This is a arithmetic progression, and the sum of seven numbers is 49. If one number is removed, the remaining numbers are divided into two groups of three. To make the sum of these two groups of numbers equal, the sum of these six numbers must be in the form of 4n+2, because the sum of each group of numbers is odd, so the number to be removed can only be 3, 7 or 1 1.

(1) Remove 3, and the sum of the remaining 6 numbers is 46, 1+9+ 13=23, 5+7+1= 23, so fill in 3 at the intersection of two circles, and 60 at both sides and one side of 3.

(2) If 7 is removed, the sum of the remaining six numbers is 42, 1+9+ 1 = 2 1, 3+5+ 13=2 1, 7 is filled at the intersection of two circles, and one side of 7 is filled with 65433.

(3) Remove 1 1, and the sum of the remaining six numbers is 38,3+7+9 =19, 1+5+ 13 = 19,/kloc-0.

* * * There are three filling methods.

Find a routine method:

1, serial number: to find a regular topic, a series of quantities arranged in a certain order are usually given, and we are required to find a general rule according to these known quantities. Find out the rule, usually the serial number of the package. Therefore, it is easier to find the mystery by comparing variables with serial numbers.

2. Fibonacci sequence method: each number is the sum of the first two numbers.

3. Arithmetic series method: the difference between every two numbers is equal.

4. Tabbing method: You can look at it every once in a while to see what the relationship between the separated numbers is, for example, 14, 1, 12, 3, 10, 5, odd items become arithmetic progression, even items also become arithmetic progression, and then you have to fill in 8.