Find common number divisible by six different numbers If there is recipe to find this - I would like to find the first common number divisible by the following six numbers- 260,
380, 460,560,760 and 960.
How does one calculate the numbers I need?
Any direction would be useful.
 A: The prime factorizations of the numbers are:
$260 = 2^2×5×13$
$380 = 2^2×5×19$
$460 = 2^2×5×23$
$560 = 2^4×5×7$
$760 = 2^3×5×19$
$960 = 2^6×3×5$
If the number needs to be divisible by all, it needs to be divisible by $2^6, 3, 5, 7, 13, 19, 23$ (Why?).
If you multiply these, you get ______.
A: $\DeclareMathOperator{\lcm}{lcm}$
As I said, you have to calculate the l.c.m. of the six numbers. The l.c.m.  is associative, so you may group some numbers when it's easier to compute. Furthermore, all these numbers are divisible by $20$, so
$$\lcm(260,380,460,560,760,960)=20\lcm(13,19,23,28,38,48)$$
On the other hand, $\lcm(19,38)=38$, $\;\lcm(28,48)=4\lcm(7,12)=4\cdot 7\cdot12)=336$, so
$$\lcm(260,\dots,960)=20\lcm(13,23,38,336)$$
Continuing, we have $\:\lcm(38, 336)=2\lcm(19,168)=2\cdot19\cdot168=6384$, $\;\lcm(13,23)=13\cdot23=299$, and finally
$$\lcm(260,\dots,960)=20\lcm(299,6384)=20\cdot299\cdot 6384=20\cdot1\,908\,520=\color{red}{38\,176\,320}$$
if my computations are correct.
A: $\begin{align} {\rm Hint\!:}\ \  &\ \, 20[19,13,23,\,28,\,38,\,48],\ \ \, \ \ \ {\rm with}\,\ \ [a,b,c] := {\rm lcm}(a,b,c)\\[.2em]
&=\, 20[13,23,2[14,19,24]]\ \ \ \ \ \ \ {\rm by}\ \ 19\mid 38\\[.2em]
&=\, 20[13,23,2[\ \ 7,19, 24]]\ \ \ \ \ \ \ {\rm by}\ \ 2^1\,||\,14,\, 2^2\mid 24;\\[.2em]
&=\, 20(13\!\cdot\! 23(2(7\cdot 19\cdot 24)))\ \ \ \ {\text{by lcm = product for pair coprimes}}\\[.2em]
&=\, 38176320\end{align}$
