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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.

Key Flex
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GRANZER
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  • @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 '18 at 04:57
  • @how did you come to that conclusion just by looking at the question? that there are just two common factors – GRANZER Jul 19 '18 at 05:00
  • why is b taken as the gcd Because if $b$ were a proper factor of the $\gcd$ then it would follow that $a$ and $c$ are not coprime. – dxiv Jul 19 '18 at 05:00
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    Denote $A=ab$, $B=bc$, then: if assume that $GCD(A,B) = b g;$ ($g> 1)$, then $A=\alpha\cdot b g, B=\beta \cdot b g$, 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 '18 at 05:04
  • @Oleg567 If a,b,c are co-prime does that make a and b also to be co-primes? and so is b and c and a and c to be coprime? – GRANZER Jul 19 '18 at 05:30
  • I was just failing with this one fact that Every common divisor of two integers divides their greatest common divisor.!! – GRANZER Jul 19 '18 at 05:37
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    @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 '18 at 05:42
  • @dxiv Of course!! Thank you. – GRANZER Jul 19 '18 at 05:58
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    @Oleg567 Did you mean $c= beta.g$ and $GCD(a,c)=g>1$? – GRANZER Jul 19 '18 at 07:18
  • @GRANZER: oops, yes, sure; thanks for correction! it should be like this: $A=ab, C=bc$, $GCD(A,C)=bg$; $A=\alpha\cdot bg, C=\gamma \cdot bg$, hence ... $a=\alpha g$, $c=\gamma g$, so $GCD(a,c)=g>1$. Thanks again. – Oleg567 Jul 19 '18 at 09:27

2 Answers2

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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.

Siong Thye Goh
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  • 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 '18 at 05:14
  • I was just failing with this one fact that Every common divisor of two integers divides their greatest common divisor. – GRANZER Jul 19 '18 at 05:37
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    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 $x,y \in \mathbb{Z}$ such that $d=xa+yb$, since $s$ divides $a$ and $b$, it divides $xa+yb$ and it divides $d$. – Siong Thye Goh Jul 19 '18 at 05:53
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    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 '18 at 06:06
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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 $.

fleablood
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