CEMC Banner

Problem of the Week
Problem D and Solution
Not as Easy as 1, 2, 3

Problem

Zephaniah places the numbers \(1\), \(2\), and \(3\) in the circles below so that each circle contains exactly one of \(1\), \(2\), and \(3\), and any two circles joined by a line do not contain the same number. He then finds the sum of the numbers in the four circles on the far right. What sums could he achieve?

Seven circles are arranged into three vertical columns with lines connecting some pairs of circles. There is one circle in the left column, two circles in the middle column, and four circles in the right column. The left circle is connected to each of the two middle circles. The two middle circles are connected to each other. The top middle circle is connected to the top two circles on the right. The top two circles on the right are connected. The bottom middle circle is connected to the bottom two circles on the right. The bottom two circles on the right are connected.

Solution

We will first find the largest possible value for the sum.
First, we note that the four circles on the right cannot all contains \(3\)s, since there are two pairs of circles that are joined. In fact, we cannot have three \(3\)s in those circles, for that means two joined circles will both contain a \(3\). Therefore, we could possibly have two \(2\)s and two \(3\)s, as shown below.

The four circles on the right contain the numbers 2, 3, 2, and 3 when read from top to bottom.

However, this would force the middle two circles to each contain a \(1\). Since these circles are joined by a line, this is not possible. Therefore, the sum of the four circles on the right cannot be \(10\).

However, we can fill in all the circles so that the sum of the four circles on the right is \(9\).

The circle on the left contains the number 3. In the middle, the top circle contains 2 and the bottom circle contains 1. On the right, the circles contain 1, 3, 2, and 3 when read from top to bottom.

Therefore the largest possible sum is \(9\).
We will now find the smallest possible sum.
We can use similar reasoning to show that the smallest possible sum is \(7\).

The four right circles cannot contain all \(1\)s or three \(1\)s. There could possibly be a \(1\) and \(2\) in the top two circles to the right and a \(1\) and a \(2\) in the bottom two circles to the right. Then the sum of the four numbers to the right would be \(6\). However, this would force the middle two circles to each contain a \(3\). Since these circles are joined by a line, this is not possible. Therefore, the sum cannot be \(6\).

However, we can fill in all the circles so that the sum of the four circles on the right is \(7\).

The circle on the left contains the number 1. In the middle, the top circle contains 3 and the bottom circle contains 2. On the right, the circles contain 1, 2, 3, and 1 when read from top to bottom.

Therefore, the smallest possible sum is \(7\).
Finally, we can also fill in all the circles so that the sum of the four circles on the right is \(8\).

The circle on the left contains the number 2. In the middle, the top circle contains 3 and the bottom circle contains 1. On the right, the circles contain 1, 2, 3, and 2 when read from top to bottom.

Therefore, he could achieve a sum of \(7\), \(8\), or \(9\).