SOLUTION: How many ways are there to re-arrange numbers 1 to 12 on a circle, so that the sum of any 3 consecutive numbers is divisible by 3? Obviously in the original arrangement (1, 2, 3,

Question 1040063: How many ways are there to re-arrange numbers 1 to 12 on a circle, so that the sum of any 3 consecutive numbers is divisible by 3?
Obviously in the original arrangement (1, 2, 3, ..., 12) this is true because the sum of any 3 consecutive integers is always divisible by 3 (n, n+1, n+2 ---> 3n+3).
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
When we are talking about multiples of 3,
all numbers are "congruent modulo 3" with either 1, 2 or 0.
We could say "congruent modulo 3 to 1, 2, or 3" instead),
There are only two ways to arrange the numbers
1, 2, and 3 (each repeated 4 times) around a circle,
so that the sum of any 3 consecutive numbers would be a multiple of 3,
and one way is the mirror image of the other:

Turning the circle around may make the arrangement look different, but it is the same arrangement.
If we read the numbers in sequence starting from 1,
we have 1,2,3,1,2,3,1,2,3,1,2,3 clockwise or counterclockwise.
To use the numbers from 1 to 12, we just replace each number with a different congruent number.
Listing 1 as the first number,
the fourth number could be 4, 7, or 10 (all congruent to 1 modulo 3);
the second, fifth, eighth, and eleventh number must be 2, 5, 8, and 11, in some order,
and the third, sixth, ninth and twelfth numbers must be 3, 6, 9, and 12 in some order.
Staring our list with number 1, there are
possible arrangements for 4, 7, and 10,
possible arrangements for 2, 5, 8, and 11,
and possible arrangements for 3, 6, 9, and 12.
All in all we could make different lists,
that can be arrange in two directions around the circle to make =

ways to arrange numbers 1 to 12 on a circle, so that the sum of any 3 consecutive numbers is divisible by 3.

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