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Oct 18, 2015 at 9:35 comment added bof @Qiaochu: I thought the correct generalization of$$\operatorname{lcm}(a,b)=\frac{a\cdot b}{\operatorname{gcd}(a,b)}$$was $$\operatorname{lcm}(a,b,c)=\frac{a\cdot b\cdot c\cdot \operatorname{gcd}(a,b,c)}{\operatorname{gcd}(a,b)\cdot \operatorname{gcd}(a,c)\cdot \operatorname{gcd}(b,c)}$$Another false belief I guess. Maybe not a common one.
May 21, 2011 at 6:25 comment added Unknown though the product is not necessarily $abc$, but it can be made so.
May 21, 2011 at 6:24 comment added Unknown @Qiaochu, glad to see that. We do have more concerning generalisations to three variables: $$gcd(lcm(a,b),lcm(a,c),lcm(b,c))=lcm(gcd(a,b),gcd(a,c),gcd(b,c))$$ $$gcd(a, lcm(b,c)) = lcm(gcd(a,b),gcd(a,c))$$ $$lcm(a, gcd(b,c)) = gcd(lcm(a,b),lcm(a,c))$$
May 8, 2011 at 23:47 comment added Qiaochu Yuan @Elohemahab: actually, the correct generalization is $\gcd(a, b, c) \text{lcm}(ab, bc, ca) = \text{lcm}(a, b, c) \gcd(ab, bc, ca) = abc$.
Feb 24, 2011 at 22:05 comment added Unknown @Qiaochu, that is a nice quick check. Let the RCF(remnant common factor) be the leftover factor that would make the above second equality true. There seems to be no interesting way of determining $RCF\left(a,b,c\right)$ so that $LCM\left(a,b,c\right)\times RCF\left(a,b,c\right) \times GCF\left(a,b,c\right) = abc$
Feb 24, 2011 at 21:08 comment added Qiaochu Yuan This kind of stuff is easy to rule out, though; it's dimensionally inconsistent. Replacing a, b, c by ka, kb, kc leads to a quick contradiction.
Nov 29, 2010 at 6:46 comment added Unknown Yes. GCF means the same: Greatest Common Factor.
Nov 27, 2010 at 19:33 comment added Zsbán Ambrus Does GCF mean gcd (greatest common divisor) here?
Nov 27, 2010 at 14:23 history answered Unknown CC BY-SA 2.5