16. (15 points) There are 6 sealed boxes, each containing red balls, white balls, and black balls. Each box contains only one color of balls, and the number of balls in each box is $15, 16, 18, 19, 20, 31$. It is known that the number of black balls is twice the number of red balls, and there is only 1 box of white balls. Answer the following questions:
（1）What color are the balls in the box with 15 balls?
（2）How many boxes contain black balls?
【Analysis】First, analyze that the number of black balls is twice the number of red balls, so the number must be a multiple of 3. The remainder of the total number of balls divided by 3 can help identify the white ball. Continue reasoning from there.

【Solution】Solution: According to the problem:
The total number of balls in the 6 boxes is $15+16+18+19+20+31=119, 119 \div 3=39 \cdots 2$,
The number of black balls is twice the number of red balls, and there is only 1 box of white balls, so the number of white balls is the remainder when divided by 3, which is 2. Therefore, the number of white balls is 20.

The total number of the remaining 5 boxes is: $119-20=99$.
The number of black balls is twice the number of red balls, so there are 66 black balls and 33 red balls. Only $15+18=33$ can be the red balls.
So the box with 15 balls contains red balls.
Among the 6 boxes, 2 are red, 1 is white, and the remaining 3 are black.
Answer: (1) The box with 15 balls contains red balls.
(2) There are 3 boxes containing black balls.