# 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?

## 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.

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 $$9$$.

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 $$7$$.

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 $$8$$.

Therefore, he could achieve a sum of $$7$$, $$8$$, or $$9$$.