72. Given three integers $x, y, z$ satisfying $x+y+z=100$, and $x<y<2z$, then the minimum value of $z$ is
Answer: 21.
Solution: Since $x, y, z$ are three integers, and $x<y<2z$, then
$$
x+2 \leq y+1 \leq 2z, \quad x \leq 2z-2, \quad y \leq 2z-1,
$$

Therefore,
$$
100 \leq (2z-2) + (2z-1) + z,
$$

which means
$$
z \geq \frac{103}{5} = 20.6,
$$

Thus, the minimum value of $z$ is 21.