The problem of finding defective products in mathematics

1: chalk 6 1 box, of which 60 boxes have the same quality, and the other box is less than other boxes 1 piece. How many times can you find this box of chalk by weighing it with a balance?

At least once: 30-30 are placed at both ends of the balance, and the result is balanced, so the last box is. This is the luckiest:)

2. "Guarantee" means that even if you are unlucky, you can find it in these times and you don't need it anymore. This assumes that after each weighing, you have to prepare for the worst: you have to weigh it again.

This kind of problem needs to divide the objects to be weighed into three equal parts, and assume that what you are looking for is in the group that needs to be weighed the most times.

If you weigh for the first time: (20, 20, 2 1), put two 20s on it, and the results are balanced. What you are looking for is in the group of 2 1. (Because it takes no less time to find one from 2 1 than from 20-provided that it can be found)

The second call: (7, 7, 7), no matter which "7" it is, it needs to be called again;

The third weighing: (2, 2, 3)-Suppose what you are looking for is at 3, if it is at 2, it is not "at least guaranteed".

The fourth weighing: (1, 1, 1) can find out whether it is balanced.

In other words, at least four times, we can guarantee to find it for you. But if you ask less, you can't "guarantee".

Note that this kind of problem looks like dichotomy, but due to the special balance principle, the weighed object is actually divided into three parts (as equal as possible).

I hope you can understand.