23. Sam wants to complete the diagram so that each of the nine circles contains one of the digits from 1 to 9 inclusive and each contains a different digit. Also, the digits in each of the three lines of four circles must have the same total. What is this total?
A 17
B 18
C 19
D 20
E 21
Solution
D We suppose that $x, a, b, c, d, e$, and $f$ are the numbers in the circles, as shown in the diagram. These numbers are the digits $1,3,4,6,7,8$, and 9 , in some order.
Let the common total of the digits in the three lines be $k$.

Then
$$
(x+a+b+5)+(x+e+f+2)+(5+c+d+2)=3 k,
$$
that is,
$$
(x+a+b+c+d+e+f)+x+14=3 k .
$$

Now
$$
x+a+b+c+d+e+f=1+3+4+6+7+8+9=38
$$
and therefore
$$
52+x=3 k .
$$

It follows that $52+x$ is divisible by 3 . Now $x$ is one of $1,3,4,6,7,8,9$, and the only one of these possible values for which $52+x$ is divisible by 3 is $x=8$. This gives $52+x=60$. Hence $60=3 k$, and so $k=20$.
In the context of the JMC, we can stop here. But to give a full mathematical solution we should satisfy ourselves that it is possible to place the digits in the circles so that each line adds up to 20 .

One way this can be done is shown in the diagram alongside.