Determine the number of ways of placing the numbers 1, 2, 3, ..... 9 in a circle, so that the sum of any three numbers in consecutive positions is divisible by 3. (Two arrangements are considered the same if one arrangement can be rotated to obtain the other.)

Guest Jul 5, 2020

#1**0 **

First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

x | x mod 3

0 0

1 1

2 2

3 0

4 1

5 2

6 0

7 1

8 2

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

0,1,2,0,1,2,0,1,2

or

2,0,1,2,0,1,2,0,1

We have 3! = 6 possible patterns. Suppose for example that we choose the pattern

0,1,2,0,1,2,0,1,2

One possible way of following this pattern would be the arrangement

3,1,2,6,4,5,9,7,8

In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).

This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have

3*3*3*2*2*2 = 216

possible ways of following a fixed pattern. Since the number of patterns was 6, we have

216*6 = 1296 possible arrangements.

Guest Jul 6, 2020