisor of $a$ and $b$ and $d$ be the largest common divisor of $a$ and $b$. By Bezout's identity there exists $x,y in mathbbZ$ such that $d=xa+yb$, since $s$ divides $a$ and $b$, it divides $xa+yb$ and it divides $d$.
  – Siong Thye Goh
  Jul 19 at 5:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $b$ is not the gcd then the gcd must have some factors that don't come from $b$. Where do they come from if they don't come from the $b $? They can't come from $a$ or $c $. What's left? They can't come from anywhere.
  – fleablood
  Jul 19 at 6:06
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry but I think I am missing some crucial point of knowledge that is making me not grasp the answer. If b is not the gcd, how would that make a and c have factors bigger than 1? Also, a,b, and c are together suppose to be coprime so that doesn't have to mean a and c should also be co-primes
  – GRANZER
  Jul 19 at 5:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I was just failing with this one fact that Every common divisor of two integers divides their greatest common divisor.
  – GRANZER
  Jul 19 at 5:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let $s$ be a common divisor of $a$ and $b$ and $d$ be the largest common divisor of $a$ and $b$. By Bezout's identity there exists $x,y in mathbbZ$ such that $d=xa+yb$, since $s$ divides $a$ and $b$, it divides $xa+yb$ and it divides $d$.
  – Siong Thye Goh
  Jul 19 at 5:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $b$ is not the gcd then the gcd must have some factors that don't come from $b$. Where do they come from if they don't come from the $b $? They can't come from $a$ or $c $. What's left? They can't come from anywhere.
  – fleablood
  Jul 19 at 6:06
  

  
  
  
  

I am sorry but I think I am missing some crucial point of knowledge that is making me not grasp the answer. If b is not the gcd, how would that make a and c have factors bigger than 1? Also, a,b, and c are together suppose to be coprime so that doesn't have to mean a and c should also be co-primes
– GRANZER
Jul 19 at 5:14

I am sorry but I think I am missing some crucial point of knowledge that is making me not grasp the answer. If b is not the gcd, how would that make a and c have factors bigger than 1? Also, a,b, and c are together suppose to be coprime so that doesn't have to mean a and c should also be co-primes
– GRANZER
Jul 19 at 5:14

I was just failing with this one fact that Every common divisor of two integers divides their greatest common divisor.
– GRANZER
Jul 19 at 5:37

I was just failing with this one fact that Every common divisor of two integers divides their greatest common divisor.
– GRANZER
Jul 19 at 5:37

1

1

Let $s$ be a common divisor of $a$ and $b$ and $d$ be the largest common divisor of $a$ and $b$. By Bezout's identity there exists $x,y in mathbbZ$ such that $d=xa+yb$, since $s$ divides $a$ and $b$, it divides $xa+yb$ and it divides $d$.
– Siong Thye Goh
Jul 19 at 5:53

Let $s$ be a common divisor of $a$ and $b$ and $d$ be the largest common divisor of $a$ and $b$. By Bezout's identity there exists $x,y in mathbbZ$ such that $d=xa+yb$, since $s$ divides $a$ and $b$, it divides $xa+yb$ and it divides $d$.
– Siong Thye Goh
Jul 19 at 5:53

1

1

If $b$ is not the gcd then the gcd must have some factors that don't come from $b$. Where do they come from if they don't come from the $b $? They can't come from $a$ or $c $. What's left? They can't come from anywhere.
– fleablood
Jul 19 at 6:06

If $b$ is not the gcd then the gcd must have some factors that don't come from $b$. Where do they come from if they don't come from the $b $? They can't come from $a$ or $c $. What's left? They can't come from anywhere.
– fleablood
Jul 19 at 6:06

add a comment | 

up vote
2
down vote

If $a,b,c $ are the numbers then $ab $ and $bc $ are the products. Those have $b $ as a common factor. But $a$ and $c $ are relatively prime and have no factors in common. So $ab$ and $bc$ can't have any factors in common that aren't a factor of $b $.

So $b$ is the greatest common factor of $ab $ and $bc $. So we can find $b $. Just divide $ab$ and $bc $ by $b$ to get $a $ and $c $.

share|cite|improve this answer

answered Jul 19 at 6:03

fleablood

60.5k22575

add a comment | 

up vote
2
down vote

If $a,b,c $ are the numbers then $ab $ and $bc $ are the products. Those have $b $ as a common factor. But $a$ and $c $ are relatively prime and have no factors in common. So $ab$ and $bc$ can't have any factors in common that aren't a factor of $b $.

So $b$ is the greatest common factor of $ab $ and $bc $. So we can find $b $. Just divide $ab$ and $bc $ by $b$ to get $a $ and $c $.

share|cite|improve this answer

answered Jul 19 at 6:03

fleablood

60.5k22575

add a comment | 

up vote
2
down vote

up vote
2
down vote

If $a,b,c $ are the numbers then $ab $ and $bc $ are the products. Those have $b $ as a common factor. But $a$ and $c $ are relatively prime and have no factors in common. So $ab$ and $bc$ can't have any factors in common that aren't a factor of $b $.

So $b$ is the greatest common factor of $ab $ and $bc $. So we can find $b $. Just divide $ab$ and $bc $ by $b$ to get $a $ and $c $.

share|cite|improve this answer

answered Jul 19 at 6:03

fleablood

60.5k22575

If $a,b,c $ are the numbers then $ab $ and $bc $ are the products. Those have $b $ as a common factor. But $a$ and $c $ are relatively prime and have no factors in common. So $ab$ and $bc$ can't have any factors in common that aren't a factor of $b $.

So $b$ is the greatest common factor of $ab $ and $bc $. So we can find $b $. Just divide $ab$ and $bc $ by $b$ to get $a $ and $c $.

share|cite|improve this answer

answered Jul 19 at 6:03

fleablood

60.5k22575

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 6:03

fleablood

60.5k22575

answered Jul 19 at 6:03

fleablood

60.5k22575

answered Jul 19 at 6:03

fleablood

60.5k22575

60.5k22575

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2856277%2fthree-numbers-which-are-co-primes-of-each-other-such-that-the-product-of-the-fir%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email