$\DeclareMathOperator{\lcm}{lcm}$There are many applications for finding the GCD or LCM of two numbers. I'm now interested in finding anything that could be used to illustrate the following mathematical relationship.

Let $r = \lcm(r_1, r_2)$. Find a divisor $c$ of $r$ such that $r/c$ is a multiple of $r_1$ but not a multiple of $r_2$ (or a multiple of $r_2$ but not a multiple of $r_1$). Notice there are cases in which no such $c$ can be found. For example, when $r_1 = r_2 = \lcm(r_1, r_2)$.

Is there any interesting motivation that [can] be put on this?

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    $\begingroup$ This doesn't quite fit, but the question somehow reminds me of Conway's Sylver Coinage game. $\endgroup$
    – Adam
    Feb 6 '20 at 14:59
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    $\begingroup$ Not sure if this helps, but it's equivalent to asking for a set $C$ such that $C \subseteq B-A$. Such an $A$, $B$, and $C$ would satisfy $C \subseteq A \cup B$ (equivalent to $c \mid r$), $A \subseteq (A \cup B) - C$ (equivalent to $r_1 \mid r/c$), but $B \not\subseteq (A \cup B)-C$ (equivalent to $r_2 \nmid r/c$). $\endgroup$
    – Aeryk
    Feb 6 '20 at 19:00
  • $\begingroup$ @Aeryk, how is "$c$ divides $r$" equivalent to "$C \subseteq A \cup B$"? $\endgroup$
    – user724963
    Feb 7 '20 at 14:52
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    $\begingroup$ Not sure if "analogous" or "equivalent" is the better choice of word. Either way, you consider a natural number to be a multiset of its prime factors and you get the translation(?) I described. $\endgroup$
    – Aeryk
    Feb 7 '20 at 15:26
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    $\begingroup$ @user724963 That is not the only case in which no $c$ can be found (though I notice now that I had a typo: I meant $r_1|r_2$, not the reverse). $\endgroup$ Feb 11 '20 at 9:39

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