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Let $m,n$ be positive integers such that $\gcd(m,n) = 6$. find $\operatorname{lcm}(4m,21n)$?

According to my approach depending upon the prime factorization of $m$ and $n$, there are different possible values for $\operatorname{lcm}$ of $4m$ and $ 21n $. for writing the complete solution should I have to discuss all the possible cases or there is some simple direct solution?

If anyone has the solution kindly post it.

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    $\begingroup$ Well, it's pretty clear that the lcm depends on the choice of $n,m$. That said, can you write down a simple formula which gives the lcm in terms of $n,m$? $\endgroup$
    – lulu
    Sep 15 '18 at 18:08
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    $\begingroup$ A fun fact about the gcd and lcm of two numbers: $m\cdot n = gcd(m,n)\cdot lcm(m,n)$ $\endgroup$ Sep 15 '18 at 18:09
  • $\begingroup$ To be clear: you do have to distinguish some cases here, according to the possible values for $\gcd(4m,21n)$. There are only finitely many things this can be, but it is not determined uniquely. $\endgroup$
    – lulu
    Sep 15 '18 at 18:19
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This is an updated edit to my previous answer, which was misguided in assuming facts not presented in the question as posed.

The numbers $6,4,21$ appear in the question. $6=2\cdot 3, 4=2^2, 21=3\cdot 7$, so to answer the question it will be necessary to keep track of the prime factors $2,3,7$ as they may appear in $m$ and $n$. In what follows, all variable names are natural numbers.

Let $m=(2^a3^b7^c\cdot d)$ and $n=(2^e3^f7^g\cdot h), \gcd(d,h)=1$. Since there is one factor of each of $2$ and $3$ in $m$ and $n$, $\min(a,e)=1, \min(b,f)=1$; that is, $2$ and $3$ must appear as factors in each of $m,n$ at least once, but in one or the other of them only once. Since $m$ and $n$ have no common factor of $7$, $\min(c,g)=0$; that is, at least one of $m,n$ must have no factor of $7$.

Next, $4m=2^{a+2}3^b7^c\cdot d$ and $21n=2^e3^{f+1}7^{g+1}\cdot h$. Let $i=\min(a+2,e), j=\min(b,f+1), k=\min(c,g+1)$. Then lcm$4m\cdot 21n=\frac{84mn}{2^i3^j7^k}$.

$i$ may take on the values $1,2,3$; $j$ may take on the values $1,2$; $k$ may take on the values $0,1$. so there are twelve possible values for the denominator $2^i3^j7^k$ (which are $6,12,18,24,36,42,72,84,126,168,252,504$) and hence twelve possible answers to the posed question. A single answer would require knowing more about the factors of $m$ and $n$, which information is not given.

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  • $\begingroup$ Remark that the answer cam be expressed concisely using gcds, e.g. see my answer. $\endgroup$ Oct 20 '18 at 19:58
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Write $\,\begin{align}(x,y)&:=\gcd(x,y)\\ [x,y]\,&:={\rm lcm}(x,y)\end{align}.\ $ Cancel $\,6\,$ from the lcm, then apply Theorem $\rm\color{#c00}T\,$ and the $\,\rm lcm * gcd\,$ law

$$[4M,21N]/6\, =\, [4m,\,21n]\, \overset{\large\color{#c00}{\rm T}}=\, [n,4]\,[m,21]\,=\smash[t]{\bbox[6px,border:1px solid orange]{ \dfrac{84\,n\,m}{(n,4)\,(m,21)}\,,}\quad \begin{align} &\, \ n,m := N/6, M/6 \\ &(n,m) = 1\end{align}}\ \ \ $$

Theorem $\, \color{#c00}{\rm T}\,\quad (a,A) = 1 = (b,B)\ \Rightarrow\ [ab,AB]\color{#0a0} = [a,B]\,[A,b]$

Proof $\ 1.\,\ \ [a,B]\,[A,b] = [ab,AB,\color{#c00}{aA},bB] = [ab,AB,\color{#0a0}{a,A},b,B] = [ab,AB]\ $ by $\ \smash[t]{\overbrace{\color{#c00}{aA} = [\color{#0a0}{a,A}]}^{\large (a,A)\,=\,1}}\,$ etc.

Proof $\ 2.\ $ Wlog we may reduce to the case $\,(a,B)=1=(A,b)\,$ by cancelling $(a,B)(A,b)$ from both sides. Then all lcm arg's are coprime so lcm = product so it reduces to $\,ab\cdot AB = (a\cdot B)(A\cdot b),\,$ which is true.

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