Let $a$ and $b$ denote integers.  To find $\frac{\mathrm{LCM}(a,b)}{a},$ we can use:  $$\frac{\mathrm{LCM}(a,b)}{a} = \frac{b}{\mathrm{HCF}(a,b)}$$ and compute $\mathrm{HCF}(a,b)$ with the Euclidean algorithm.     Question. Is there a similar formula that allows us to find $$\frac{\mathrm{LCM}(a,b,c)}{a}$$ using only arithmetical operations and highest common factors?
45388
If I know the GCD of 20 numbers, and know the 20 numbers, is there a formula such that I input the 20 numbers and input their GCD, and it outputs their LCM? I know that $$\frac{\left| a\cdot b\right|}{\gcd(a,b)} = \text{lcm}(a,b).$$ So is it$$\frac{\left| a\cdot b\cdot c\cdot d\cdot e\cdot f\right|}{\gcd(a,b,c,d,e,f)}?$$If not, what is it?
94419
I'm trying to prove that if $a$ divides $bc$ and $\gcd(a,b) = d$ then $\frac a d$ divides c. I tried using Bezout identity but couldn't get anywhere.
63121
I have to show the following:"If the sum of two irreducible fractions with positive denominators is an integer, then the denominators are equal." $$\frac{a}{b}+\frac{c}{d}=k, \text{ where k an integer }$$ Since the fractions are irreducible, $(a,b)=1$ and $(c,d)=1$. Right? But how can I continue??
106815
I need to show that the $$\gcd\left(\frac a {\gcd(a,b)},\frac b {\gcd(a,b)}\right) =1$$  I'm not sure exactly how to approach this .. I tried using the fact that $\gcd(a,b)=d$ so $d=ma+nb$ but I didn't get too far  Could anyone suggest where to start?
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Suppose I know that I can find the $\gcd$ of two integers with the Euclidean algorithm. What is the 'extension' so I can use this method also for polynomials?  Edit: I found an easy answer with less rigor which is approachable without knowledge of matrices:    also the first 3 subchapters of
10195
If $a, b, c \in \mathbb{N}$, and $abc = 1$, prove that: $$S = \frac{2}{(a+1)^2+b^2+1} + \frac{2}{(b+1)^2+c^2+1} + \frac{2}{(c+1)^2+a^2+1} \le 1$$  Here is my try: $$\begin{align}\frac{2}{(a+1)^2+(b^2+1)} &amp; \le \frac{1}{\sqrt{(a+1)^2 \cdot (b^2+1)}}\\ &amp; = \frac{1}{(a+1) \cdot \sqrt{b^2+1}}\\ &amp; \le \frac{1}{(a+1) \cdot b}\\ &amp; \le \frac{1}{ab} \end{align}$$  So, $$S \le \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{a+b+c}{1}$$  Can you help me to continue it, please? Thanks!
109275
Prove that if $a,b,c$ are integers and both $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ and $\frac{a}{c} + \frac{b}{a} + \frac{c}{b}$ are integers, then $|a|=|b|=|c|$.   Well this is what I have done so far:  From the fact that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is an integer, we get  $$abc \mid ab^2 + bc^2 + ca^2 \tag{1}$$  In the same way, we also have  $$abc \mid a^2b + b^2c + c^2a \tag{2}$$  so  $$\begin{align} abc \mid a(ab^2 + bc^2 + ca^2) - b(a^2b + b^2c + c^2a) &amp;\Rightarrow abc \mid c(a^3-b^3) \tag{3}\\ &amp;\Rightarrow ab \mid a^3-b^3 \tag{4}\\ &amp;\Rightarrow ab \mid b^3(a^3-b^3) \tag{5} \end{align}$$ and so $a \mid b^6$. In the same way, we can also get $b \mid c^6$ and $c \mid a^6$.      But what should I do after this?   Any help is surely appreciated! Thanks!
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Let $a,b$ be two group elements of finite order that commute.   What can be said about the order of $ab$?   I thought that $|ab| = \text{lcm}(|a|,|b|)$. My proof was that $(ab)^n = a^n b^n =e$ if and only if $a^n = b^n = e$ if and only if $|a|,|b|$ both divide $n$. The smallest $n$ such that both orders divide it is the least common multiple of $|a|$ and $|b|$.  By chance I came across . It has an upvote so it's clearly correct. (?)  But it contradicts what I think: It's clear that   $$ (ab)^{\text{lcm}(|a|,|b|)} = e$$  hence $|ab| \mid \text{lcm}(|a|,|b|)$.  It seems to me that this is saying more than $|ab| \mid |a| |b|$. Is it not?     Now my question is: what is the precisest statement that can be made about $|ab|$?  Is there anything more than $|ab| \mid \text{lcm}(|a|,|b|)$ that can be said about $|ab|$?
