A farmer harvested 100kg of millet and 56kg of maize. He wants to pour them into containers of equal mass. What is the greatest mass of a container he can have?

Masses: 100 kg, 150 kg, 200 kg.

The greatest mass $m$ kg per new container must divide each original mass exactly (using all maize from each without mixing sources).

Thus, $m = \gcd(100, 150, 200)$.

$$150 = 1 \times 100 + 50$$ $$100 = 2 \times 50 + 0$$

So, $\gcd(100, 150) = 50$ kg.

$$200 = 4 \times 50 + 0$$

So, $\gcd(50, 200) = 50$ kg.

$$100 \div 50 = 2 \text{ containers}$$ $$150 \div 50 = 3 \text{ containers}$$ $$200 \div 50 = 4 \text{ containers}$$

All integers, uses all maize.

No larger common divisor exists.

The greatest mass of maize per container is $\boxed{50 \text{ kg}}$.