Key points: If you want to find out which of the three balls (abc) is unqualified, suppose to compare A and B first:

With luck, as long as 1 time, a=b, then C is unqualified.

Bad luck is twice, A>B (A <; B, the same) because I don't know the weight, I need to compare with C again. (If I compare with A), a=c, B is unqualified, and a>c and A are unqualified.

Easily extended to 10:

10 is divided into four groups (from upstairs), group A: 3, group B: 3, group C: 3 and group D: 1.

The first case: D is unqualified. (Of course, I didn't know until later, so let's compare ABC first. ) A=B, B=C, and confirm that ABC is qualified.

The second case: D is qualified. Compare ABC first. According to the above three examples, determine which group in ABC needs 1-2 times.

It takes 1-2 times to determine which of a group of three is unqualified. The final possible situations are: 1+ 1, 1+2, 2+ 1, 2+2; That is, 2, 3 and 4 times.

I found a mistake when I wrote here. In order to mislead everyone, I still send out the wrong ideas for everyone to find fault with.

But the final answer is 2, 3, 4.