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Show that the following conditions are equivalent:  i) There exist positive integers $a, b$ such that $\gcd(a,b)=d$ and $\operatorname{lcm}(a,b)=m$.  ii) $d\mid m$
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Let $R$ be a UFD. Then for any two elements $a,b \in R \setminus \{0\}$ $ab \sim \gcd (a,b){\operatorname{lcm} (a,b)}$.   My attempt :  Let $d = \gcd (a,b)$.Then $d|a$ and $d|b$.Then $\exists x,y \in R$ such that $a = dx$ and $b = dy$.Now it is to be shown that $c = dxy$ is the lcm of $a,b$.Which I find difficulty to show.It is clear that $a|c$ and $b|c$. Now if $u$ is any common multiple of $a$ and $b$.Then $dx|u$ and also since $y|b$ we have $y|u$ i.e. $dxy|u^2$ $\implies c|u^2$.But I have to show $c|u$ to complete the proof which I fail to prove.Please help me to complete the proof.  Thank you in advance.
93378
$$\frac{1}{a} + \frac{1}{b} = \frac{1}{12}$$  I'm trying to find all integer solutions to this equation. I've played with this equation algebraically to see if I can figure out the restrictions on $a$ and $b$. For example, I know at least one of them must be divisible by $12$ because  $$   a + b = \frac{ab}{12}  $$  I'm wondering if there are more general approaches.
54212
Find $a,b \in\mathbb{N}$ such that:      Remainder of $a$ divided by $b$ equals $5$       $\mathrm{lcm}(a,b)=12\gcd(a,b)$   I don't know how or where to start. The most similar problem I found was  Still I can't figoure out how to solve my problem, any help will be appreciated, thanks.
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Some observations I made is for $\frac{a}{b}+\frac{b}{a}$, is either:   the denominator has to be one, the numerator has to be a multiple of the denominator or  the numerator and denominator have to be the same.    Obviously, with the given conditions case 3 is eliminated since  $a \neq b$.  For case 1, if $b=1$, then $a=1$ which is a contradiction to the given condition that says $a \neq b$.  For case 2, I would think of examples. For example if $b=2$ then a multiple is $4$, so $a=4$. Then we have:  $$\frac{a}{b}+\frac{b}{a}=\frac{4}{2}+\frac{2}{4}=\frac{10}{4}$$  which is not an integer, but how would i proceed to show this case for all integers.
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A method to finding the LCM of two numbers is to use prime factoring. I know how to do this, but logically why does this work in finding the lcm
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Suppose that $m,n \in \mathbb{N}$. Using the fundamental theorem of arithmetic it is easy to show that there exist $c,d \in \mathbb{N}$ such that $(c,d) = 1$, $cd = (m,n)$ and $(\frac{m}{c},\frac{n}{d}) = 1$.   Is there any quick way to prove this without using the prime factorizations of $m$ and $n$, i.e. only the basic properties of the gcd, lcm, etc.?
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I try to prove the following formula:  $$n!=\prod_{k=1}^n \operatorname{lcm}(1,2,...,\lfloor n/k \rfloor)$$  I noticed that  $\upsilon_{p}(\operatorname{lcm}(1,2,...,\lfloor n/k \rfloor)) = s$ iff $\lfloor n/(p^{s+1}) \rfloor &lt; k \leq \lfloor n/(p^{s}) \rfloor $
15053
We want to show that if $a,b\in G$ where $G$ is a finite Abelian group, we have  $\operatorname{LCM}(|a|,|b|) = |ab|$ given that $ab \neq e$.   How I approached this question was by saying let $\operatorname{LCM}(|a|,|b|) = L$. Then if we show that $L$ divides $|ab|$ and $|ab|$ divides $L$ then $|ab| = L$. Showing that $|ab|$ divides $L$ was fine. But then I am having trouble with the second part that is showing $L$ divides $|ab|$. For this, so far I have:  Consider $(ab)^{|ab|} = e = a^{|ab|}b^{|ab|} $  since we know $a \neq b^{-1}$ by assumption then we can conclude that $|a|$ divides $|ab|$ and $|b|$ divides $|ab|$. We know that $L = \frac{|a||b|}{\operatorname{gcd}(|a|,|b|)}$ from here can we conclude that $L$ divides $|ab|$?