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We know that $LCM(a,b)= [\frac{ab}{\ GCD(a,b)}]$. What about $LCM (a,b,c)$? Can anyone help us because our instructors doesn't know the ways and she just lay the problem on us. Thanks.

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    $\begingroup$ $LCM(a,b,c) = LCM(LCM(a,b),c)$ $\endgroup$ – achille hui Jan 12 '15 at 8:00
  • $\begingroup$ How can we prove this? We introduce this to our instructors but she argued and reject our idea. How can we defend it? $\endgroup$ – whatever Jan 12 '15 at 8:01
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For 3 numbers, we have the relation

$$ LCM(a,b,c) = \frac{abc}{GCD(ab,bc,ca)} $$

The proof is as follows:

\begin{align} a | n \text{ and } b | n \text{ and } c|n & \iff abc | nab \text{ and } abc | nbc \text{ and } abc | nca\\ & \iff abc | GCD(nab,nbc,nca)\\ & \iff \frac{abc}{GCD(ab,bc,ca)} | n \end{align}

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  • $\begingroup$ Making use of this still requires figuring out what $GCD(a,b,c)$ means, though. $GCD(a,b,c)=GCD(GCD(a,b),c)$, sure, but how would you prove that in a way that doesn't immediately lead to $LCM(a,b,c)=LCM(LCM(a,b),c)$ the same way? $\endgroup$ – hvd Jan 12 '15 at 15:22
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By considering prime factorizations, the statement $\text{lcm}(a,b,c)=\text{lcm}(\text{lcm}(a,b),c)$ reduces to: $$ \max\{x,y,z\}=\max\{\max\{x,y\},z\}\qquad(\star) $$ We prove this in two steps. First of all, it is clear that $\max\{x,y\}\leq \max\{x,y,z\}$ and $z\leq \max\{x,y,z\}$. Thus $$ \max\{x,y,z\}\geq\max\{\max\{x,y\},z\}. $$ On the other hand, we observe that either $\max\{x,y,z\}=\max\{x,y\}$ or $\max\{x,y,z\}=z$. Hence in either case, $$ \max\{x,y,z\}\leq\max\{\max\{x,y\},z\}. $$ Thus we've shown both inequalities, so equality $(\star)$ follows.


If you don't see why the statement $\max\{\max\{x,y\},z\}$ implies the $\text{lcm}$ statement, write out the prime factorizations of $a,b,c$. Fix any prime $p$, and compare the exponents appearing in $a,b,c$. Then the $\text{lcm}$ operation corresponds to taking the maximum of the exponents appearing in $a,b,c$ for the prime $p$.

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For me, the easiest applied way (which can also easily be programmed in computer) to find it is using the prime factorization:

$$a_1,...,a_k\in\Bbb N\;,\;\;a_i=\prod_{m=1}^{n_i} p_{im}^{\beta_{im}}\;\;,\;p_{im}\;\;\text{primes,}\;,\;\;\beta_{im}\in\Bbb N$$

then

$$l.c.m.(a_1,...,a_n)=\prod_{i,m}p_{im}^{\max\limits_i(\beta_{i1},...,\beta_{ik})}$$

For example, if we have $\;15\;,\;\;72\;,\;\;32\;$ , then

$$\begin{cases}15=3\cdot5\\72=2^3\cdot3^2\\32=2^5\end{cases}\implies\;l.c.m.(15,72,32)=2^5\cdot3^2\cdot5$$

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