Japan 2013 Initial Round

44 Since $2100=2^2\cdot 3\cdot 5^2\cdot 7$, if we take $x=5^2=25$, $y=7$, $z=2^2\cdot 3=12$, then the least common multiple of $x$, $y$, $z$ is $2100$ and we get $x+y+z=44$.

Now let us show that $44$ is the desired minimum value. Assume that the least common multiple of $x$, $y$, $z$ is $2100$, and $x+y+z<44$ is satisfied. Then, at least one of $x$, $y$, $z$ must be a multiple of $5^2$. By symmetry we may assume without loss of generality that one such number is $x$. Since $2\cdot 5^2>44$, we must have $x=5^2=25$. Then, we see that $y+z<19$, and among $y$ and $z$, we must have a multiple of $2^2$, a multiple of $3$ and a multiple of $7$. By symmetry we may assume that $y$ is a multiple of $7$. Then, since $2^2\cdot 7>19$ and $3\cdot 7>19$, we see that $y$ can be a multiple neither of $2^2$ nor of $3$. Therefore, we see that $z$ must be a multiple of $2^2\cdot 3=12$. But then, we get $x+y+z\ge 25+7+12=44$, which shows that it is impossible to have $x+y+z<44$, and this establishes the claim that $44$ is the desired minimum value.