An Interesting Clock Game

Number Theory Level 4

Nick feels boring in a day, and he looks at the clock. He figures out a game to spend his time.

He draws a big circle on paper and 12 squares. The 12 squares are equally placed in the circle (just like how a clock put 1-12).

Now he wants to fill in the 12 squares with 1-12 in such a way that the sum of any 3 adjacent numbers is divisible by 3.

How many different clocks can he obtain?

Clarify: orientation does matter. For example, starting at the top and go clockwise, we count the orientation 1 − 2 − 3 − 4 − 5 − 6 − 7 − 8 − 9 − 10 − 11 − 12 1-2-3-4-5-6-7-8-9-10-11-12 and 2 − 3 − 4 − 5 − 6 − 7 − 8 − 9 − 10 − 11 − 12 − 1 2-3-4-5-6-7-8-9-10-11-12-1 as 2 solutions (even though we can rotate one to get another) because Nick doesn't recognize numbers when they are not orientated properly.


The answer is 82944.

Raymond Chan by Raymond Chan , 20, USA

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1 solution

Numbers, from 1 1 to 12 12 can be classified in three forms 3 k , 3 k + 1 , 3 k − 1 3k, 3k+1, 3k-1 . The sum of a triple is divisible by 3 if the triple is of any of the following forms (ignoring the order).

( 3 t , 3 k , 3 r ) , ( 3 t , 3 k + 1 , 3 r − 1 ) (3t,3k,3r) , (3t,3k+1,3r-1)

A sequence of the 12 12 numbers can have the required property if its of the ordered, repeated, formats

( 3 t , 3 k + 1 , 3 r − 1 ) (3t,3k+1,3r-1)

or

( 3 t , 3 k − 1 , 3 r + 1 ) (3t,3k-1,3r+1)

note that, having a sequence that has the required property, numbers of similar type, with respect to the remainder when divided by 3 3 , can permute, while the required property is preserved.

We might wanna fix number 3 3 at the origin of the clock and have the sequence format ( 3 t , 3 k + 1 , 3 r − 1 ) (3t,3k+1,3r-1) . Then there would be 4 ! × 4 ! × 3 ! 4! \times 4! \times 3! possibilities. foreach of the possibilities, we can rotate the sequence to have 3 3 at a position different from the origin. So, there would be a total of 12 × 4 ! × 4 ! × 3 ! 12 \times 4! \times 4! \times 3! .

The same as the previous paragraph is done, while the format of the sequence is ( 3 t , 3 k − 1 , 3 r + 1 ) (3t,3k-1,3r+1) . there would be 12 × 4 ! × 4 ! × 3 ! 12 \times 4! \times 4! \times 3! possible sequences.

Finally, there would be a total of 2 × 12 × 4 ! × 4 ! × 3 ! = 82944 2 \times 12 \times 4! \times 4! \times 3!= 82944 valid sequences.

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