9. The numbers from 1 to 6 are to be placed at the intersections of three circles, one number in each of the six squares. The number 6 is already placed.
Which number must replace $x$, so that the sum of the four numbers on each circle is the same?
A 1
B 2
C 3
D 4
E 5
Solution
A

Every number is placed on two circles, so the total of all three circles combined is $2 \times(1+$ $2+3+4+5+6)=42$. Thus each circle has total $42 \div 3=14$. Hence the three numbers that appear in a circle with the 6 must add to $14-6=8$. There are only two ways to get a total of 8 from three of the other numbers: $1+2+5$ or $1+3+4$. Since $x$ appears on both of the circles with 6 , and the only number that appears in both $1+2+5$ and $1+3+4$ is 1 , we have $x=1$.