Question

Find the number of ways of arranging the numbers

Answer 1

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

$\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}$

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

$3\cdot 3\cdot 3\cdot 2 \cdot 2 \cdot 2 = 216$

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

$216\cdot 6 = 1296$

possible arrangements.

Answer 2

Answer: 432

Step-by-step explanation:

Each number is a multiple of 3, or 1 more than a multiple of 3, or one less than a multiple of 3.  These number we will call x's, y's, and z's.

If we add x + y + z, then we get x, because y and z will even each other out, since y is 1 more than multiple of 3 and z is one less.

Because of this, we can conclude that we have to arrange the numbers in a repeating pattern, like x, y, z, x, y, z, ... or z, y, x, z, y, x, ...

In this set 1, 2, 3, ..., 9, there are 3 x's, 3 y's, and 3 z's.

To fill up a pattern, there are 3 choices for the first x, y, z, 2 choices for second, then 1 choice.

3^3 * 2^3 * 1^3 = 27*8*1 = 216

This is equal for both patterns, so there are 216 * 2 = 432 ways to arrange the numbers.