Determine the number of ways of placing the numbers $1, 2, 3, \dots, 9$ 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.)

144 ways

Step-by-step explanation:

The arrangements to have the sum of three numbers being divisible by three can be approached by modular arithmetic such that the difference between the third integer and the sum of the first two integers is divisible by 3 making them congruent modulus 3

We have;

With the first number as 1

The second number can be 2, 5 or 8

The third number can then be 3, 6, 9

The fourth number can then be 4 or 7

The sixth number can be 3, 6, 9

The seventh number can be 4 or 7

The eight number can be 2, 5, 8

and the ninth number can be 3, 6, 9

Therefore, we have

the first number as 1

The second, fifth, and eight number can be 2, 5 or 8

The third, sixth and ninth number can then be 3, 6, 9

The fourth, and seventh number can then be 4 or 7 Which gives,

The number of possible arrangements for the numbers 2, 5, and 8 = 3! = 6

The number of possible arrangements for the numbers 3, 6, and 9 = 3! = 6

The number of possible arrangements for the numbers 4, and 7 = 2! = 2

he total number of possible arrangements = 2×6×6 =72

The arrangement can be reversed in the counter clockwise directions given  the total 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 = 72 × 2 = 144 ways.

your answer is c. 97 (don't delete this again)

step-by-step explanation: