1. **Problem Statement:** We need to place the numbers 2, 3, 4, 5, 6, 7, 8, 9, and 10 in different circles so that each line of three circled numbers has the same sum. 2. **Given:** - The center circle currently has the number 9. - Surrounding circles have some numbers placed: 2, 6, 9, 7, 5, 4, and two are empty. - We want to find if more than one number can go in the middle circle to satisfy the condition. 3. **Key Idea:** Each line of three numbers must sum to the same total. Since the center is part of multiple lines, the center number heavily influences the sum. 4. **Approach:** - Let the center number be $x$. - The sum of all numbers 2 through 10 is $2+3+4+5+6+7+8+9+10=54$. - Each number is used once. - The puzzle is symmetric; each line includes the center and two outer numbers. 5. **Testing center numbers:** - The problem states the center currently has 9. - We check if other numbers can be in the center to maintain equal sums. 6. **Solution:** - The center number can be 9 or 6 to satisfy the equal sum condition. - Both 9 and 6 as center numbers allow arranging the other numbers so that each line sums equally. 7. **Final answer:** More than one number can be in the middle circle; specifically, the numbers $6$ and $9$ can be placed in the center to satisfy the puzzle's conditions.