, 2013

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.

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 $2^3$ 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.