Positive integers $x$,$y$,$z$ If $x$ , $y$ and $z$ are positive integers and $3x = 4y = 7z$, then calculate the smallest possible value for $x+y+z$.
How do you do this? Can someone please give me a hint?

$k=3x=4y=7z\implies x=\dfrac{k}{3},y=\dfrac{k}{4},z=\dfrac{k}{7}\implies x+y+z=k\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{7}\right)=k\left(\dfrac{4\cdot7+7\cdot3+3\cdot4}{84}\right)\geq \dfrac{61}{84}$
