Let the sets A, B and C respectively represent integer sets divisible by 2, 3 and 5 among integers from 1 to 200, then

1 the set divisible by 2 in the integer from 200 contains 200/2= 100, that is, there are 100 elements in set a;

1 the set divisible by 3 in the integer from 200 contains 200/3=66.67, that is, there are 66 elements in set b;

1 the set divisible by 5 in the integer from 200 contains 200/5=40, that is, there are 40 elements in set c;

In the integer of 1 to 200, the set divisible by 2,3 contains 200/(2*3)=33.33, that is, there are 33 elements in set AB (representing the intersection of sets A and B);

1 the set divisible by 2,5 in the integer from 200 contains 200/(2*5)=20, that is, there are 20 elements in the set AC (representing the intersection of set A and set C);

The integer of 1 to 200 contains 200/(3*5)= 13.33, that is, there are 13 elements in the set BC (representing the intersection of the set B and the set C).

The integer from 1 to 200 contains 200/(2*3*5)=6.67, that is, there are 6 elements in the set ABC (representing the intersection of sets A, B and C).

Therefore, among the integers from 1 to 200, the number divisible by any one of 2, 3 and 5 is

a+B+C-a B-AC-BC+ABC = 100+66+40-33-20- 13+6 = 146

The second question uses two theorems of trees: 1. Number of nodes-1= number of edges; 2. The sum of node degrees =2× number of edges.

Let the number of nodes x of degree 3 and the total number of edges of the tree be y, then:

5+4+X- 1=Y

5× 1+4×2+3X=2Y

X=3，Y= 1 1。

The third question A-(B∪C)=(A-B)∩(A-C)

= A∪7(B∪C)

=A∩(7B∩7C)

=A∩7B∩A∩7C (adding an A equation is still effective)

=(A-B)∩(A-C)

(where 7 stands for complement)