13. (15 points) Fifteen boxes, each containing a white ball or a black ball, and there are no more than 12 white balls. You can choose any three boxes to ask: “Do these three boxes contain any white balls?” and receive a truthful answer. How many times do you need to ask at minimum to find one or more white balls?

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【Analysis】First, number the 15 boxes, and ask a question about the boxes numbered 1, $m$, and $n$ once, a total of 91 times. The answers are all “yes” or “no”, and the solution is derived by discussing different cases based on the answers.

【Solution】Solution: Number all the boxes: $1,2,3, \cdots, 15$. Ask 91 times as follows: select the boxes numbered 1, $m$, and $n$ and ask once, $2 \leqslant m \leqslant n$, a total of: $14 \times(14-1) \div 2=91$ (times). If all the answers are “yes”, then the ball in box 1 is white, because among boxes 2 to 15, 2 boxes contain black balls; if the ball in box 1 is black, the answer to the combination of it with any other two boxes must be “no”; if there are “no” answers among the 91 answers, then the ball in box 1 is black. Among the 91 answers, all the three boxes (including box 1) that answered “no” can be removed, and the balls in the remaining boxes are all white.

Below is the proof that fewer than 91 times may not necessarily find the white ball:
If the ball in box 1 is black, fewer than 91 times means that there are boxes numbered $k(>1)$ and $m(>1)$ that have not been combined with box 1 for questioning. If the balls in the other 12 boxes are all white, the answers received will all be “yes”, and saying that the ball in box 1 is white would be incorrect.