16. (3 points) $A, B, C, D$ four boxes contain $8, 6, 3, 1$ balls respectively. The first child finds the box with the fewest balls, then takes one ball from each of the other boxes and puts it into this box; the second child also finds the box with the fewest balls, and then also takes one ball from each of the other boxes and puts it into this box, ..., after the 50th child has finished, the number of balls in box $A$ is $\qquad$ .
A $\cdots)$, the 6th child repeats with the 2nd, meaning a cycle of 4 groups; thus, by analogy: $(50-1) \div 4=12 \cdots 1$ (times);

That is: Excluding the first irregular group, there should be 49 repeated groups, with one remaining, so after the 50th child takes, the four boxes $A B C D$ should contain: $6,4,5,3$ balls respectively.

【Solution】Solution: As analyzed: The 6th child repeats with the 2nd, meaning a cycle of 4 groups; thus, by analogy: $(50-1) \div 4=12 \cdots 1$ (times);

After the 50th child takes, the four boxes $A B C D$ should contain: $6,4,5,3$ balls respectively;
Answer: After 50 children have placed their balls, the number of balls in box $A$ is 6;
Therefore, the answer is: 6.