There are only three questions in the final of the multiple intelligence competition. It is known that: (1) 25 students from a school participated in the competition, and each student solved at least one problem; (2) Among all the students who have not solved the first question, the students who have solved the second question are twice as many as the third question: (3) There are more students who have only solved the first question than the rest1; (4) Half of the students who only solved one problem did not solve the first problem, so the number of students who only solved the second problem is ()

The number of people in each category is a 1, a2, a3, a 12, a 13, a23, a 123.

According to (1), a1+a2+a3+a12+a13+a23+a123 = 25 ... ①.

From (2): A2+A23 = (A3+A23) × 2 ...

According to (3), a12+a13+a123 = a1-1... ③.

From (4): A 1 = A2+A3...④

From ②, A23 = A2-A3× 2...⑤

Then a12+a13+a123 = a2+a3-16 is obtained from ③ ④.

Then substitute ④ ⑤ ⑤ into ① and sort it out.

a2×4+a3=26

Since a2 and a3 both represent the number of people, we can find their integer solutions:

When A2 = 6,5,4,3,2 and 1, A3 = 2,6, 10, 14, 18 and 22.

According to A23 = A2-A3× 2...⑤, we can know: a2 & gta3.

Therefore, only A2 = 6 and A3 = 2 are eligible.

Then we can deduce A 1 = 8, a12+a13+a123 = 7, A23 = 2, and the total number of people = 8+6+2+7+2 = 25, and check that all conditions are equal.

Therefore, the number of students who only solved the second problem A2 = 6.