Problem description:

1 There are 10 people in the room, all of whom are under 60 years old (all ages are integers), but at least not less than 1 year old.

1: After verification, you can always find two groups of people (two groups do not contain the same person), and the sum of each group's ages is equal.

2. If possible, can 10 in the question be changed to a smaller number?

Based on the principle of pigeon nest (pigeon hole principle), I can't think of a solution, and I want to ask you.

Analysis:

1. First of all, remember that this 10 person is composed of * * * A = {A 1, A2, ..., A 10}

There are two cases where 10- 1 = 1023 took at least one person out of this 10; The total age of the selected person is 5 1+52+ ... the maximum is +60=555, and the minimum is 1, that is, there are at most 555 different values; According to the pigeon hole principle, there must be two s: S 1 equals S2; Let the * * * of the person corresponding to S 1 be M, and the * * * of the person corresponding to S2 be N. If the intersection of M and N is empty, then M and N are two groups of people respectively; If the intersection of m and n is not empty, remove their common elements in m and n respectively, and the remaining two groups are these two groups of people.

2. OK ... If it is changed to 9, at least one person should be selected in the case of 2 9-1= 51; The total age of these people is 52+53+ ... the maximum +60=504 and the minimum 1, that is, there are at most 504 different values, which can be proved by the above method; That is, it is enough to change it to 9; If you change it to 8, you can't prove it by the above method, but you can't guarantee 8. It may be possible to prove 8 in other ways (but I can't think of it ~ ~ ~).